Consider dissolving sodium acetate in water:
$$NaCH_3CO_2(s)⇋Na^+(aq)+CH_3CO_2^-(aq)$$
The sodium ion does not undergo appreciable acid or base ionization and has no effect on the solution pH. This may seem obvious from the ion’s formula, which indicates no hydrogen or oxygen atoms, but some dissolved metal ions function as weak acids, as addressed later in this section.
The acetate ion, $CH_3CO_2^-$ is the conjugate base of acetic acid, CH3CO2H, and so its base ionization (or base hydrolysis) reaction is represented by
$$CH_3CO_2^-(aq)+H_2O(l)⇋CH_3CO_2H(aq)+OH^-(aq)\qquad K_b=K_w/K_a$$
Because acetic acid is a weak acid, its Ka is measurable and Kb > 0 (acetate ion is a weak base).
Dissolving sodium acetate in water yields a solution of inert cations (Na+) and weak base anions ($CH_3CO_2^-$) resulting in a basic solution.
Equilibrium in a Solution of a Salt of a Weak Acid and a Strong Base
Determine the acetic acid concentration in a solution with $[CH_3CO_2^-]=0.050\;M\text{ and }[OH^-]=2.5×10^{-6}\;M$ at equilibrium. The reaction is:
SolutionThe provided equilibrium concentrations and a value for the equilibrium constant will permit calculation of the missing equilibrium concentration. The process in question is the base ionization of acetate ion, for which
$$K_b\;(\text{for }CH_3CO_2^-)=\frac{K_w}{K_a\;\text{for }CH_3CO_2H}=\frac{1.0×10^{-14}}{1.8×10^{-5}}=5.6×10^{-10}$$Substituting the available values into the Kb expression gives
$$K_b=\frac{[CH_3CO_2H][OH^-]}{[CH_3CO_2^-]}=5.6×10^{-10}$$ $$=\frac{[CH_3CO_2H](2.5×10^{-6}}{(0.050)}=5.6×10^{-10}$$Solving the above equation for the acetic acid molarity yields [CH3CO2H] = $1.1×10^{-5}\;M$.
Check Your LearningWhat is the pH of a 0.083 M solution of NaCN?
Answer:
11.11