a) m(CH₃COONa) = 20 g.
n(CH₃COONa) = m(CH₃COONa) ÷ M(CH₃COONa).
n(CH₃COONa) = 20 g ÷ 82.034 g/mol.
n(CH₃COONa) = 0.244 mol; amount of substance.
V(CH₃COONa) = 500 ml ÷ 1000 ml/L.
V(CH₃COONa) = 0.5 L.
c(CH₃COONa) = n(CH₃COONa) ÷ V(CH₃COONa).
c(CH₃COONa) = 0.244 mol ÷ 0.5 L.
c(CH₃COONa) = 0.488 M; molarity of sodium acetate.
c(CH₃COOH) = 0.150 M; molarity of acetic acid.
Ka(CH₃COOH) = 1,8·10⁻⁵.
pKa = -logKa = 4,75.
Henderson–Hasselbalch equation: pH = pKa + log(cs/ck).
pH = 4.75 + log(0.488M / 0.150M).
pH = 5.262; pH of buffer solution.
b) Chemical reactions:
1) HCl(aq) → H⁺(aq) + Cl⁻(aq).
2) CH₃COONa(aq) → CH₃COO⁻(aq) + Na⁺(aq).
3) CH₃COO⁻(aq) + H⁺(aq) ⇄ CH₃COOH.
Sum: CH₃COONa + HCl ⇄ CH₃COOH + NaCl.