Sunday, October 17, 2010
Strong Acids and Bases: Names and Formulas
The STRONG ACIDS are:
Hydrochloric acid: HCl
Hydrobromic acid: HBr
Hydroiodic acid: HI
Sulfuric acid: H2SO4
Nitric acid: HNO3
Perchloric acid: HClO4
The STRONG BASES are:
Lithium hydroxide: LiOH
Sodium hydroxide: NaOH
Potassium hydroxide: KOH
Calcium hydroxide: Ca(OH)2
Strontium hydroxide: Sr(OH)2
Barium hydroxide: Ba(OH)2
17.2-17.6: Solubility and the Common Ion Effect-Complex Ions and Solubility
Solubility and the Common Ion Effect:
If you have a salt in a solution of another salt having the same cation or anion, the salt will be less soluble in this solution than it would be in pure water. For example, when adding calcium oxalate, CaC2O4, to a solution of calcium chloride, CaCl2, both salts will contribute a Ca2+ ion. This change in solubility is explained by Le Chatelier's principle:
CaC2O4 (s) ⇌ Ca2+ (aq) + C2O42- (aq)
Calcium chloride will also contribute Ca2+ ion, which will act as a stress on the right hand side of the equation and the equilibrium will shift to the left, and calcium oxalate will precipitate from the solution.
Example:
MgC2O4is added to a 0.437 M solution of Na2C2O4. What is the solubility of MgC2O4 in this solution?
MgC2O4 ⇌ Mg2+ (aq) + C2O42- (aq)
Notice that there is an initial value for C2O42-, contributed by the Na2C2O4.
Ksp = (x)(0.437 + x) = 8.5 x 10-5
(always check, but the approximation should hold for Ksp problems)
x = 1.95 x 10-4 M
Remember the question asked for solubility, not molar solubility. So we have to convert mols to grams using the periodic table:
( 1.95 x 10-4 mol / 1 L) x (112 g MgC2O4 / 1 mol MgC2O4) = 0.022 g/L
Example:
Find the molar solubility for Ca3(PO4)2 in 0.115 M solution of Ca(NO3)2.
(Ksp for Ca3(PO4)2 = 1 x 10-26)
Ca3(PO4)2 (s) ⇌ 3Ca2+ (aq) + 2PO43- (aq)
Ksp = (0.115 + 3x)3(2x)2 → approximation holds
Ksp = (0.115)3(4x2) = 1 x 10-26
x = 1.28 x 10-12
Precipitation Reactions:
To try to find out whether or not precipitation will occur in a reaction, we need to compare the Qc to the Kc, or in this case, the Ksp. (We went over Qc and comparing it to Kc in Chapter 14). The Qc, or ion product as it is called when considering solubility, is found the same way as the Ksp, except the Ksp is found at equilibrium and the Qc is not.
Example:
AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)
[Ag+][Cl-] = Ksp at equilibrium
[Ag+][Cl-] = Qc when not at equilibrium
When Qc is less than Ksp, there is not enough product yet to be at equilibrium, so the reaction will shift to the right, and no ppt (precipitate) will form.
When Qc is greater than Ksp, there is too much product, so the reaction will shift to the left, and a ppt will form.
When Qc is equal to Ksp, the reaction is at equilibrium.
Example:
Will there be a ppt if 50.0mL of 0.450 M AgNO3 is mixed with 50.0mL of 0.350 M NaCl?
The first step is to think about what the possible ppt might be from these compounds. Write out what happens when these two compounds break up:
AgNO3 (s) ⇌ Ag+ (aq) + NO3-(aq)
NaCl (s) ⇌ Na+ (aq) + Cl- (aq)
When these four ions are floating around in the solution, the only possible ppt is AgCl.
Because we are mixing, the first thing we need to do is convert to mols then divide by the new volume to find the new concentrations of AgNO3 and NaCl:
(0.500 L)(0.450 mol / 1 L) = (0.0225 mol) / (0.1 L) = 0.225 M AgNO3
(0.500 L)(0.350 mol / 1 L) = (0.0175 mol) / (0.1 L) = 0.175 M NaCl
Qc = [Ag+][Cl-] = (0.225)(0.175) = 0.0394
Ksp = 1.8 x 10-10
Qc is greater than Ksp, the reaction will shift to the left and a ppt will form.
Fractional Precipitation:
A technique of separating two or more ions from a solution by adding a reactant that precipitates first one ion, then another, and so forth.
Example:
Na2C2O4 (s) ⇌ 2Na+ (aq) + C2O42- (aq)
Three compounds with low solubility will form: CdC2O4, CaC2O4, MgC2O4.
Ksp for CdC2O4 = 1.5 x 10-8
Ksp for CaC2O4= 2.3 x 10-9
Ksp for MgC2O4 = 8.5 x 10-5
A precipitate will form when Qc is greater than Ksp. In other words, just past the equilibrium. So an easy way to see when a ppt will form is to set Qc = Ksp. To see which ppt will form first, look at which Ksp will be reached first, in other words, compare to see which Ksp is smaller.
In this case, CaC2O4 has the smallest Ksp, so it will be reached first, and CaC2O4 will form ppt first. The second largest is CdC2O4, so this will ppt out second, and MgC2O4 last.
So this question is extremely easy! When asked from a list of compounds which will precipitate out first, or the order of precipitation, look at the Ksp values and go from smallest to largest.
A harder question would be one that asks when each of the ions fall out of solution. In other words, at what [C2O42-] does each metal fall out?
First look at each compound and build the Ksp equation. Solve for the unknown, [C2O42-].
Example:
For CaC2O4:
Ksp = [Ca2+][C2O42-]
2.3 x 10-9 = (0.200)[C2O42-]
[C2O42-] = 1.15 x 10-8 M
Note that the concentration for Ca2+ was given at the beginning of the problem.
For CdC2O4:
Ksp = [Cd2+][C2O42-]
1.5 x 10-8 = (0.200)[C2O42-]
[C2O42-] = 7.5 x 10-8
For MgC2O4:
Ksp = [Mg2+][C2O42-]
8.5 x 10-5 = (0.200)[C2O42-]
[C2O42-] = 4.25 x 10-4
Example:
In this same situation, what is the percent Ca2+ ion left when Cd2+ begins to ppt?
Remember that we figured out earlier than Cd2+ begins to ppt when [C2O42-] = 7.5 x 10-8. To see how much Ca2+ has fallen out at this point, substitute this value for [C2O42-] into the following equation and solve for Ca2+:
Ksp = [Ca2+][C2O42-]
2.3 x 10-9 = [Ca2+](7.5 x 10-8)
[Ca2+] = 0.0307 M
To find the percent Ca2+ that has precipitated out at this point, divide this concentration by the original concentration of Ca2+ and multiply by 100:
(0.0307 M) / (0.200 M) x 100 = 15.3 %
Effect of pH on Solubility:
What happens to the solubility of a compound when a strong acid is added?
Consider a solution of the solids CaSO4 and CaCO3 (according to solubility rules, these are insoluble, so imagine them as solids resting at the bottom of a solution while a tiny percent of their ions escape into solution).
CaSO4 (s) ⇌ Ca2+ (aq) + SO42-(aq)
CaCO3 (s) ⇌ Ca2+ (aq) + CO32- (aq)
If a strong acid is added, H3O+ ion is added to the solution. The anions will react with it:
SO42- (aq) + H3O+ (aq) ⇌ HSO4- (aq) + H2O (l)
The small amount of free SO42- ions will react with H3O+. According to Le Chatelier's, this will drive more SO42- ion out of the CaSO4, but only by a very small margin. The other compound will be more effected by the change in [H3O+]:
CO32- (aq) + H3O+ (aq) ⇌ HCO3- (aq) + H2O (l)
HCO3- reacts with more of the H3O+:
HCO3- (aq) + H3O+ (aq) ⇌ H2CO3 (aq) + H2O (l)
H2CO3 is unstable and reacts further:
H2CO3 (aq) → H2O (l) + CO2 (g)
If the container is open, the CO2 gas will escape, and according to Le Chatelier's, the HCO3- will continue to replenish out of the CaCO3. This compound will be more soluble in the presence of a strong acid.
Complex Ion Formation:
In Chapter 15 we learned the definition of a Lewis Acid/Base:
Lewis Acid: a species that can form a covalent bond by accepting an electron pair from another species.
Lewis Base: a species that can form a covalent bond by donating an electron pair to another species.
And in previous classes we learned that a coordinate covalent bond is a form of bonding between two atoms in which both electrons shared in the bond come from the same atom.
Complex Ion: an ion formed from a metal ion with a Lewis base attached to it by a coordinate covalent bond. A complex is a compound containing complex ions.
Example:
Ligand: a Lewis base that bonds to a metal ion to form a complex ion.
Example:
The overall equation for when aqueous silver ions form a complex ion with ammonia:
Ag+ (aq) + 2NH3 (aq) ⇌ Ag(NH3)2+ (aq)
Ag+ is the Lewis acid. NH3 is the Lewis base, and ligand.
The formation constant, or stability constant: Kf, is yet another form of Kc. The equilibrium constant for the formation of the complex ion from aqueous metal ion and the ligands.
For Ag(NH3)2+:
Kf = [Ag(NH3)2+] / [Ag+][NH3]2
Kf values are usually very large. For example,
Kf for Ag(NH3)2+ = 1.7 x 107
When doing calculations with Kf values, it would be harder to work with these values, because the approximation method will never hold. An easy way around this is to remember the mathematical rules that apply to Kc.
When a reaction is reversed so that the reactants are products and the products are reactants, the new Kc is the inverse (1 / Kc) of the original Kc.
Dissociation Constant: Kd, the inverse of Kf.
Example:
Reverse the reaction and take the new Kf, which will be Kd:
Ag(NH3)2+ (aq) ⇌ Ag+ (aq) + 2NH3 (aq)
Kd = [Ag+][NH3]2 / [Ag(NH3)2+] =( 1 / Kf)
Example:
For the formation of the complex ion:
Cu2+ (aq) + 4NH3 (aq) ⇌ Cu(NH3)42+ (aq), Kf = 4.8 x 1012
What is the [Cu2+] at equilibrium when you start with 0.100 M Cu(NO3)2 and 1.00 M NH3?
Note that the approximation method will not hold for these numbers, and we'll be left to work through some unnecessarily tedious math. Or, we could use some shortcuts and make it much easier:
First, assume that the reaction goes to completion:
Cu2+ (aq) + 4NH3 → Cu(NH3)42+ (aq)
Note that only 0.100 M is subtracted in the second line of the table because there is an excess of NH3, and less Cu2+.
