Buffers and Titration

The Danger of Antifreeze

Antifreeze contains ethylene glycol ( $HOCH2CH2OH$ ), also known as 1,2-ethanediol.
Ethylene glycol has a sweet taste and initially causes drunkenness.
In the liver, ethylene glycol is metabolized to glycolic acid ( $HOCH_2COOH$ ), also known as α-hydroxyethanoic acid.
Glycolic acid is toxic because, at high concentrations, it overwhelms the buffering ability of $HCO_3^-$ in the blood, causing the blood pH to drop.
This drop in pH compromises the blood's ability to carry O2, leading to acidosis:HbH^+(aq) + O2(g) \rightleftharpoons HbO2(aq) + H^+(aq) $.$
A treatment involves administering ethyl alcohol, which has a higher affinity for the enzyme that metabolizes ethylene glycol.

Buffers

Buffers resist changes in pH when an acid or base is added.
They neutralize added acid or base.
Buffers have a limit to their capacity.
Many buffers are made by mixing a weak acid with a soluble salt containing its conjugate base anion.
Blood contains a mixture of $H2CO3$ and $HCO_3^-$ .

Acid Buffers: How Addition of Base Works

Buffers apply Le Châtelier’s Principle to weak acid equilibrium: $HA(aq) + H2O(l) \rightleftharpoons A^−(aq) + H3O^+(aq)$ .
Buffer solutions contain significant amounts of the weak acid molecules, HA.
These HA molecules react with added base to neutralize it: $HA(aq) + OH^−(aq) → A^−(aq) + H_2O(l)$ .
Alternatively, $H3O^+$ combines with $OH^−$ to make $H2O$ , and the equilibrium shifts to replace the consumed $H_3O^+$ .

Acid Buffers: How Addition of Acid Works

The buffer solution contains significant amounts of the conjugate base anion, A−.
These A− ions combine with added acid to make more HA: $H^+(aq) + A^−(aq) → HA(aq)$ .
After the equilibrium shifts, the concentration of $H_3O^+$ is kept constant.

Common Ion Effect

Adding a salt containing the anion NaA, which is the conjugate base of the acid (the common ion), shifts the equilibrium to the left: $HA(aq) + H2O(l) \rightleftharpoons A^−(aq) + H3O^+(aq)$ .
This causes the pH to be higher than the pH of the acid solution by lowering the $H_3O^+$ ion concentration.

Example 16.1: pH of a Buffer

Calculate the pH of a buffer that is 0.100 M $HC2H3O2$ and 0.100 M $NaC2H3O2$ .
Write the reaction for the acid with water: $HC2H3O2 + H2O \rightleftharpoons C2H3O2^− + H3O^+$ .
Construct an ICE table:
- Initial: [HA] = 0.100, [A−] = 0.100, [H3O+] ≈ 0
- Change: [HA] = −x, [A−] = +x, [H3O+] = +x
- Equilibrium: [HA] = 0.100 − x, [A−] = 0.100 + x, [H3O+] = x
$Ka$ for $HC2H3O2 = 1.8 \times 10^{−5}$ .
Approximate: [HA]eq ≈ [HA]init and [A−]eq ≈ [A−]init because $K_a$ is very small.
Solve for x: $Ka = \frac{[C2H3O2^−][H3O^+]}{[HC2H3O2]} = \frac{(0.100 + x)(x)}{(0.100 − x)} ≈ \frac{(0.100)(x)}{(0.100)} = x$ .
$x = 1.8 \times 10^{−5}$ .
Check if the approximation is valid: x < 5% of [ $HC2H3O_2$ ]init:
- \frac{1.8 \times 10^{−5}}{0.100} \times 100\% = 0.018\% < 5\%, the approximation is valid.
Substitute x into the equilibrium concentration definitions: [ $H_3O^+$ ] = $1.8 \times 10^{−5}$ .
Substitute [ $H_3O^+$ ] into the formula for pH: $pH = −log(1.8 \times 10^{−5}) = 4.74$ .
Check by substituting the equilibrium concentrations back into the $K_a$ expression.

