A buffer solution contains 0.20 M #CH_3COOH# and 0.30 M #CH_3COONa# at #25^oC#? (#Ka=1.8 x10^-5#). What is the pH after the addition of (a) 20.0 mL of 0.050 M #NaOH# or (b) 20.0 mL of 0.050 M #HCl# to 80.0 mL of this buffer solution?

Question

Naomi Aronson · Accepted Answer

Buffer pH = 4.93
after adding 20 ml of 0.05M NaOH => pH = 4.96
after adding 20 ml of 0.05M HCl => pH = 4.88

Buffer = 0.20M HOAc + 0.30M NaOAc
Ka(HOAc) = 1.8 x #10^-5# => pKa = -log#Ka# = -log(1.8 x #10^-5#) = 4.75

pH(Bfr) = p#K_a#+ log#[[OAc^-]]/[[HOAc]]# =#4.75 + log((0.30)/(0.20))#= 4.93

Avery Adams · Answer

undefined (a) pH after the addition of 20.0 mL of 0.050 M NaOH:
Initial moles of CH3COOH: 0.20 M × 0.080 L = 0.016 mol
Initial moles of CH3COONa: 0.30 M × 0.080 L = 0.024 mol
Moles of NaOH added: 0.050 M × 0.020 L = 0.001 mol
After the reaction, moles of CH3COOH remaining: 0.016 mol - 0.001 mol = 0.015 mol
After the reaction, moles of CH3COONa formed: 0.024 mol + 0.001 mol = 0.025 mol
Calculate the concentration of CH3COOH: [CH3COOH] = moles / total volume = 0.015 mol / 0.100 L = 0.15 M
Calculate the concentration of CH3COO-: [CH3COO-] = moles / total volume = 0.025 mol / 0.100 L = 0.25 M
Calculate the concentration of OH-: [OH-] = moles / total volume = 0.001 mol / 0.100 L = 0.010 M
Use the equilibrium expression for CH3COOH to find the concentration of H+:
Ka = [H+][CH3COO-] / [CH3COOH]
[H+] = Ka × [CH3COOH] / [CH3COO-] = (1.8 × 10^-5) × (0.15 M) / (0.25 M) = 1.08 × 10^-5 M
Calculate pH: pH = -log[H+] = -log(1.08 × 10^-5) ≈ 4.97

(b) pH after the addition of 20.0 mL of 0.050 M HCl:
The addition of HCl will react with CH3COO- to form CH3COOH.
Moles of HCl added: 0.050 M × 0.020 L = 0.001 mol
After the reaction, moles of CH3COOH formed: 0.001 mol
Calculate the concentration of CH3COOH: [CH3COOH] = (initial moles + moles formed) / total volume = (0.016 mol + 0.001 mol) / 0.100 L = 0.17 M
Calculate the concentration of CH3COO-: [CH3COO-] = (initial moles - moles formed) / total volume = (0.024 mol - 0.001 mol) / 0.100 L = 0.23 M
Use the equilibrium expression for CH3COOH to find the concentration of H+:
Ka = [H+][CH3COO-] / [CH3COOH]
[H+] = Ka × [CH3COOH] / [CH3COO-] = (1.8 × 10^-5) × (0.17 M) / (0.23 M) ≈ 1.33 × 10^-5 M
Calculate pH: pH = -log[H+] = -log(1.33 × 10^-5) ≈ 4.88