Now reverse the reaction to let some of the complex ion fall apart and convert Kf to Kd:
Cu(NH3)42+ (aq) ⇌ Cu2+ (aq) + 4NH3 (aq)
Kd = 1 / Kf = 2.08 x 10-13
2.08 x 10-13 = (x)(0.600 + 4x)4 / (0.100 – x)
Approximation method holds,
2.08 x 10-13 = (x)(0.600)4 / (0.100)
x = 1.60 x 10-13
Complex Ions and Solubility:
Example:
If you have a salt in a solution of another salt having the same cation or anion, the salt will be less soluble in this solution than it would be in pure water. For example, when adding calcium oxalate, CaC2O4, to a solution of calcium chloride, CaCl2, both salts will contribute a Ca2+ ion. This change in solubility is explained by Le Chatelier's principle:
CaC2O4 (s) ⇌ Ca2+ (aq) + C2O42- (aq)
Calcium chloride will also contribute Ca2+ ion, which will act as a stress on the right hand side of the equation and the equilibrium will shift to the left, and calcium oxalate will precipitate from the solution.
Example:
MgC2O4is added to a 0.437 M solution of Na2C2O4. What is the solubility of MgC2O4 in this solution?
MgC2O4 ⇌ Mg2+ (aq) + C2O42- (aq)
Notice that there is an initial value for C2O42-, contributed by the Na2C2O4.
Ksp = (x)(0.437 + x) = 8.5 x 10-5
(always check, but the approximation should hold for Ksp problems)
x = 1.95 x 10-4 M
Remember the question asked for solubility, not molar solubility. So we have to convert mols to grams using the periodic table:
( 1.95 x 10-4 mol / 1 L) x (112 g MgC2O4 / 1 mol MgC2O4) = 0.022 g/L
Example:
Find the molar solubility for Ca3(PO4)2 in 0.115 M solution of Ca(NO3)2.
(Ksp for Ca3(PO4)2 = 1 x 10-26)
Ca3(PO4)2 (s) ⇌ 3Ca2+ (aq) + 2PO43- (aq)
Ksp = (0.115 + 3x)3(2x)2 → approximation holds
Ksp = (0.115)3(4x2) = 1 x 10-26
x = 1.28 x 10-12
Precipitation Reactions:
To try to find out whether or not precipitation will occur in a reaction, we need to compare the Qc to the Kc, or in this case, the Ksp. (We went over Qc and comparing it to Kc in Chapter 14). The Qc, or ion product as it is called when considering solubility, is found the same way as the Ksp, except the Ksp is found at equilibrium and the Qc is not.
Example:
AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)
[Ag+][Cl-] = Ksp at equilibrium
[Ag+][Cl-] = Qc when not at equilibrium
When Qc is less than Ksp, there is not enough product yet to be at equilibrium, so the reaction will shift to the right, and no ppt (precipitate) will form.
When Qc is greater than Ksp, there is too much product, so the reaction will shift to the left, and a ppt will form.
When Qc is equal to Ksp, the reaction is at equilibrium.
Example:
Will there be a ppt if 50.0mL of 0.450 M AgNO3 is mixed with 50.0mL of 0.350 M NaCl?
The first step is to think about what the possible ppt might be from these compounds. Write out what happens when these two compounds break up:
AgNO3 (s) ⇌ Ag+ (aq) + NO3-(aq)
NaCl (s) ⇌ Na+ (aq) + Cl- (aq)
When these four ions are floating around in the solution, the only possible ppt is AgCl.
Because we are mixing, the first thing we need to do is convert to mols then divide by the new volume to find the new concentrations of AgNO3 and NaCl:
(0.500 L)(0.450 mol / 1 L) = (0.0225 mol) / (0.1 L) = 0.225 M AgNO3
(0.500 L)(0.350 mol / 1 L) = (0.0175 mol) / (0.1 L) = 0.175 M NaCl
Qc = [Ag+][Cl-] = (0.225)(0.175) = 0.0394
Ksp = 1.8 x 10-10
Qc is greater than Ksp, the reaction will shift to the left and a ppt will form.
Fractional Precipitation:
A technique of separating two or more ions from a solution by adding a reactant that precipitates first one ion, then another, and so forth.
Example:
Na2C2O4 (s) ⇌ 2Na+ (aq) + C2O42- (aq)
Three compounds with low solubility will form: CdC2O4, CaC2O4, MgC2O4.
Ksp for CdC2O4 = 1.5 x 10-8
Ksp for CaC2O4= 2.3 x 10-9
Ksp for MgC2O4 = 8.5 x 10-5
A precipitate will form when Qc is greater than Ksp. In other words, just past the equilibrium. So an easy way to see when a ppt will form is to set Qc = Ksp. To see which ppt will form first, look at which Ksp will be reached first, in other words, compare to see which Ksp is smaller.
In this case, CaC2O4 has the smallest Ksp, so it will be reached first, and CaC2O4 will form ppt first. The second largest is CdC2O4, so this will ppt out second, and MgC2O4 last.
So this question is extremely easy! When asked from a list of compounds which will precipitate out first, or the order of precipitation, look at the Ksp values and go from smallest to largest.
A harder question would be one that asks when each of the ions fall out of solution. In other words, at what [C2O42-] does each metal fall out?
First look at each compound and build the Ksp equation. Solve for the unknown, [C2O42-].
Example:
For CaC2O4:
Ksp = [Ca2+][C2O42-]
2.3 x 10-9 = (0.200)[C2O42-]
[C2O42-] = 1.15 x 10-8 M
Note that the concentration for Ca2+ was given at the beginning of the problem.
For CdC2O4:
Ksp = [Cd2+][C2O42-]
1.5 x 10-8 = (0.200)[C2O42-]
[C2O42-] = 7.5 x 10-8
For MgC2O4:
Ksp = [Mg2+][C2O42-]
8.5 x 10-5 = (0.200)[C2O42-]
[C2O42-] = 4.25 x 10-4
Example:
In this same situation, what is the percent Ca2+ ion left when Cd2+ begins to ppt?
Remember that we figured out earlier than Cd2+ begins to ppt when [C2O42-] = 7.5 x 10-8. To see how much Ca2+ has fallen out at this point, substitute this value for [C2O42-] into the following equation and solve for Ca2+:
Ksp = [Ca2+][C2O42-]
2.3 x 10-9 = [Ca2+](7.5 x 10-8)
[Ca2+] = 0.0307 M
To find the percent Ca2+ that has precipitated out at this point, divide this concentration by the original concentration of Ca2+ and multiply by 100:
(0.0307 M) / (0.200 M) x 100 = 15.3 %
Effect of pH on Solubility:
What happens to the solubility of a compound when a strong acid is added?
Consider a solution of the solids CaSO4 and CaCO3 (according to solubility rules, these are insoluble, so imagine them as solids resting at the bottom of a solution while a tiny percent of their ions escape into solution).
CaSO4 (s) ⇌ Ca2+ (aq) + SO42-(aq)
CaCO3 (s) ⇌ Ca2+ (aq) + CO32- (aq)
If a strong acid is added, H3O+ ion is added to the solution. The anions will react with it:
SO42- (aq) + H3O+ (aq) ⇌ HSO4- (aq) + H2O (l)
The small amount of free SO42- ions will react with H3O+. According to Le Chatelier's, this will drive more SO42- ion out of the CaSO4, but only by a very small margin. The other compound will be more effected by the change in [H3O+]:
CO32- (aq) + H3O+ (aq) ⇌ HCO3- (aq) + H2O (l)
HCO3- reacts with more of the H3O+:
HCO3- (aq) + H3O+ (aq) ⇌ H2CO3 (aq) + H2O (l)
H2CO3 is unstable and reacts further:
H2CO3 (aq) → H2O (l) + CO2 (g)
If the container is open, the CO2 gas will escape, and according to Le Chatelier's, the HCO3- will continue to replenish out of the CaCO3. This compound will be more soluble in the presence of a strong acid.
Complex Ion Formation:
In Chapter 15 we learned the definition of a Lewis Acid/Base:
Lewis Acid: a species that can form a covalent bond by accepting an electron pair from another species.
Lewis Base: a species that can form a covalent bond by donating an electron pair to another species.
And in previous classes we learned that a coordinate covalent bond is a form of bonding between two atoms in which both electrons shared in the bond come from the same atom.
Complex Ion: an ion formed from a metal ion with a Lewis base attached to it by a coordinate covalent bond. A complex is a compound containing complex ions.
Example:
Ligand: a Lewis base that bonds to a metal ion to form a complex ion.
Example:
The overall equation for when aqueous silver ions form a complex ion with ammonia:
Ag+ (aq) + 2NH3 (aq) ⇌ Ag(NH3)2+ (aq)
Ag+ is the Lewis acid. NH3 is the Lewis base, and ligand.
The formation constant, or stability constant: Kf, is yet another form of Kc. The equilibrium constant for the formation of the complex ion from aqueous metal ion and the ligands.
For Ag(NH3)2+:
Kf = [Ag(NH3)2+] / [Ag+][NH3]2
Kf values are usually very large. For example,
Kf for Ag(NH3)2+ = 1.7 x 107
When doing calculations with Kf values, it would be harder to work with these values, because the approximation method will never hold. An easy way around this is to remember the mathematical rules that apply to Kc.
When a reaction is reversed so that the reactants are products and the products are reactants, the new Kc is the inverse (1 / Kc) of the original Kc.
Dissociation Constant: Kd, the inverse of Kf.
Example:
Reverse the reaction and take the new Kf, which will be Kd:
Ag(NH3)2+ (aq) ⇌ Ag+ (aq) + 2NH3 (aq)
Kd = [Ag+][NH3]2 / [Ag(NH3)2+] =( 1 / Kf)
Example:
For the formation of the complex ion:
Cu2+ (aq) + 4NH3 (aq) ⇌ Cu(NH3)42+ (aq), Kf = 4.8 x 1012
What is the [Cu2+] at equilibrium when you start with 0.100 M Cu(NO3)2 and 1.00 M NH3?
Note that the approximation method will not hold for these numbers, and we'll be left to work through some unnecessarily tedious math. Or, we could use some shortcuts and make it much easier:
First, assume that the reaction goes to completion:
Cu2+ (aq) + 4NH3 → Cu(NH3)42+ (aq)
Note that only 0.100 M is subtracted in the second line of the table because there is an excess of NH3, and less Cu2+.