Practice Problem

What is the pH of a buffer that is 0.14 M HF ( $pK_a = 3.15$ ) and 0.071 M KF?
Write the reaction for the acid with water: $HF + H2O \rightleftharpoons F^− + H3O^+$ .
Construct an ICE table:
- Initial: [HA] = 0.14, [A−] = 0.071, [H3O+] ≈ 0
- Change: [HA] = −x, [A−] = +x, [H3O+] = +x
- Equilibrium: [HA] = 0.14 − x, [A−] = 0.071 + x, [H3O+] = x
Determine the value of $Ka$ : $Ka = 10^{−3.15} = 7.0 \times 10^{−4}$ .
Approximate: [HA]eq ≈ [HA]init and [A−]eq ≈ [A−]init because $K_a$ is very small.
- Solve for x: $Ka = \frac{[F^−][H3O^+]}{[HF]} = \frac{(0.071 + x)(x)}{(0.14 − x)} ≈ \frac{(0.071)(x)}{(0.14)} = x$ .
$x = 1.4 \times 10^{−3}$ .
Check if the approximation is valid: x < 5% of [HF]init:
- \frac{1.4 \times 10^{−3}}{0.14} \times 100\% = 1\% < 5\%, the approximation is valid.
Substitute x into the equilibrium concentration definitions: [ $H_3O^+$ ] = $1.4 \times 10^{−3}$ .
Substitute [ $H_3O^+$ ] into the formula for pH: $pH = −log(1.4 \times 10^{−3}) = 2.85$ .
Check by substituting the equilibrium concentrations back into the $K_a$ expression.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch Equation simplifies the calculation of the pH of a buffer solution using the $K_a$ expression.
Calculates the pH of a buffer from the $pK_a$ and initial concentrations of the weak acid and salt of the conjugate base.
Valid as long as the “x is small” approximation is valid.

Derivation of the Henderson-Hasselbalch Equation

Example 16.2: Using the Henderson-Hasselbalch Equation

What is the pH of a buffer that is 0.050 M $HC7H5O2$ and 0.150 M $NaC7H5O2$ ?
$Ka$ for $HC7H5O2 = 6.5 \times 10^{−5}$ .
Assume [HA] and [A−] equilibrium concentrations are the same as the initial concentrations.
Substitute into the Henderson-Hasselbalch equation: $pH = pK_a + log(\frac{[A^−]}{[HA]})$ .
$pH = −log(6.5 \times 10^{−5}) + log(\frac{0.150}{0.050}) = 4.19 + log(3) = 4.67$ .
Check the “x is small” approximation: $HC7H5O2 + H2O \rightleftharpoons C7H5O2^− + H3O^+$ .

Practice Problem: Henderson-Hasselbalch Equation

What is the pH of a buffer that is 0.14 M HF ( $pK_a = 3.15$ ) and 0.071 M KF?
Find the $pKa$ from the given $Ka$ .
Assume the [HA] and [A−] equilibrium concentrations are the same as the initial.
Substitute into the Henderson-Hasselbalch equation: $pH = pK_a + log(\frac{[A^−]}{[HA]})$ .
$pH = 3.15 + log(\frac{0.071}{0.14}) = 3.15 + log(0.507) = 2.85$ .
Check the “x is small” approximation: $HF + H2O \rightleftharpoons F^− + H3O^+$ .

Full Equilibrium Analysis vs. Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is generally good enough when the “x is small” approximation is applicable.
The “x is small” approximation will work when both of the following are true:
- The initial concentrations of acid and salt are not very dilute
- The $K_a$ is fairly small
For most problems, the initial acid and salt concentrations should be over 100 to 1000x larger than the value of $K_a$ .

Change in pH of a Buffer Upon Adding Acid or Base

Calculating the new pH after adding acid or base requires breaking the problem into two parts:
1. Stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer.
  - Added acid reacts with the A− to make more HA
  - Added base reacts with the HA to make more A−
2. Equilibrium calculation of [ $H_3O^+$ ] using the new initial values of [HA] and [A−].

