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.

15.1-15.8: All of Chapter 15

Strong Acids and Bases:

(If it is an acid or a base and it is not in this table, then it is a weak acid or base. An indicator that something is an acid is that it will start with a hydrogen, ex: H₃PO₃, H₃PO₄).

Three Definitions:

1. Arrhenius Concept of Acids and Bases:

-Acid: a substance that, when dissolved in water, increases the concentration of hydronium ion, H₃O⁺ (aq). H⁺ (aq) is interchangeable with H₃O⁺ (aq), but remember that it is not just a lone proton, but a proton chemically bonded to water (and with other water molecules via hydrogen bonding).

-Strong Acid: a substance that completely ionizes in aqueous solution to give H₃O⁺ (aq) and an anion.

Example:

HCl (aq) + H₂O (l) → H₃O⁺ (aq) + Cl^- (aq)
Hydrochloric acid donates a hydrogen to the water molecule to yield a hydronium ion. Hydrochloric acid thus increases the hydronium ion concentration of the solution and is considered an Arrhenius acid.

-Base: a substance that, when dissolved in water, increases the concentration of hydroxide ion, OH^- (aq).

-Strong Base: a substance that completely ionizes in aqueous solution to give OH^- and a cation.

-Weak Acids and Bases: are not completely ionized in solution and exist in reversible reaction with the corresponding ions.

Example:

NaOH (aq) → Na⁺ (aq) + OH^- (aq)
Sodium hydroxide dissociates in water and increases the concentration of hydroxide ion. It is thus considered an Arrhenius base.

2. BrØnsted-Lowery Concept of Acids and Bases:

Independently, these two guys came up with a more flexible definition of acids and bases.

Acid: the species donating a proton in a proton-transfer reaction.

Base: the species accepting the proton in a proton-transfer reaction.

Example:

HCl (aq) + H₂O (l) → H₃O⁺ (aq) + Cl^- (aq)
Hydrochloric acid donates a proton to water to yield H₃O⁺, and water accepts a proton to become H₃O⁺. So here, Hydrochloric acid is the acid and water is the base.

Example:

NH₃ (aq) + H₂O (l) → NH₄⁺ (aq) + OH^- (aq)
Ammonia accepts a proton from water to yield an ammonium ion and a hydroxide ion. So here, water is the acid, and ammonia is the base.

Conjugate Acid-Base Pair: consists of two species in an acid-base reaction, one acid and one base, that differ by the loss or gain of a proton. The acid of the pair is called the conjugate acid of the base, and the base is called the conjugate base of the acid. (ex: NH₃ and NH₄⁺, remember when you add the hydrogen to add one positive charge)


-Amphiprotic/Amphoteric Species: a species that can act as either an acid or a base (it can gain or lose a proton) depending on the other reaction. (ex: OH^– ← H₂O → H₃O⁺).

3. Lewis Concept of Acids and Bases:

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.

Example:



Relative Strengths of Acids and Bases:


As seen in the above table, the strongest acids have the weakest conjugate bases, and the strongest bases have the weakest conjugate acids. This table can be used to predict the direction of an acid-base reaction: the normal direction of reaction is from the stronger acid and base to the weaker acid and base.

Molecular Structure and Acid Strength:

-The strength of an acid depends on how easily the proton, H⁺, is lost or removed from an H-X bond in the acid species. (ex: H-X = H-Cl, H-X = H-Br, etc.)

1) In going down a column of elements of the periodic table, the size of atom X increases (have more and more electrons), the H-X bond strength decreases, thus increases the strength of the acid.

Example: Group VIIA elements: HF, HCl, HBr, HI = HI > HBr > HCl > HF

2) Going across a row of elements of the periodic table, the electronegativity increases, the H-X bond polarity increases, and the acid strength increases.

3) An oxoacid has the structure H-O-Y- where the acidic H atom is always attached to an O atom, which, in turn, is attached to an atom Y. Other groups may be attached to Y.

-For a series of oxoacids of the same structure, differing only in the atom Y, the acid strength increases with the electronegativity of Y.

Example: HClO, HBrO, and HIO. Consider the electronegativities of Cl, Br, and I.
HClO > HBrO > HIO

4) For a series of oxoacids with the structure H-O-Y-O_n, with the same Y, more O's = higher acid strength.



5) The acid strength of a polyprotic acid and its anions decreases with increasing negative charge.

Example: H₃PO₄ > H₂PO₄^- > HPO₄^2- (the negative charge acts on the proton to hold it in tighter and make it less acidic).


Self Ionization of Water:

Self-Ionization/Autoionization: a reaction in which two like molecules react to give ions.

H₂O (l) + H₂O (l) ⇌ H₃O⁺ (aq) + OH^– (aq)

Ion-Product Constant for Water: (a different version of K_c)

K_w = [H₃O⁺][OH^-] = 1.00 x 10^-14 (at 25 degrees Celsius)
-Just like K_c, no units of measure
-Can be rearranged to find hydronium and hydroxide ion concentrations

[H₃O⁺] = 1.00 x 10^-14 / [OH^-]
[OH^-] = 1.00 x 10^-14 / [H₃O⁺]

Example:

What is the [OH^-] for 0.150 M HCl in the following reaction?

HCl (aq) + H₂O (l) → H₃O⁺ (aq) + Cl^- (aq)



We know the concentration of water is constant because it is a pure liquid, and we know that HCl dissociates completely (to 0M) because it is a strong acid.

So now we know the [H₃O⁺] is 0.150 M, so:

[OH^-] = 1.00 x 10^-14 / 0.150M = 6.67 x 10^-14 M

-In a neutral solution, [H₃O⁺] = [OH^-]
-In an acidic solution [H₃O⁺] > [OH^-]
-In a basic solution [H₃O⁺] < [OH^-]

The pH of a Solution:

-Defined as the negative logarithm of the molar hydronium ion concentration. (pH = percent hydrogen).

pH = -log [H₃O⁺]
or
[H₃O+] = 10^-pH

(Note: it is possible for pH to be larger than 14 and less than 0, it's just very uncommon)

-The lower the pH, the more acidic; the higher the pH, the more basic. Can also be used with [OH^-].

pOH = -log [OH^-]
or
[OH^-] = 10^-pOH

Also:

pH + pOH = 14