Now reverse the reaction to let some of the complex ion fall apart and convert Kf to Kd:
Cu(NH3)42+ (aq) ⇌ Cu2+ (aq) + 4NH3 (aq)
Kd = 1 / Kf = 2.08 x 10-13
2.08 x 10-13 = (x)(0.600 + 4x)4 / (0.100 – x)
Approximation method holds,
2.08 x 10-13 = (x)(0.600)4 / (0.100)
x = 1.60 x 10-13
Complex Ions and Solubility:
Example:
Calculate the molar solubility of AgCl in 1.0 M NH3.
This problem involves the solubility equilibrium for AgCl and the complex ion equilibrium:
AgCl (s) ⇌ Ag+ (aq) + Cl- (aq), Ksp = 1.8 x 10-10
Ag+ (aq) + 2NH3 (aq) ⇌ Ag(NH3)2+ (aq), Kf = 1.7 x 107
The overall equation for the process can be obtained by applying Hess's Law and adding them together:
AgCl (s) + Ag+ (aq) + 2NH3 (aq) ⇌ Ag+ (aq) + Cl- (aq) + Ag(NH3)2+ (aq)
The Ag+ will cancel on each side and the equation will be:
AgCl (s) + 2NH3 (aq) ⇌ Cl- (aq) + Ag(NH3)2+ (aq)
And remember from Chapter 14 that when you add two reactions, you obtain the new equilibrium constant by multiplying their equilibrium constants.
Kc = Ksp x Kf =
(1.8 x 10-10)(1.7 x 107) = 0.00306
0.00306 = (x2) / (1.00 – 2x)2 → approximation holds
0.00306 = (x2) / (1.00)2
x = 0.0553 M = [Ag(NH3)2+] = [AgCl]
17.1: The Solubility Product Constant
Remember these? Solubility rules:
However, the compounds that we have referred to as “insoluble” in the past are realistically “slightly soluble.” Everything is at least slightly soluble in water.
Note: Every reaction considered here is assumed to be at 25oC.
Solubility Constant = Ksp = really just another version of Kc. It is the equilibrium constant for the solubility equilibrium of a slightly soluble (or nearly insoluble) ionic compound.
Example:
AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)
Kc = [Ag+][Cl-] = Ksp
Remember not to include the [AgCl] as part of the Ksp because it is a solid, so just like before, it is ignored.
It's important to be able to recognize which ions an ionic compound will break up into. Once the equilibrium is written, it's easy to find the Ksp. Here's a reminder of the polyatomic ions:
Examples:
CaSO4 (s) ⇌ Ca2+ (aq) + SO42- (aq)
Ksp = [Ca2+][SO42-]
AgC2H3O2 (s) ⇌ Ag+ (aq) + C2H3O2- (aq)
Ksp = [Ag+][C2H3O2-]
CaF2 (s) ⇌ Ca2+ (aq) + 2F- (aq)
Ksp = [Ca2+][F-]2
Ca3(PO4)2 (s) ⇌ 3Ca2+ (aq) + 2PO43- (aq)
Ksp = [Ca2+]3[PO43-]2
Zn(OH)2 (s) ⇌ Zn2+ (aq) + 2OH- (aq)
Ksp = [Zn2+][OH-]2
Solubility is generally expressed in grams per liter: g/L. Molar solubility is expressed in mols per liter: mols/L. Note that this is the same thing as molarity.
Example:
Solubility of AgCl = 1.9 x 10-3 g/L
What is the Ksp?
The values that are needed to find the Ksp are expressed in Molarity, which is mols per liter, and we have grams per liter. So the first step is to convert from grams to mols using the periodic table.
For reference:
[AgCl] = (1.9 x 10-3 g / 1 L) x (1 mol AgCl / 143 g AgCl) = 1.33 x 10-5 mol/L AgCl
AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)
Ksp = x2
Since we know that 1.33 x 10-5 M of AgCl will dissociate into 1.33 x 10-5 M of Ag+ and 1.33 x 10-5 M of Cl-, we know that x = 1.33 x 10-5 mol/L.
Ksp = (1.33 x 10-5)2 = 1.77 x 10-10
In some problems, we won't be given the solubility directly when we are asked to find the Ksp.
Example:
One liter of a solution saturated with TcF is evaporated, leaving 5.32 x 10-5g of TcF. What is the Ksp?
TcF (s) ⇌ Tc+ (aq) + F- (aq)
Ksp = [Tc+][F-]
We have been given the information that for one liter of solution, we have 5.32 x 10-5g of TcF, which is solubility in g/L. So the first step is to convert from grams to mols:
( 5.32 x 10-5 g / 1L) x (1 mol TcF / 117 g TcF) = 4.55 x 10-7 M TcF
And we don't need to build a table to see that 4.55 x 10-7 M of TcF will dissociate into 4.55 x 10-7 M of Tc+ and 4.55 x 10-7 M of F-. Also, that Ksp = (4.55 x 10-7)2 = 2.07 x 10-13
Everything up to this point has been pretty simple because everything has been 1:1. Remember that when ions have coefficients associated with them, these coefficients show up not only in the calculation for Ksp as exponents, but also in the table we'll build to solve for x.
Example:
One liter of a solution saturated with W2S3 is evaporated, leaving 9.35 x 10-4g of W2S3. What is the Ksp?
The first step is to convert from grams to mols:
(9.35 x 10-4 g / L) x (1 mol W2S3 / 464 g W2S3) = 2.02 x 10-6 M W2S3
W2S3 (s) ⇌ 2W3+ (aq) + 3S2- (aq)
Ksp = [W3+]2[S2-]3
Ksp = (2x)2(3x)3
We know 2.02 x 10-6 M W2S3 dissociates, so we know x = 2.02 x 10-6
Ksp = (2 x 2.02 x 10-6)2(3 x 2.02 x 10-6)3 = 3.63 x 10-27
Other problems may give the Ksp and ask for the solubility.
Example:
Find the molar solubility and the solubility of Pb3(AsO4)2. The Ksp for Pb3(AsO4)2 = 4 x 10-36
Pb3(AsO4)2 (s) ⇌ 3Pb2+ (aq) + 2AsO43- (aq)
Ksp = [Pb2+]3[AsO43-]2 = (3x)3(2x)2 = (27x3)(4x2) = 108x5 = 4 x 10-36
(divide by 108 and take the fifth root)
x = 3.26 x 10-8 M = molar solubility
To find solubility, use the periodic table to convert this value to grams per liter:
(3.26 x 10-8 mols / L) x (899 g Pb3(AsO4)2/ mol Pb3(AsO4)2) =
2.93 x 10-5 g / L Pb3(AsO4)2
However, the compounds that we have referred to as “insoluble” in the past are realistically “slightly soluble.” Everything is at least slightly soluble in water.
Note: Every reaction considered here is assumed to be at 25oC.
Solubility Constant = Ksp = really just another version of Kc. It is the equilibrium constant for the solubility equilibrium of a slightly soluble (or nearly insoluble) ionic compound.
Example:
AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)
Kc = [Ag+][Cl-] = Ksp
Remember not to include the [AgCl] as part of the Ksp because it is a solid, so just like before, it is ignored.
It's important to be able to recognize which ions an ionic compound will break up into. Once the equilibrium is written, it's easy to find the Ksp. Here's a reminder of the polyatomic ions:
Examples:
CaSO4 (s) ⇌ Ca2+ (aq) + SO42- (aq)
Ksp = [Ca2+][SO42-]
AgC2H3O2 (s) ⇌ Ag+ (aq) + C2H3O2- (aq)
Ksp = [Ag+][C2H3O2-]
CaF2 (s) ⇌ Ca2+ (aq) + 2F- (aq)
Ksp = [Ca2+][F-]2
Ca3(PO4)2 (s) ⇌ 3Ca2+ (aq) + 2PO43- (aq)
Ksp = [Ca2+]3[PO43-]2
Zn(OH)2 (s) ⇌ Zn2+ (aq) + 2OH- (aq)
Ksp = [Zn2+][OH-]2
Solubility is generally expressed in grams per liter: g/L. Molar solubility is expressed in mols per liter: mols/L. Note that this is the same thing as molarity.
Example:
Solubility of AgCl = 1.9 x 10-3 g/L
What is the Ksp?
The values that are needed to find the Ksp are expressed in Molarity, which is mols per liter, and we have grams per liter. So the first step is to convert from grams to mols using the periodic table.
For reference:
[AgCl] = (1.9 x 10-3 g / 1 L) x (1 mol AgCl / 143 g AgCl) = 1.33 x 10-5 mol/L AgCl
AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)
Ksp = x2
Since we know that 1.33 x 10-5 M of AgCl will dissociate into 1.33 x 10-5 M of Ag+ and 1.33 x 10-5 M of Cl-, we know that x = 1.33 x 10-5 mol/L.
Ksp = (1.33 x 10-5)2 = 1.77 x 10-10
In some problems, we won't be given the solubility directly when we are asked to find the Ksp.
Example:
One liter of a solution saturated with TcF is evaporated, leaving 5.32 x 10-5g of TcF. What is the Ksp?
TcF (s) ⇌ Tc+ (aq) + F- (aq)
Ksp = [Tc+][F-]
We have been given the information that for one liter of solution, we have 5.32 x 10-5g of TcF, which is solubility in g/L. So the first step is to convert from grams to mols:
( 5.32 x 10-5 g / 1L) x (1 mol TcF / 117 g TcF) = 4.55 x 10-7 M TcF
And we don't need to build a table to see that 4.55 x 10-7 M of TcF will dissociate into 4.55 x 10-7 M of Tc+ and 4.55 x 10-7 M of F-. Also, that Ksp = (4.55 x 10-7)2 = 2.07 x 10-13
Everything up to this point has been pretty simple because everything has been 1:1. Remember that when ions have coefficients associated with them, these coefficients show up not only in the calculation for Ksp as exponents, but also in the table we'll build to solve for x.
Example:
One liter of a solution saturated with W2S3 is evaporated, leaving 9.35 x 10-4g of W2S3. What is the Ksp?
The first step is to convert from grams to mols:
(9.35 x 10-4 g / L) x (1 mol W2S3 / 464 g W2S3) = 2.02 x 10-6 M W2S3
W2S3 (s) ⇌ 2W3+ (aq) + 3S2- (aq)
Ksp = [W3+]2[S2-]3
Ksp = (2x)2(3x)3
We know 2.02 x 10-6 M W2S3 dissociates, so we know x = 2.02 x 10-6
Ksp = (2 x 2.02 x 10-6)2(3 x 2.02 x 10-6)3 = 3.63 x 10-27
Other problems may give the Ksp and ask for the solubility.
