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

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)



K_a = (x²) / (0.0857 – x) → approximation holds
K_a= (x²) / (0.0857)

We need to find the K_a for NH₄⁺, so first find K_b of the conjugate base, NH₃, and plug it in:

K_a = K_w / K_b

K_b for NH₃ = 1.8 x 10^-5
K_a 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)



K_b = (x²) / (0.140 – x) → approximation holds
K_b = (x²) / (0.140)

We need to find the K_b for ClO^-, so first find K_a of the acid, HClO, and plug it in:

K_b = K_w / K_a

K_a for HClO = 3.5 x 10^-8
K_b 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.)

Also note that this equation only works when the approximation method holds, when C_a / K_a is greater than 100.

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)



K_a = 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.

K_a = 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)



K_a = 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: K_a 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^-




K_a = 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^-




K_a = 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)



K_a = 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.

16.3-16.4: 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 = K_b (yet another version of K_c)



Example:

In a solution of 2.35 M NH₃, what is the pH and pOH?

NH₃ (aq) + H₂O (l) ⇌ NH₄⁺ (aq) + OH^- (aq)



K_b = (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 K_b 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 K_a 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 K_a of the cation with the K_b of the anion. If K_a is larger, the solution is acidic. If K_b is larger, the solution is basic.

Example:
Salt: NH₄F

NH₄F → NH₄⁺ + F^-

NH₄⁺ + H₂O ⇌ H₃O⁺ + NH₃
What is the K_a for NH₃? (A hint that we need to find K_a and not K_b is H₃O⁺ is a product)
NH₃ is not on the list of K_a, but it is on the list of K_b in the text. And we know that:
K_w = 1 x 10^-14 = K_a x K_b, so: K_a = K_w / K_b

For NH₃, K_a = (1 x 10^-14) / (1.8 x 10^-5) = 5.56 x 10^-10

F^- + H₂O ⇌ OH^- + HF
What is the K_b for HF? (Need to find the K_b because OH^- is a product)
HF is not on the list of K_b in the text, but it is on the list of K_a.
K_b = K_w / K_a

For HF, K_b = (1 x 10^-14) / (6.8 x 10^-4) = 1.47 x 10^-11

K_a is greater than K_b here, so solution will be acidic.

16.1: Acid-Ionization Equilibria

Recap:

In pure water, [H₃0⁺] = [OH^-] = 1 x 10^-7 M and pH 7 is neutral.

Acidic Solution:
[H₃0⁺] is greater than [OH^-]
[H₃0⁺] is greater than 1 x 10^-7 M
[OH^-] is less than 1 x 10^-7 M
pH is less than 7

Basic Solution:
[OH^-] is greater than [H₃0⁺]
[H₃0⁺] is less than 1 x 10^-7 M
[OH^-] is greater than 1 x 10^-7 M
pH is greater than 7

Weak Acids/Bases:

Example:

A strong acid, HCl dissociates in water 100%, so a concentration of 0.0100 M HCl would yield +0.0100 M H₃O⁺, with a pH of -log(0.0100) = 2.
A weak acid dissociates less than 100%. So 0.0100 M of a weak acid would yield less than 0.0100 M H₃O⁺. For a weak acid, the concentrations of ions in solution are determined from the acid-ionization constant.

Acid-ionization constant: the equilibrium constant for the ionization of a weak acid. Denoted K_a.

Shorthand for a weak acid is HA (where H is the hydrogen and A is the rest) for a monoprotic acid,
H₂A for a diprotic acid, and H₃A for a triprotic acid. So:

HA (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + A^- (aq)




where [H₃0⁺] and [H⁺] are interchangeable.

Example:

HC₆H₄NO₂ (aq) + H₂0 (l) ⇌ H₃O⁺ (aq) + C₆H₄NO₂^- (aq)
is the same as
HA (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + A^- (aq)

[HA] = 0.012 M
pH = 3.39
What is the K_a and what is the degree of ionization?

Degree of Ionization: the fraction of molecules that react with water to give ions. Given as a percentage it is called the percent ionization.




Since we know the pH is 3.39, we can find [H₃O⁺].

[H₃O⁺] = 10^-pH
[H₃O⁺] = 4.07 x 10^-4

So, x = 4.07 x 10^-4

K_a = (4.07 x 10^-4)(4.07 x 10^-4) / (0.0120)
K_a = 1.38 x 10^-5

To obtain the degree of ionization, notice that 4.07 x 10^-4 mol out of 0.0120 mol of acid ionizes, so

degree of ionization = (4.07 x 10^-4) / (0.0120) = 0.0339
percent ionization = 0.0339 x 100 = 3.39 %

Approximation Method:

Let C_a represent the initial concentration of the weak acid.

There will be less than a 5% error with the approximation method if C_a / K_a is greater than 100.

Example:

What is the pH for 0.500 M of acetic acid? K_a = 1.7 x 10^-5




K_a = 1.7 x 10^-5 = (x²) / (0.500 - x)

To solve this, we need to use the quadratic equation. Or, we can test to see if we are able to apply the approximation method and use that instead, since it is much quicker and easier.

Is C_a / K_a greater than 100?

(0.500) / (1.7 x 10^-5) = 29411.8

Yes.

Here is the Approximation: in the term (0.500 - x), the x will be so small that subtracting it from 0.500 will be negligible, and the term is roughly the same as simply 0.500. This approximation greatly simplifies the equation:

1.7 x 10^-5 = (x²) / (0.500)
x = 0.00292
pH = -log(0.00292) = 2.54

Remember, this substitution only works if C_a / K_a is greater than 100. If it's not, then the quadratic equation must be used.