Example 16.3: Adding Base to a Buffer

What is the pH of a buffer that has 0.100 mol $HC2H3O2$ and 0.100 mol $NaC2H3O2$ in 1.00 L that has 0.010 mol NaOH added to it?
Write a reaction for $OH^−$ with HA: $HC2H3O2 + OH^− \rightleftharpoons C2H3O2^− + H_2O$ .
Construct a stoichiometry table:
- Initial moles: [HA] = 0.100, [A−] = 0.100, [ $OH^−$ ] = 0.010
- Change: [HA] = −0.010, [A−] = +0.010, [ $OH^−$ ] = −0.010
- Moles after: [HA] = 0.090, [A−] = 0.110, [ $OH^−$ ] ≈ 0
New molarities: [HA] = 0.090 M, [A−] = 0.110 M.
Write the reaction for the acid with water: $HC2H3O2 + H2O \rightleftharpoons C2H3O2^− + H3O^+$ .
Construct an ICE table:
- Initial: [HA] = 0.090, [A−] = 0.110, [H3O+] ≈ 0
- Change: [HA] = −x, [A−] = +x, [H3O+] = +x
- Equilibrium: [HA] = 0.090 − x, [A−] = 0.110 + x, [H3O+] = x
$Ka$ for $HC2H3O2 = 1.8 \times 10^{−5}$ .
Approximate: [HA]eq ≈ [HA]init and [A−]eq ≈ [A−]init because $K_a$ is very small
- Solve for x: $Ka = \frac{[C2H3O2^−][H3O^+]}{[HC2H3O2]} = \frac{(0.110 + x)(x)}{(0.090 − x)} ≈ \frac{(0.110)(x)}{(0.090)} = x$ .
$x = 1.47 \times 10^{−5}$ .
Check if the approximation is valid: x < 5% of [HC2H3O2]init:
- \frac{1.47 \times 10^{−5}}{0.090} \times 100 = 0.016\% < 5\%, approximation is valid.
Substitute x into the equilibrium concentration definitions: [ $H_3O^+$ ] = $1.47 \times 10^{−5}$ .
Substitute [ $H_3O^+$ ] into the formula for pH: $pH = −log(1.47 \times 10^{−5}) = 4.83$ .
Check by substituting the equilibrium concentrations back into the $K_a$ expression.

Example 16.3: Using Henderson-Hasselbalch Equation

pH = 4.745 + log(0.110/0.090) = 4.745 + 0.087 = 4.832

Practice Problem: Adding Acid to a Buffer

What is the pH of a buffer that has 0.140 moles HF ( $pK_a = 3.15$ ) and 0.071 moles KF in 1.00 L of solution when 0.020 moles of HCl is added? (The “x is small” approximation is valid)
Write a reaction for $H3O^+$ with A−: $F^− + H3O^+ \rightleftharpoons HF + H_2O$ .
Construct a stoichiometry table:
- Initial moles: [F−] = 0.071, [ $H_3O^+$ ] = 0.020, [HF] = 0.140
- Change: [F−] = −0.020, [ $H_3O^+$ ] = −0.020, [HF] = +0.020
- Moles after: [F−] = 0.051, [ $H_3O^+$ ] ≈ 0, [HF] = 0.160
New molarities: [F−] = 0.051 M, [HF] = 0.160 M.
HF + H2O  F− + H3O+
Assume the [HA] and [A−] equilibrium concentrations are the same as the initial.
Substitute into the Henderson-Hasselbalch equation: $pH = pK_a + log(\frac{[F^−]}{[HF]})$ .
Substitute [ $H_3O^+$ ] into the formula for pH: $pH = 3.15 + log(\frac{0.051}{0.160}) = 3.15 − 0.497 = 2.65$ .

Basic Buffers

Buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, $H:B^+Cl^−$ : $H2O(l) + NH3 (aq) \rightleftharpoons NH_4^+ (aq) + OH^− (aq)$ .
B: + H2O  H:B+ + OH−
To apply the Henderson-Hasselbalch equation, the chemical equation of the basic buffer must be looked at like an acid reaction H:B^+ + H2O  B: + H3O^+.

Henderson-Hasselbalch Equation for Basic Buffers

$pOH = pK_b + log(\frac{[B]}{[HB^+]})$ .
$pH = 14 − pOH$ .