Example:
Find the molar solubility and the solubility of Pb3(AsO4)2. The Ksp for Pb3(AsO4)2 = 4 x 10-36
Pb3(AsO4)2 (s) ⇌ 3Pb2+ (aq) + 2AsO43- (aq)
Ksp = [Pb2+]3[AsO43-]2 = (3x)3(2x)2 = (27x3)(4x2) = 108x5 = 4 x 10-36
(divide by 108 and take the fifth root)
x = 3.26 x 10-8 M = molar solubility
To find solubility, use the periodic table to convert this value to grams per liter:
(3.26 x 10-8 mols / L) x (899 g Pb3(AsO4)2/ mol Pb3(AsO4)2) =
2.93 x 10-5 g / L Pb3(AsO4)2
Friday, October 15, 2010
16.7: Acid-Base Titration Curves
Acid-Base Titrations:
An acid-base titration is a procedure for determining the amount of acid (or base) in a solution by determining the volume of base (or acid) of known concentration that will completely react with it.
Can titrate:
1) A strong acid with a strong base
2) A strong acid with a weak base
3) A weak acid with a strong base
4) A weak acid with a weak base → this kind of titration will not be on the exam.
An acid-base titration curve is a plot of the pH of a solution of acid (or base) against the volume of added base (or acid). Here is one for a strong acid titrated with a strong base:
The change in pH as the base is added will be slow at first, then it will reach a point where the pH increases drastically before leveling out again.
The middle of this rapid incline is called the equivalence point: the point in a titration when the smallest amount of titrant has been added that is sufficient to fully neutralize or react with the acid or base (the point when all the acid and base have reacted).
Example:
The case of a strong acid and a strong base:
Note: The pH at the equivalence point for a strong acid and a strong base will always be 7.
What is the pH when 27.0mL of 0.450 M HCl is titrated (read as mixed!) with 45.0mL of 0.500 M NaOH?
Remember with mixing/titrating, the first thing we do is convert to moles:
mols HCl = (0.0270 L)(0.450 mols/L) = 0.0122 mol HCl
mols NaOH = (0.0450 L)(0.500 mols/L) = 0.0225 mol NaOH
The acid and base react to form a salt:
HCl (aq) + NaOH (aq) → H2O (l) + NaCl (aq)
Convert to concentration by dividing by new volume:
NaOH = (0.0103 mol) / (0.072 L) = 0.143 M NaOH = 0.143 M OH-
pOH = -log(0.143) = 0.844
pH = 14 – pOH = 13.2
Example:
The case of a strong acid and a weak base:
What is the pH at the equivalence point when 0.150 M of HCl titrates 20 mL of 0.200 M NH3?
Remember that at the equivalence point, mols of HCl = mols of NH3.
Since we're dealing with a titration, first convert to mols:
mols NH3= (0.200 mols/L)(0.0200 L) = 4.00 x 10-3 mols NH3
and mols NH3 = mols HCl = 4.00 x 10-3 mols
To get the final volume after titration, we'll need to know the volume of HCl. Since we know the mols and molarity, we can use what we know to find the volume.
Concentration = Mols x Volume,
Volume = Mols / Concentration
Volume HCl = (4.00 x 10-3 mols) / (0.150 M) = 0.0267 L HCl
So total volume of titration is 0.0267L + 0.0200 L = 0.0467 L
Concentrations of HCl and NH3 = (4.00 x 10-3 mols) / (0.0467 L) = 0.0857 M
So 0.0857 M of the acid and 0.0857 M of the base react to form 0.0857 M of a salt:
HCl + NH3 → NH4Cl
NH4Cl dissociates in water and leaves NH4+ and Cl- ions. Cl- ions form a strong acid which dissociates 100% in water, so no reaction there. When NH4+ ions react with water:
NH4+ (aq) + H2O (l) ⇌ H3O+ (aq) + NH3 (aq)
Ka = (x2) / (0.0857 – x) → approximation holds
Ka= (x2) / (0.0857)
We need to find the Ka for NH4+, so first find Kb of the conjugate base, NH3, and plug it in:
Ka = Kw / Kb
Kb for NH3 = 1.8 x 10-5
Ka for NH4+= (1 x 10-14) / (1.8 x 10-5) = 5.56 x 10-10
5.56 x 10-10 = (x2) / (0.0857)
x = 6.9 x 10-6 = [H3O+]
pH = -log(6.9 x 10-6) = 5.16
Example:
The case of a weak acid and a strong base:
What is the pH when 35.0mL of 0.215 M HClO is titrated by 0.400 M LiOH to the equivalence point?
Remember that at equivalence, mols HClO = mols LiOH.
And since we're titrating, remember the first thing to do is convert to mols:
mols HClO = (0.350 L)(0.215 mols/L) = 0.00753 mols HClO
and mols HClO = mols LiOH = 0.00753 mols
To get the final volume after titration, we'll need to know the volume of LiOH. Since we know the mols and molarity, we can use what we know to find the volume.
Concentration = Mols x Volume,
Volume = Mols / Concentration
Volume LiOH = ( 0.00753 mols) / (0.400 M) = 0.0188 L LiOH
So total volume of titration is 0.0350L + 0.0188 L = 0.0538 L
Concentrations of HClO and LiOH = (0.00753 mols) / (0.0538 L) = 0.140 M
So 0.140 M of the acid and 0.140 M of the base react to form 0.140 M of a salt:
HClO + LiOH → LiClO
LiClO dissociates in water and leaves Li+ and ClO- ions. Li+ ions form a strong base which dissociates 100% in water, so no reaction there. When ClO- ions react with water:
ClO- (aq) + H2O (l) ⇌ HClO (aq) + OH- (aq)
Kb = (x2) / (0.140 – x) → approximation holds
Kb = (x2) / (0.140)
We need to find the Kb for ClO-, so first find Ka of the acid, HClO, and plug it in:
Kb = Kw / Ka
Ka for HClO = 3.5 x 10-8
Kb for ClO- = (1 x 10-14) / (3.5 x 10-8) = 2.86 x 10-7
2.86 x 10-7 = (x2) / (0.140)
x = 2.00 x 10-4 = [OH-]
pOH = -log(2.00 x 10-4) = 3.7
pH = 14 – 3.7 = 10.3
An acid-base titration is a procedure for determining the amount of acid (or base) in a solution by determining the volume of base (or acid) of known concentration that will completely react with it.
Can titrate:
1) A strong acid with a strong base
2) A strong acid with a weak base
3) A weak acid with a strong base
4) A weak acid with a weak base → this kind of titration will not be on the exam.
An acid-base titration curve is a plot of the pH of a solution of acid (or base) against the volume of added base (or acid). Here is one for a strong acid titrated with a strong base:
The change in pH as the base is added will be slow at first, then it will reach a point where the pH increases drastically before leveling out again.
The middle of this rapid incline is called the equivalence point: the point in a titration when the smallest amount of titrant has been added that is sufficient to fully neutralize or react with the acid or base (the point when all the acid and base have reacted).
Example:
The case of a strong acid and a strong base:
Note: The pH at the equivalence point for a strong acid and a strong base will always be 7.
What is the pH when 27.0mL of 0.450 M HCl is titrated (read as mixed!) with 45.0mL of 0.500 M NaOH?
Remember with mixing/titrating, the first thing we do is convert to moles:
mols HCl = (0.0270 L)(0.450 mols/L) = 0.0122 mol HCl
mols NaOH = (0.0450 L)(0.500 mols/L) = 0.0225 mol NaOH
The acid and base react to form a salt:
HCl (aq) + NaOH (aq) → H2O (l) + NaCl (aq)
Convert to concentration by dividing by new volume:
NaOH = (0.0103 mol) / (0.072 L) = 0.143 M NaOH = 0.143 M OH-
pOH = -log(0.143) = 0.844
pH = 14 – pOH = 13.2
Example:
The case of a strong acid and a weak base:
What is the pH at the equivalence point when 0.150 M of HCl titrates 20 mL of 0.200 M NH3?
Remember that at the equivalence point, mols of HCl = mols of NH3.
Since we're dealing with a titration, first convert to mols:
mols NH3= (0.200 mols/L)(0.0200 L) = 4.00 x 10-3 mols NH3
and mols NH3 = mols HCl = 4.00 x 10-3 mols
To get the final volume after titration, we'll need to know the volume of HCl. Since we know the mols and molarity, we can use what we know to find the volume.
Concentration = Mols x Volume,
Volume = Mols / Concentration
Volume HCl = (4.00 x 10-3 mols) / (0.150 M) = 0.0267 L HCl
So total volume of titration is 0.0267L + 0.0200 L = 0.0467 L
Concentrations of HCl and NH3 = (4.00 x 10-3 mols) / (0.0467 L) = 0.0857 M
So 0.0857 M of the acid and 0.0857 M of the base react to form 0.0857 M of a salt:
HCl + NH3 → NH4Cl
NH4Cl dissociates in water and leaves NH4+ and Cl- ions. Cl- ions form a strong acid which dissociates 100% in water, so no reaction there. When NH4+ ions react with water:
NH4+ (aq) + H2O (l) ⇌ H3O+ (aq) + NH3 (aq)
Ka = (x2) / (0.0857 – x) → approximation holds
Ka= (x2) / (0.0857)
We need to find the Ka for NH4+, so first find Kb of the conjugate base, NH3, and plug it in:
Ka = Kw / Kb
Kb for NH3 = 1.8 x 10-5
Ka for NH4+= (1 x 10-14) / (1.8 x 10-5) = 5.56 x 10-10
5.56 x 10-10 = (x2) / (0.0857)
x = 6.9 x 10-6 = [H3O+]
pH = -log(6.9 x 10-6) = 5.16
Example:
The case of a weak acid and a strong base:
What is the pH when 35.0mL of 0.215 M HClO is titrated by 0.400 M LiOH to the equivalence point?
Remember that at equivalence, mols HClO = mols LiOH.
And since we're titrating, remember the first thing to do is convert to mols:
mols HClO = (0.350 L)(0.215 mols/L) = 0.00753 mols HClO
and mols HClO = mols LiOH = 0.00753 mols
To get the final volume after titration, we'll need to know the volume of LiOH. Since we know the mols and molarity, we can use what we know to find the volume.
