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.