Aim # 12: What is a Buffer Solution? H.W. # 12 Study pp. 715-727 Ans. ques. p. 751 # 29,31,33,37,41,47 Use the Henderson – Hasselbach equation where possible. It’s faster. Do Now: Calculate the formate ion concentration and pH of a solution that is 0.05 M in formic acid, HCHO2 (Ka = 1.8 x 10-4), and 0.100 M in HNO3.

A buffer is prepared by mixing a weak acid (base) with a I Buffer Solution – a solution that is resistant to a change in its pH when small amounts of acid or base are added to it. A buffer is prepared by mixing a weak acid (base) with a salt of its conjugate base (acid). e.g. HC2H3O2 and NaC2H3O2 (which forms C2H3O2-) H2CO3 and KHCO3 (which forms HCO3-) NH4Cl and NH3 (which forms NH4+)

Problem: What is the pH of a buffer solution that is 0 Problem: What is the pH of a buffer solution that is 0.75 M in lactic acid (HC3H5O3, Ka = 1.4 x 10-4) and 0.25 M in sodium lactate (NaC3H5O3). Ans: Species in solution Na+, C3H5O3-, HC3H5O3, H+, H2O HC3H5O3(aq) H+(aq) + C3H5O3-(aq) Ka = 1.4 x 10-4 = [H+][C3H5O3-] [HC3H5O3]

HC3H5O3(aq) H+ + C3H5O3-(aq) I 0.75 M 0 M 0.25 M C -x +x E (0.75 – x) M x M (0.25 + x) M 1.4 x 10-4 = (x)(0.25 + x) ≈ (x)(0.25) (0.75 – x) (0.75) 1.4 x 10-4 = .33x [H+] = x = 4.24 x 10-4 pH = -log(4.24 x 10-4) = 3.37

II Buffering: How does it work? For HA(aq) ↔ H+(aq) +A-(aq) Ka = [H+][A-] [HA] Rearranging, [H+] = Ka[HA] [A-] Also, NaA(aq) → Na+(aq) + A-(aq)

[HA] ↑ and [A-] ↓ [H+] = Ka [HA] [A-] If [HA]0 >> Δ [HA] If we add H+ H+(aq) + A-(aq) → HA(aq) [HA] ↑ and [A-] ↓ If [HA]0 >> Δ [HA] and [A-] >> Δ [A-] then [HA] doesn’t change much [A-] nor does [H+] and pH

[H+] = Ka [HA] [A-] If we add OH- OH-(aq) + HA(aq) → H2O(ℓ) + A-(aq) [HA] ↓ and [A-] ↑ If [HA]0 >> Δ [HA] and [A-] >> Δ [A-] then [HA] doesn’t change much [A-] nor does [H+] and pH

III The Henderson-Hasselbach equation [H+] = Ka [HA] [A-] -log[H+] = -log Ka – log [HA] [A-] pH = pKa – log [HA] [A-] OR pH = pKa + log [A-] ͘ [HA]

Problem: Find the pH of the lactic acid/sodium lactate buffer solution using the Henderson-Hasselbach equation. Ans: pH = pKa + log [C3H5O3-] ͘ [HC3H5O3] pH = -log(1.4 x 10-4) + log (0.25 + x) ͘ (0.75 – x) pH ≈ -log(1.4 x 10-4) + log (0.25) ͘ (0.75) pH = 3.85 - .48 pH = 3.37 Note: This calculation allows us to neglect [H+]

IV The effect of adding a strong acid (or base) to a buffer solution Problem: A buffer is made by adding 0.500 mol NaC2H3O2 to 0.500 mol HC2H3O2 in enough H2O to make 1.00 L of solution. (Ka = 1.8 x 10-5) a) What is the pH of the buffer? b) What will be the pH if 0.050 mol of NaOH is added to the buffer? (Assume no volume change.) c) What will be the pH if 0.050 mol HCl is added to the original buffer? (Assume no volume change.)