Concentration = Mols x Volume,
Volume = Mols / Concentration
Volume LiOH = ( 0.00753 mols) / (0.400 M) = 0.0188 L LiOH
So total volume of titration is 0.0350L + 0.0188 L = 0.0538 L
Concentrations of HClO and LiOH = (0.00753 mols) / (0.0538 L) = 0.140 M
So 0.140 M of the acid and 0.140 M of the base react to form 0.140 M of a salt:
HClO + LiOH → LiClO
LiClO dissociates in water and leaves Li+ and ClO- ions. Li+ ions form a strong base which dissociates 100% in water, so no reaction there. When ClO- ions react with water:
ClO- (aq) + H2O (l) ⇌ HClO (aq) + OH- (aq)
Kb = (x2) / (0.140 – x) → approximation holds
Kb = (x2) / (0.140)
We need to find the Kb for ClO-, so first find Ka of the acid, HClO, and plug it in:
Kb = Kw / Ka
Ka for HClO = 3.5 x 10-8
Kb for ClO- = (1 x 10-14) / (3.5 x 10-8) = 2.86 x 10-7
2.86 x 10-7 = (x2) / (0.140)
x = 2.00 x 10-4 = [OH-]
pOH = -log(2.00 x 10-4) = 3.7
pH = 14 – 3.7 = 10.3
16.6: Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation:
An equation that relates the pH of a buffer for different concentrations of conjugate acid and base:
or
And thank you wikipedia for making it easy to see how it relates to things we've already learned:
(Note: I know wikipedia is shunned for it's user contributed content, but every wikipedia article has a detailed list of linked references at the bottom of the page that can be checked to verify that an article's content isn't just made up bullshit. When in doubt, check references.)
An equation that relates the pH of a buffer for different concentrations of conjugate acid and base:
or
And thank you wikipedia for making it easy to see how it relates to things we've already learned:
(Note: I know wikipedia is shunned for it's user contributed content, but every wikipedia article has a detailed list of linked references at the bottom of the page that can be checked to verify that an article's content isn't just made up bullshit. When in doubt, check references.)
Also note that this equation only works when the approximation method holds, when Ca / Ka is greater than 100.
Sunday, October 10, 2010
16.5-16.6: The Common Ion Effect-Buffers
Common Ion Effect: the shift in an ionic equilibrium caused by the addition of a solute that provides an ion that takes part in the equilibrium.
Example:
The problem we did in class, at least from what I wrote down, does not follow what the book says. According to Le'Chatelier's principle, adding a strong acid to a weak acid should decrease the degree of ionization for the weak acid, since adding more product (H3O+) will cause the reaction to shift to the left. For the problem we did in class, what I wrote down has the reaction shifting to the right. So instead, see the example from the book, pages 672-673.
Example:
What is the pH for a solution of 0.5 M HClO and 1.5 M HBr?
The common ion here is H+ (H3O+). Since we know HBr is a strong acid, we also know it will dissociate 100% in water, adding 1.5M H3O+ to the solution.
HClO (aq) + H2O (l) ⇌ H3O+ (aq) + ClO- (aq)
Ka = x(1.50 + x) / (0.50 – x) → approximation holds
3.5 x 10-8 = (1.50x) / (0.50)
x = 1.17 x 10-8
[H3O+] = 1.50 + 1.17 x 10-8 = 1.50 M
pH = -log(1.50) = -0.18
Note: It is possible to have a pH greater than 14 and less than 0. For example, if you have greater than 1M of a strong acid, the pH will be less than zero.
Also, the pH of a solution containing a weak acid and a strong acid will generally be equal to the pH of the strong acid alone.
Buffers:
(Basically the same as a common ion problem.)
Buffer: a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it.
Buffers contain either a weak acid and its conjugate base in a salt or a weak base and its conjugate acid in a salt. Consider a buffer that contains approximately equal concentrations of a weak acid HA and it's conjugate base A-. When a strong acid is added to the buffer, it supplies hydronium ions that react with the base A- :
H3O+ (aq) + A- (aq) → HA (aq) + H2O (l)
When a strong base is added to the buffer it supplies hydroxide ions which react with the acid HA:
OH- (aq) + HA (aq) ⇌ H2O (l) + A- (aq)
A buffer solution resists changes in pH through its ability to combine with both H3O+ and OH- ions.
Examples of buffers:
HCN and NaCN (notice the common ion CN-)
NH3 and NH4Cl
Example:
What is the pH of a solution of 0.400 M HC2H3O2 and 0.200 M NaC2H3O2?
HC2H3O2 → weak acid → HA
NaC2H3O2 → salt containing conjugate base → C2H3O2- → A-
Note: Na+ doesn't do much but show up. Na+ is like the person C2H3O2- brought to the party then didn't hang out with once they got there. Which is just mean. So C2H3O2- is kind of an asshole (A-). Maybe that's an easy way to remember it.
But we know 0.200 M NaC2H3O2 will dissociate to yield 0.200 M C2H3O2-.
HC2H3O2 (aq) + H2O (l) ⇌ H3O+ (aq) + C2H3O2- (aq)
Notice that there is a nonzero initial concentration of A-. This is from the salt.
Ka = x(0.200 + x) / (0.400 – x) → approximation holds
1.7 x 10-5 = x(0.200) / (0.400)
x = 3.4 x 10-5 = [H3O+]
pH = -log(3.4 x 10-5) = 4.47
Example:
What is the pH of a solution that is 0.45 M HF and 0.115 M NaF?
HF (aq) + H2O (l) ⇌ H3O+ (aq) + F- (aq)
Ka = x(0.115 + x) / (0.45 – x) → approximation holds
6.8 x 10-4 = x(0.115) / (0.45)
x = 0.00266 = [H3O+]
pH = -log(0.00266) = 2.57
Example:
For a buffer containing HC2H3O2 and NaC2H3O2, which will react with NaOH?
Consider NaOH to be OH-.
HC2H3O2 + OH- → H2O + C2H3O2-
Which will react with HCl?
Consider HCl to be H+ (H3O+).
C2H3O2- + H3O+ → HC2H3O2 + H2O
or A- + H3O+ → HA + H2O
Note: When acids and bases are added to a buffer, the pH will change, but only slightly. The pH does not stay exactly the same.
Example:
THIS IS THAT HUGE 3 PART PROBLEM REGARDING BUFFERS
(Note: Ka for HC3H5O2 = 1.3 x 10-5)
What is the pH for a buffer made from MIXING 47.0 mL of 0.678 M HC3H5O2 and 33.0 mL of 0.533 M NaC3H5O2?
Note: In this problem, we are making a buffer.
HC3H5O2 → weak acid → HA
NaC3H5O2 → salt with conjugate base → NaA → A-
Whenever you see the words MIXING or TITRATING it is a clue that the volume is going to change, so molarity will change. The first step in these problems is always to convert to moles:
In order to do that we need to convert mL to L:
(47.0 mL) x (1 L / 1000mL) = 0.047L
(33.0 mL) x (1L / 1000mL) = 0.033L
(0.047L) x (0.678 mols / L HA) = 0.0319 mols HA
(0.033L) x (0.533 mols / L A-) = 0.0176 mols A-
Now divide by the new volume to get the new molarity:
New volume = 0.047L + 0.033 L = 0.080 L
0.0319 mols HA / 0.080 L = 0.399 M HA
0.0176 mols A- / 0.080 L = 0.220 M A-
HA + H2O ⇌ H3O+ + A-
Ka = x(0.220 + x) / (0.399 – x) → approximation holds
1.3 x 10-5 = x(0.220) / (0.399)
x = 2.36 x 10-5 = [H3O+]
pH = -log(2.63 x 10-5) = 4.63
Now find the pH when 95.0 mL of the buffer we've just created is mixed with 5.00 mL of 0.100 M HCl.
We know that it is the conjugate base in our buffer that will react with the HCl (the A-).
We also know that HCl is a strong acid that dissociates 100 % in water.
When mixing, convert to moles:
(0.399 M HA / L)(0.095 L) = 0.379 mols HA
(0.220 M A- / L)(0.095 L) = 0.0209 mols A-
(0.100 M HCl / L)(0.005 L) = 5 x 10-4 mols HCl
Now we need to look at how the conjugate base of the buffer and the HCl are going to react and how much HA they are going to produce:
HCl + A- → HA + Cl-
We subtract the mols of HCl from the left side of the reaction because the HCl will be used up before the A- is used up, there is less of it (limiting). And since this is what we are subtracting from the left side, it is also what we are adding to the right side. So now:
A- = 0.0204 mols
HA = 0.0384 mols
Remember we're still in mols, so we need to divide by our new total volume to get molarity:
A- = 0.0204 mols / 0.1 L = 0.204 M
HA = 0.0384 mols / 0.1 L = 0.384 M
0.384 M is our new acid concentration. What happens when it dissociates in water? We get our final hydronium ion concentration from which we can calculate pH:
HA + H2O ⇌ H3O+ + A-
Ka = x(0.204 + x) / (0.384 – x) → approximation holds
1.3 x 10-5 = x(2.04) = (0.384)
x = 2.45 x 10-5 = [H3O+]
pH = -log(2.45 x 10-5) = 4.61
Now find the pH when 95.0 mL of the buffer we initially created is mixed with 5.00 mL of 0.350 M NaOH:
Since we're dealing with a changing volume, first thing to do is convert to moles:
(0.399 mols / 1 L HA)(0.0950L) = 0.0379 mols HA
(0.220 mols/ 1 L A-)(0.0950L) = 0.0209 mols A-
(0.350 mols/ 1 L NaOH)(0.005L) = 0.00175 mols OH-
Which component of the buffer will the NaOH react with? The HA.
OH- + HA → H2O + A-
We know to subtract 0.00175 mols because there is less of it and everything is 1:1. Now take the new values for HA and A-, which are in mols, and divide them by the new volume to get the concentration in molars:
HA = 0.0362 mols / 0.1 L = 0.362 M HA
A- = 0.0227 mols / 0.1 L = 0.227 M A-
Now this acid will dissociate in water to yield H3O+, from which we can find the new pH:
HA (aq) + H2O (l) ⇌ A- (aq) + H3O+ (aq)
Ka = x(0.227 + x) / (0.362 – x) → approximation holds
1.3 x 10-5 = x(0.227) / (0.362)
x = 2.07 x 10-5 = [H3O+]
pH = -log(2.07 x 10-5) = 4.68
Notice that the pH of the original buffer was 4.63. When we added a strong acid, the pH only dropped to 4.61, and when we added a strong base, the pH went up to 4.68. So although the pH didn't stay exactly the same, it stayed pretty stable.
Example:
The problem we did in class, at least from what I wrote down, does not follow what the book says. According to Le'Chatelier's principle, adding a strong acid to a weak acid should decrease the degree of ionization for the weak acid, since adding more product (H3O+) will cause the reaction to shift to the left. For the problem we did in class, what I wrote down has the reaction shifting to the right. So instead, see the example from the book, pages 672-673.
Example:
What is the pH for a solution of 0.5 M HClO and 1.5 M HBr?
The common ion here is H+ (H3O+). Since we know HBr is a strong acid, we also know it will dissociate 100% in water, adding 1.5M H3O+ to the solution.
HClO (aq) + H2O (l) ⇌ H3O+ (aq) + ClO- (aq)
Ka = x(1.50 + x) / (0.50 – x) → approximation holds
3.5 x 10-8 = (1.50x) / (0.50)
x = 1.17 x 10-8
[H3O+] = 1.50 + 1.17 x 10-8 = 1.50 M
pH = -log(1.50) = -0.18
Note: It is possible to have a pH greater than 14 and less than 0. For example, if you have greater than 1M of a strong acid, the pH will be less than zero.
Also, the pH of a solution containing a weak acid and a strong acid will generally be equal to the pH of the strong acid alone.
Buffers:
(Basically the same as a common ion problem.)
Buffer: a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it.
Buffers contain either a weak acid and its conjugate base in a salt or a weak base and its conjugate acid in a salt. Consider a buffer that contains approximately equal concentrations of a weak acid HA and it's conjugate base A-. When a strong acid is added to the buffer, it supplies hydronium ions that react with the base A- :
H3O+ (aq) + A- (aq) → HA (aq) + H2O (l)
When a strong base is added to the buffer it supplies hydroxide ions which react with the acid HA:
OH- (aq) + HA (aq) ⇌ H2O (l) + A- (aq)
A buffer solution resists changes in pH through its ability to combine with both H3O+ and OH- ions.
Examples of buffers:
HCN and NaCN (notice the common ion CN-)
NH3 and NH4Cl
Example:
What is the pH of a solution of 0.400 M HC2H3O2 and 0.200 M NaC2H3O2?
HC2H3O2 → weak acid → HA
NaC2H3O2 → salt containing conjugate base → C2H3O2- → A-
Note: Na+ doesn't do much but show up. Na+ is like the person C2H3O2- brought to the party then didn't hang out with once they got there. Which is just mean. So C2H3O2- is kind of an asshole (A-). Maybe that's an easy way to remember it.
But we know 0.200 M NaC2H3O2 will dissociate to yield 0.200 M C2H3O2-.
HC2H3O2 (aq) + H2O (l) ⇌ H3O+ (aq) + C2H3O2- (aq)
Notice that there is a nonzero initial concentration of A-. This is from the salt.
Ka = x(0.200 + x) / (0.400 – x) → approximation holds
1.7 x 10-5 = x(0.200) / (0.400)
x = 3.4 x 10-5 = [H3O+]
pH = -log(3.4 x 10-5) = 4.47
Example:
What is the pH of a solution that is 0.45 M HF and 0.115 M NaF?
HF (aq) + H2O (l) ⇌ H3O+ (aq) + F- (aq)
Ka = x(0.115 + x) / (0.45 – x) → approximation holds
6.8 x 10-4 = x(0.115) / (0.45)
x = 0.00266 = [H3O+]
pH = -log(0.00266) = 2.57
Example:
For a buffer containing HC2H3O2 and NaC2H3O2, which will react with NaOH?
Consider NaOH to be OH-.
HC2H3O2 + OH- → H2O + C2H3O2-
Which will react with HCl?
Consider HCl to be H+ (H3O+).
C2H3O2- + H3O+ → HC2H3O2 + H2O
or A- + H3O+ → HA + H2O
Note: When acids and bases are added to a buffer, the pH will change, but only slightly. The pH does not stay exactly the same.
Example:
THIS IS THAT HUGE 3 PART PROBLEM REGARDING BUFFERS
(Note: Ka for HC3H5O2 = 1.3 x 10-5)
What is the pH for a buffer made from MIXING 47.0 mL of 0.678 M HC3H5O2 and 33.0 mL of 0.533 M NaC3H5O2?
Note: In this problem, we are making a buffer.
HC3H5O2 → weak acid → HA
NaC3H5O2 → salt with conjugate base → NaA → A-
Whenever you see the words MIXING or TITRATING it is a clue that the volume is going to change, so molarity will change. The first step in these problems is always to convert to moles:
In order to do that we need to convert mL to L:
(47.0 mL) x (1 L / 1000mL) = 0.047L
(33.0 mL) x (1L / 1000mL) = 0.033L
(0.047L) x (0.678 mols / L HA) = 0.0319 mols HA
(0.033L) x (0.533 mols / L A-) = 0.0176 mols A-
Now divide by the new volume to get the new molarity:
New volume = 0.047L + 0.033 L = 0.080 L
0.0319 mols HA / 0.080 L = 0.399 M HA
0.0176 mols A- / 0.080 L = 0.220 M A-
HA + H2O ⇌ H3O+ + A-
Ka = x(0.220 + x) / (0.399 – x) → approximation holds
1.3 x 10-5 = x(0.220) / (0.399)
x = 2.36 x 10-5 = [H3O+]
pH = -log(2.63 x 10-5) = 4.63
Now find the pH when 95.0 mL of the buffer we've just created is mixed with 5.00 mL of 0.100 M HCl.
We know that it is the conjugate base in our buffer that will react with the HCl (the A-).
We also know that HCl is a strong acid that dissociates 100 % in water.
When mixing, convert to moles:
(0.399 M HA / L)(0.095 L) = 0.379 mols HA
(0.220 M A- / L)(0.095 L) = 0.0209 mols A-
(0.100 M HCl / L)(0.005 L) = 5 x 10-4 mols HCl
Now we need to look at how the conjugate base of the buffer and the HCl are going to react and how much HA they are going to produce:
HCl + A- → HA + Cl-
We subtract the mols of HCl from the left side of the reaction because the HCl will be used up before the A- is used up, there is less of it (limiting). And since this is what we are subtracting from the left side, it is also what we are adding to the right side. So now:
A- = 0.0204 mols
HA = 0.0384 mols
Remember we're still in mols, so we need to divide by our new total volume to get molarity:
A- = 0.0204 mols / 0.1 L = 0.204 M
HA = 0.0384 mols / 0.1 L = 0.384 M
0.384 M is our new acid concentration. What happens when it dissociates in water? We get our final hydronium ion concentration from which we can calculate pH:
HA + H2O ⇌ H3O+ + A-
Ka = x(0.204 + x) / (0.384 – x) → approximation holds
1.3 x 10-5 = x(2.04) = (0.384)
x = 2.45 x 10-5 = [H3O+]
pH = -log(2.45 x 10-5) = 4.61
Now find the pH when 95.0 mL of the buffer we initially created is mixed with 5.00 mL of 0.350 M NaOH:
Since we're dealing with a changing volume, first thing to do is convert to moles:
(0.399 mols / 1 L HA)(0.0950L) = 0.0379 mols HA
(0.220 mols/ 1 L A-)(0.0950L) = 0.0209 mols A-
(0.350 mols/ 1 L NaOH)(0.005L) = 0.00175 mols OH-
Which component of the buffer will the NaOH react with? The HA.
OH- + HA → H2O + A-
We know to subtract 0.00175 mols because there is less of it and everything is 1:1. Now take the new values for HA and A-, which are in mols, and divide them by the new volume to get the concentration in molars:
HA = 0.0362 mols / 0.1 L = 0.362 M HA
A- = 0.0227 mols / 0.1 L = 0.227 M A-
Now this acid will dissociate in water to yield H3O+, from which we can find the new pH:
HA (aq) + H2O (l) ⇌ A- (aq) + H3O+ (aq)
Ka = x(0.227 + x) / (0.362 – x) → approximation holds
1.3 x 10-5 = x(0.227) / (0.362)
x = 2.07 x 10-5 = [H3O+]
pH = -log(2.07 x 10-5) = 4.68
Notice that the pH of the original buffer was 4.63. When we added a strong acid, the pH only dropped to 4.61, and when we added a strong base, the pH went up to 4.68. So although the pH didn't stay exactly the same, it stayed pretty stable.
16.4: Acid-Base Properties of Salt Solutions
Predicting Whether a Salt Solution is Acidic, Basic, or Neutral:
Example:
Salt: NaCl
Parents: NaOH/HCl
pH: neutral (a strong acid and a strong base)
Salt: NaF
Parents: NaOH/HF
pH: basic (strong base, weak acid)
Salt: NH4Br
Parents: NH3/HBr
pH: acidic (strong acid, weak base)
Salt: NH4NO2
Parents: NH3/HNO2
pH: must find Ka, Kb and compare
But what do we need to find the Ka and Kb for? When dealing with a salt, follow solubility rules. All salts of ammonium are soluble, so the salt will break apart in water:
NH4NO2 (aq) → NH4+ (aq) + NO2- (aq)
These ions will go on to react with water, where one will produce OH- (giving a basic property) and one will produce H3O+ (to produce an acidic quality). Whether the solution itself is acidic or basic depends on whether more OH- is produced or more H3O+ is produced. In other words, we can find whether the solution is acidic or basic by seeing which is larger, the Ka or the Kb:
NH4+ (aq) + H2O (l) ⇌ H3O+ (aq) + NH3 (aq)
NO2- (aq) + H20 (l) ⇌ OH– (aq) + HNO2 (aq)
Because the reaction with NH4+ produces H3O+, we know we need to find the Ka for this reaction. The table doesn't list the Ka for NH4+. How do we find it?
The product of Ka and Kb for conjugate acid-base pairs equals Kw.
Ka x Kb = Kw
So to find the Ka for NH4+, we find the Kb of the conjugate base, and plug these numbers into the above equation to solve for Ka.
KaNH4+ = Kw / KbNH3 = (1 x 10-14) / (1.8 x 10-5) = 5.56 x 10-6
Because the reaction with NO2- produces OH-, we know we need to find the Kb for this reaction. The table doesn't list the Kb for NO2-. How do we find it?
Again, find the Ka for the conjugate acid, and plug these numbers into the above equation and solve for Kb.
KbNO2- = Kw / KaHNO2= (1 x 10-14) / (4.5 x 10-4) = 2.22 x 10-11
Here, Ka is greater than Kb, so the solution will be acidic.
Example:
What is the pH for 0.250 M KClO?
KClO is a salt. According to solubility rules, it is soluble:
KClO (aq) → K+ (aq) + ClO-(aq)
These ions will go on to react with water, where one will produce OH- (giving a basic property) and one will produce H3O+ (to produce an acidic quality).