Ans: a) pH = pKa + log [C2H3O2-] ͘ [HC2H3O2] pH = -log(1.8 x 10-5) + log (.500-x) ͘ (.500+x) assuming x << .500 M, pH = -log(1.8 x 10-5) = log (.500) ͘ (.500) pH = 4.74 + 0 = 4.74

addition 0.450 M 0.550 M change equilibrium b) Note: The strong base reacts completely with the weak acid. Upon addition of the NaOH: stoichiometry HC2H3O2 + OH- → C2H3O2- + H2O before addition 0.500 M 0.050 M ----- addition -0.050 M +0.050 M after addition 0.450 M 0 M 0.550 M After neutralization of the base: equilibrium HC2H3O2 H+ + C2H3O2- Initial 0.450 M 0 M 0.550 M change - x M + x M equilibrium (0.450 – x) M x M (0.550 + x) M

Ka = 1.8 x 10-5 = (x)(0.550 + x) ≈ (x)(0.550) (0.450 – x) (0.450) [H+] = x = 1.8 x 10-5(0.450) = 1.47 x 10-5 (0.550) pH = 4.83 Using Henderson-Hasselbach pH = pKa + log [C2H3O2-] ͘ [HC2H3O2] pH = 4.74 + log (0.550 + x) (0.450 – x)

pH ≈ 4.74 + log (0.550) (0.450) pH = 4.74 + .087 pH = 4.83

c) upon addition of HCl: stoichiometry C2H3O2- + H+ → HC2H3O2- before addition 0.500 M 0.050 M addition -0.500 M -0.050 M +0.050 M after addition 0.450 M 0 M 0.550 M After neutralization of the acid: equilibrium HC2H3O2 H+ + C2H3O2- initial 0.550 M 0 M 0.450 M change - x M +x M equilibrium (0.550 – x) x M (0.450 + x) M

Ka = 1.8 x 10-5 = (x)(0.450 + x) ≈ (x)(0.450) (0.550 – x) (0.550) [H+] = x = 1.8 x 10-5(0.550) = 2.2 x 10-5 (0.450) pH = -log(2.2 x 10-5) = 4.65 Using Henderson-Hasselbach pH = pKa + log (0.450 + x) (0.550 – x) pH = 4.74 + log (0.450) (0.550) pH = 4.74 - .087 = 4.65

Problem: What would be the change in pH if 0 Problem: What would be the change in pH if 0.050 mol of HCl was added to 0.500 M HC2H3O2? Ans: Before addition: Ka = 1.8 x 10-5 = x2 ≈ x2 ͘ .500 – x .500 [H+] = x = 3.0 x 10-3 pH = -log(3.0 x 10-3) = 2.52 After addition: [H+] = .050 + .003 = .053 M pH = -log(.053) = 1.28

V Preparing a buffer solution A. Buffering Capacity – the amount of acid or base a buffer can neutralize before the pH begins to change significantly. Note: As [HA] AND [A-] increases, the buffering capacity increases. B. The pH of a buffer solution depends upon the ratio [HA] [A-] Remember, [H+] = Ka [HA] [A-] Note: Buffers work best when [HA] = [A-] ••• To prepare a buffer solution, choose an acid with a pKa close to the pH desired.

Problem: Calculate the concentration of sodium benzoate that must be present in a buffer solution of 0.20 M benzoic acid, HC7H5O2, to produce a pH of 4.00. Ka = 6.5 x 10-5 Ans: HC7H5O2(aq) H+(aq) + C7H5O2-(aq) pH = pKa + log [C7H5O-] ͘ [HC7H5O2] pKa = -log(6.5 x 10-5) = 4.19 4.00 = 4.19 + log (x) ͘ (.20) -.19 = log (x) ͘ (.20) .646 = x ͘ .20 x = [NaC7H5O2] = .13 M

Problem: Which of the following acids and its sodium salt would be best to prepare a buffer solution with a pH of 3.00? HClO2 Ka = 1.2 x 10-2 HF Ka = 7.2 x 10-4 HC2H3O2 Ka = 1.8 x 10-5 Ans: The acids pKa values are HClO pKa = -log(1.2 x 10-2) = 1.92 HF pKa = 3.14 HC2H3O2 pKa = 4.74 HF would be best suited fro the necessary buffer. Use equal concentr5ations of HF and NaF.

Practice Problems Zumdahl (8th ed.) p. 737 # 34-36,38,40,41