K+ (aq) +H2O (l) ⇌ KOH (aq) + H3O+ (aq)
ClO- (aq) + H2O (l) ⇌ HClO (aq) OH- (aq)
But there's a catch. KOH is a strong base, and we know that strong bases dissociate (ionize) 100% in water. So really,
K+ (aq) + H2O (l) → No Reaction
We also know that we had 0.250 M KClO to begin with. The KClO dissociated into K+ and ClO- where everything is 1:1, so it dissociates to form 0.250 M of K+ (that produces no reaction) and 0.250 M ClO-, which will go on to react with water. This is a little more obvious when we build a table:
Now that we know we have 0.250M ClO-, we can build our table for the reaction with water:
ClO- (aq) + H2O (l) ⇌ HClO (aq) OH- (aq)
Hint: we know we need to find Kb because OH- is a product
Kb = (x2) / (0.250 – x) → approximation method holds
How do we find the Kb for ClO-?
KbClO- = Kw / KaHClO = (1 x 10-14) / (3.5 x 10-8) = 2.89 x 10-7
So: 2.89 x 10-7 = (x2) / 0.250
x = 2.69 x 10-4 = [OH-]
pOH = -log(2.69 x 10-4) = 3.57
pH = 14 – 3.57 = 10.4
Example:
What is the pH for 0.135M NH4Cl?
NH4Cl is a salt, which ions will it form when it dissociates?
NH4+ and Cl-
NH4Cl → NH4+ + Cl-
These ions with go on to react with water. We know Cl- will react to form HCl, which is a strong acid, and as such it will dissociate 100% in water, so overall, there is no reaction. So what happens to the NH4+? It is the conjugate acid of a weak base, so:
NH4+ (aq) + H2O (l) ⇌ NH3 (aq) + H3O+ (aq)
Again, it is easy to see that since we start out with 0.135 M of NH4Cl, we will get 0.135 M of each of the ions (1:1), so when we build a table for the reaction of the ammonium ion with water, we know to start with 0.135 M.
Hint: we know we need to find Ka because H3O+ is a product
Ka = (x2) / (0.135 – x) → approximation holds
How do we find the Ka for NH4+?
KaNH4+ = Kw / KbNH3 = (1 x 10-14) / (1.8 x 10-5) = 5.56 x 10-10
So: 5.56 x 10-10 = (x2) / (0.135)
x = 8.66 x 10-6 = [H3O+]
pH = -log(8.66 x 10-6) = 5.06
Example:
Salt: NaCl
Parents: NaOH/HCl
pH: neutral (a strong acid and a strong base)
Salt: NaF
Parents: NaOH/HF
pH: basic (strong base, weak acid)
Salt: NH4Br
Parents: NH3/HBr
pH: acidic (strong acid, weak base)
Salt: NH4NO2
Parents: NH3/HNO2
pH: must find Ka, Kb and compare
But what do we need to find the Ka and Kb for? When dealing with a salt, follow solubility rules. All salts of ammonium are soluble, so the salt will break apart in water:
NH4NO2 (aq) → NH4+ (aq) + NO2- (aq)
These ions will go on to react with water, where one will produce OH- (giving a basic property) and one will produce H3O+ (to produce an acidic quality). Whether the solution itself is acidic or basic depends on whether more OH- is produced or more H3O+ is produced. In other words, we can find whether the solution is acidic or basic by seeing which is larger, the Ka or the Kb:
NH4+ (aq) + H2O (l) ⇌ H3O+ (aq) + NH3 (aq)
NO2- (aq) + H20 (l) ⇌ OH– (aq) + HNO2 (aq)
Because the reaction with NH4+ produces H3O+, we know we need to find the Ka for this reaction. The table doesn't list the Ka for NH4+. How do we find it?
The product of Ka and Kb for conjugate acid-base pairs equals Kw.
Ka x Kb = Kw
So to find the Ka for NH4+, we find the Kb of the conjugate base, and plug these numbers into the above equation to solve for Ka.
KaNH4+ = Kw / KbNH3 = (1 x 10-14) / (1.8 x 10-5) = 5.56 x 10-6
Because the reaction with NO2- produces OH-, we know we need to find the Kb for this reaction. The table doesn't list the Kb for NO2-. How do we find it?
Again, find the Ka for the conjugate acid, and plug these numbers into the above equation and solve for Kb.
KbNO2- = Kw / KaHNO2= (1 x 10-14) / (4.5 x 10-4) = 2.22 x 10-11
Here, Ka is greater than Kb, so the solution will be acidic.
Example:
What is the pH for 0.250 M KClO?
KClO is a salt. According to solubility rules, it is soluble:
KClO (aq) → K+ (aq) + ClO-(aq)
These ions will go on to react with water, where one will produce OH- (giving a basic property) and one will produce H3O+ (to produce an acidic quality).
K+ (aq) +H2O (l) ⇌ KOH (aq) + H3O+ (aq)
ClO- (aq) + H2O (l) ⇌ HClO (aq) OH- (aq)
But there's a catch. KOH is a strong base, and we know that strong bases dissociate (ionize) 100% in water. So really,
K+ (aq) + H2O (l) → No Reaction
We also know that we had 0.250 M KClO to begin with. The KClO dissociated into K+ and ClO- where everything is 1:1, so it dissociates to form 0.250 M of K+ (that produces no reaction) and 0.250 M ClO-, which will go on to react with water. This is a little more obvious when we build a table:
Now that we know we have 0.250M ClO-, we can build our table for the reaction with water:
ClO- (aq) + H2O (l) ⇌ HClO (aq) OH- (aq)
Hint: we know we need to find Kb because OH- is a product
Kb = (x2) / (0.250 – x) → approximation method holds
How do we find the Kb for ClO-?
KbClO- = Kw / KaHClO = (1 x 10-14) / (3.5 x 10-8) = 2.89 x 10-7
So: 2.89 x 10-7 = (x2) / 0.250
x = 2.69 x 10-4 = [OH-]
pOH = -log(2.69 x 10-4) = 3.57
pH = 14 – 3.57 = 10.4
Example:
What is the pH for 0.135M NH4Cl?
NH4Cl is a salt, which ions will it form when it dissociates?
NH4+ and Cl-
NH4Cl → NH4+ + Cl-
These ions with go on to react with water. We know Cl- will react to form HCl, which is a strong acid, and as such it will dissociate 100% in water, so overall, there is no reaction. So what happens to the NH4+? It is the conjugate acid of a weak base, so:
NH4+ (aq) + H2O (l) ⇌ NH3 (aq) + H3O+ (aq)
Again, it is easy to see that since we start out with 0.135 M of NH4Cl, we will get 0.135 M of each of the ions (1:1), so when we build a table for the reaction of the ammonium ion with water, we know to start with 0.135 M.
Hint: we know we need to find Ka because H3O+ is a product
Ka = (x2) / (0.135 – x) → approximation holds
How do we find the Ka for NH4+?
KaNH4+ = Kw / KbNH3 = (1 x 10-14) / (1.8 x 10-5) = 5.56 x 10-10
So: 5.56 x 10-10 = (x2) / (0.135)
x = 8.66 x 10-6 = [H3O+]
pH = -log(8.66 x 10-6) = 5.06
Sunday, October 3, 2010
16.3-16.4: Base Ionization Equilibria-Acid/Base Properties of Salt Solutions
Base-Ionization Equilibria:
Equilibria involving weak bases are treated similarly to equilibria involving weak acids.
Example:
Ammonia ionizes in water as follows:
NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)
In the general form, the weak base is replaced with B (remember acids were replaced with HA) as follows:
B (aq) + H2O (l) ⇌ HB+ (aq) + OH- (aq)
Base-Ionization Constant = Kb (yet another version of Kc)
Example:
In a solution of 2.35 M NH3, what is the pH and pOH?
NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)
Kb = (x2) / (2.35 - x) = 1.8 x 10-5
Approximation holds, so:
x = 0.00650 = [OH-]
pOH = -log(0.00650) = 2.19
pH = 14-pOH = 11.81
Acid-Base Properties of Salt Solutions:
A salt is the product of a neutralization (acid/base) reaction.
Parent Acid + Parent Base → Salt + Water
Example:
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
NaCl parents are HCl and NaOH (the cation is Na+ and the anion is Cl-. Add OH- to the cation and H+ to the anion to get the parents).
Example:
Salt: KC2H3O2
Cation: K+
Anion: C2H3O2-
Parents: KOH, HC2H3O2
Example:
Salt: MgBr2
Cation: Mg2+
Anion: Br-
Parents: Mg(OH)2, 2HBr
Example:
Salt: LiCN
Cation: Li+
Anion: CN-
Parents: LiOH, HCN
Consider a solution of sodium cyanide: NaCN
The second reaction can also be seen as a base ionization, and the Kb can easily be written from it.
The sodium ion is unreactive with water, but the cyanide ion reacts to produce HCN and OH-. From the BrØnsted-Lowry point of view, CN- acts as a base because it accepts a proton from H2O. And since OH- is a product, the solution will have a basic pH. This reaction is the hydrolysis of CN-.
Hydrolysis: the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion.
Consider a solution with ammonium ion:
NH4+ (aq) + H2O (l) ⇌ NH3 (aq) + H3O+ (aq)
The ammonium ion acts as an acid, donating a proton to H2O and forming NH3. Since H3O+ is a product, the solution will be acidic.
This reaction can also be seen as an acid ionization, and the Ka can easily be written from it.
Is the salt acidic, basic, or neutral? (How does the pH change?)
1. A salt of a strong base and a strong acid will be neutral, pH 7. The salt dissociates into ions in water that then react with water to form strong acids and bases, and we know from before that strong acids and bases dissociate 100% in water, so there is no net reaction and pH doesn't change.
2. A salt of a strong base and a weak acid will be basic, pH above 7.
3. A salt of a weak base and a strong acid will be acidic, pH below 7.
4. A salt of a weak base and weak acid is not as easy to predict. Both ions hydrolyze and the acidity or basicity depends on the relative strengths of the two ions. Must compare the Ka of the cation with the Kb of the anion. If Ka is larger, the solution is acidic. If Kb is larger, the solution is basic.
Example:
Salt: NH4F
NH4F → NH4+ + F-
NH4+ + H2O ⇌ H3O+ + NH3
What is the Ka for NH3? (A hint that we need to find Ka and not Kb is H3O+ is a product)
NH3 is not on the list of Ka, but it is on the list of Kb in the text. And we know that:
Kw = 1 x 10-14 = Ka x Kb, so: Ka = Kw / Kb
For NH3, Ka = (1 x 10-14) / (1.8 x 10-5) = 5.56 x 10-10
F- + H2O ⇌ OH- + HF
What is the Kb for HF? (Need to find the Kb because OH- is a product)
HF is not on the list of Kb in the text, but it is on the list of Ka.
Kb = Kw / Ka
For HF, Kb = (1 x 10-14) / (6.8 x 10-4) = 1.47 x 10-11
Ka is greater than Kb here, so solution will be acidic.
Equilibria involving weak bases are treated similarly to equilibria involving weak acids.
Example:
Ammonia ionizes in water as follows:
NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)
In the general form, the weak base is replaced with B (remember acids were replaced with HA) as follows:
B (aq) + H2O (l) ⇌ HB+ (aq) + OH- (aq)
Base-Ionization Constant = Kb (yet another version of Kc)
Example:
In a solution of 2.35 M NH3, what is the pH and pOH?
NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)
Kb = (x2) / (2.35 - x) = 1.8 x 10-5
Approximation holds, so:
x = 0.00650 = [OH-]
pOH = -log(0.00650) = 2.19
pH = 14-pOH = 11.81
Acid-Base Properties of Salt Solutions:
A salt is the product of a neutralization (acid/base) reaction.
Parent Acid + Parent Base → Salt + Water
Example:
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
NaCl parents are HCl and NaOH (the cation is Na+ and the anion is Cl-. Add OH- to the cation and H+ to the anion to get the parents).
Example:
Salt: KC2H3O2
Cation: K+
Anion: C2H3O2-
Parents: KOH, HC2H3O2
Example:
Salt: MgBr2
Cation: Mg2+
Anion: Br-
Parents: Mg(OH)2, 2HBr
Example:
Salt: LiCN
Cation: Li+
Anion: CN-
Parents: LiOH, HCN
Consider a solution of sodium cyanide: NaCN
The second reaction can also be seen as a base ionization, and the Kb can easily be written from it.
The sodium ion is unreactive with water, but the cyanide ion reacts to produce HCN and OH-. From the BrØnsted-Lowry point of view, CN- acts as a base because it accepts a proton from H2O. And since OH- is a product, the solution will have a basic pH. This reaction is the hydrolysis of CN-.
Hydrolysis: the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion.
Consider a solution with ammonium ion:
NH4+ (aq) + H2O (l) ⇌ NH3 (aq) + H3O+ (aq)
The ammonium ion acts as an acid, donating a proton to H2O and forming NH3. Since H3O+ is a product, the solution will be acidic.
This reaction can also be seen as an acid ionization, and the Ka can easily be written from it.
Is the salt acidic, basic, or neutral? (How does the pH change?)
1. A salt of a strong base and a strong acid will be neutral, pH 7. The salt dissociates into ions in water that then react with water to form strong acids and bases, and we know from before that strong acids and bases dissociate 100% in water, so there is no net reaction and pH doesn't change.
2. A salt of a strong base and a weak acid will be basic, pH above 7.
3. A salt of a weak base and a strong acid will be acidic, pH below 7.
4. A salt of a weak base and weak acid is not as easy to predict. Both ions hydrolyze and the acidity or basicity depends on the relative strengths of the two ions. Must compare the Ka of the cation with the Kb of the anion. If Ka is larger, the solution is acidic. If Kb is larger, the solution is basic.
Example:
Salt: NH4F
NH4F → NH4+ + F-
NH4+ + H2O ⇌ H3O+ + NH3
What is the Ka for NH3? (A hint that we need to find Ka and not Kb is H3O+ is a product)
NH3 is not on the list of Ka, but it is on the list of Kb in the text. And we know that:
Kw = 1 x 10-14 = Ka x Kb, so: Ka = Kw / Kb
For NH3, Ka = (1 x 10-14) / (1.8 x 10-5) = 5.56 x 10-10
F- + H2O ⇌ OH- + HF
What is the Kb for HF? (Need to find the Kb because OH- is a product)
HF is not on the list of Kb in the text, but it is on the list of Ka.
Kb = Kw / Ka
For HF, Kb = (1 x 10-14) / (6.8 x 10-4) = 1.47 x 10-11
Ka is greater than Kb here, so solution will be acidic.
Saturday, October 2, 2010
16.2: Polyprotic Acids
Polyprotic Acids:
-Polyprotic Acid: an acid that yields two or more H3O+ per molecule.
Example:
H3PO4 is a triprotic acid, it can lose three hydrogens to yield three H3O+ in aqueous solution. But in that solution, H3PO4, H2PO4-, HPO42-, and PO43- are all present.
1) H3PO4 (aq) + H2O (l) ⇌ H3O+ (aq) + H2PO4- (aq)
Ka1 = [H3O+][H2PO4-] / [H3PO4]
2) H2PO4- (aq) + H2O (l) ⇌ H3O+ (aq) + HPO42- (aq)
Ka2 = [H3O+][HPO42-] / [H2PO4-]
3) HPO42- (aq) + H2O (l) ⇌ H3O+ (aq) + PO43- (aq)
Ka3 = [H3O+][PO43-] / [HPO42-]
Ka1, the first ionization constant, is much larger than the second one. For a triprotic acid, Ka2, the second ionization constant is much larger than the third one.
Example:
For 1.25 M of H3PO4 at 25oC, find pH, [H3PO4], [H2PO4-], [HPO42-], [PO43-], and [H3O+].
(Ka for H3PO4 = 6.9 x 10-3, Ka for H2PO4- = 6.2 x 10-8, Ka for HPO42- = 4.8 x 10-13)
Since this reaction takes place in parts, lets look at the first part and see what information we get from it. The first reaction is:
H3PO4 (aq) + H2O (l) ⇌ H3O+ (aq) + H2PO4- (aq)
(x2) / (1.25 - x) = 6.9 x 10-3
Approximation method holds, so:
(x2) / (1.25) = 6.9 x 10-3
x = 0.0929
According to the table, [H3O+] = x = 0.0929
and pH = -log[H3O+] = 1.03
[H2PO4-] = x = 0.0929, but it is used in the next reaction, so we don't know the final concentration yet.
The second reaction is:
H2PO4- (aq) + H2O (l) ⇌ H3O+ (aq) + HPO42- (aq)
Notice that the initial concentration of H2PO4- is the 0.0929M we had left over from the first reaction. And the initial concentration of H3O+ is also 0.0929M, which was yielded in the first reaction also.
(0.0929 + x)(x) / (0.0929 - x) = 6.2 x 10-8
Approximation method holds, so:
(0.0929)(x) / (0.0929) = 6.2 x 10-8
x = 6.2 x 10-8
According to the table, [H2PO4-] = 0.0929M - 6.2 x 10-8 = 0.0929M
[HPO42-] = x = 6.2 x 10-8
[H3O+] = 0.0929M - 6.2 x 10-8 = 0.0929M
We know that Ka2 is supposed to be much smaller than Ka1. When we subtract this very small value from 0.0929, the difference is nearly negligible, and it rounds back to 0.0929. So really, we found these values after the first reaction.
The third reaction is:
HPO42- (aq) + H2O (l) ⇌ H3O+ (aq) + PO43- (aq)
4.8 x 10-13 = (0.0929 + x)(x) / (6.2 x 10-8 - x)
Approximation method holds, so:
4.8 x 10-13 = (0.0929)(x) / (6.2 x 10-8)
x = 3.20 x 10-19 M
According to the table, [PO43-] = x = 3.20 x 10-19 M
(Note that there is no Ka for strong acids, because they ionize 100% in water).
-Polyprotic Acid: an acid that yields two or more H3O+ per molecule.
Example:
H3PO4 is a triprotic acid, it can lose three hydrogens to yield three H3O+ in aqueous solution. But in that solution, H3PO4, H2PO4-, HPO42-, and PO43- are all present.
1) H3PO4 (aq) + H2O (l) ⇌ H3O+ (aq) + H2PO4- (aq)
Ka1 = [H3O+][H2PO4-] / [H3PO4]
2) H2PO4- (aq) + H2O (l) ⇌ H3O+ (aq) + HPO42- (aq)
Ka2 = [H3O+][HPO42-] / [H2PO4-]
3) HPO42- (aq) + H2O (l) ⇌ H3O+ (aq) + PO43- (aq)
Ka3 = [H3O+][PO43-] / [HPO42-]
Ka1, the first ionization constant, is much larger than the second one. For a triprotic acid, Ka2, the second ionization constant is much larger than the third one.
Example:
For 1.25 M of H3PO4 at 25oC, find pH, [H3PO4], [H2PO4-], [HPO42-], [PO43-], and [H3O+].
(Ka for H3PO4 = 6.9 x 10-3, Ka for H2PO4- = 6.2 x 10-8, Ka for HPO42- = 4.8 x 10-13)
Since this reaction takes place in parts, lets look at the first part and see what information we get from it. The first reaction is:
H3PO4 (aq) + H2O (l) ⇌ H3O+ (aq) + H2PO4- (aq)
(x2) / (1.25 - x) = 6.9 x 10-3
Approximation method holds, so:
(x2) / (1.25) = 6.9 x 10-3
x = 0.0929
According to the table, [H3O+] = x = 0.0929
and pH = -log[H3O+] = 1.03
[H2PO4-] = x = 0.0929, but it is used in the next reaction, so we don't know the final concentration yet.
The second reaction is:
H2PO4- (aq) + H2O (l) ⇌ H3O+ (aq) + HPO42- (aq)
Notice that the initial concentration of H2PO4- is the 0.0929M we had left over from the first reaction. And the initial concentration of H3O+ is also 0.0929M, which was yielded in the first reaction also.
(0.0929 + x)(x) / (0.0929 - x) = 6.2 x 10-8
Approximation method holds, so:
(0.0929)(x) / (0.0929) = 6.2 x 10-8
x = 6.2 x 10-8
According to the table, [H2PO4-] = 0.0929M - 6.2 x 10-8 = 0.0929M
[HPO42-] = x = 6.2 x 10-8
[H3O+] = 0.0929M - 6.2 x 10-8 = 0.0929M
We know that Ka2 is supposed to be much smaller than Ka1. When we subtract this very small value from 0.0929, the difference is nearly negligible, and it rounds back to 0.0929. So really, we found these values after the first reaction.
The third reaction is:
HPO42- (aq) + H2O (l) ⇌ H3O+ (aq) + PO43- (aq)
4.8 x 10-13 = (0.0929 + x)(x) / (6.2 x 10-8 - x)
Approximation method holds, so:
4.8 x 10-13 = (0.0929)(x) / (6.2 x 10-8)
x = 3.20 x 10-19 M
According to the table, [PO43-] = x = 3.20 x 10-19 M
(Note that there is no Ka for strong acids, because they ionize 100% in water).
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