Question 18.12: Determining the pH of a Solution of A^− Problem Sodium aceta......

Plan We know the initial concentration of Ac^− (0.25 M) and the K_a of HAc (1.8×10^{−5}), and we have to find the pH of the Ac^− solution, knowing that Ac^– acts as a weak base in water. We write the base-dissociation equation and K_b expression. If we can find [OH^−], we can use K_w to find [H_3O^+] and convert it to pH. To solve for [OH^−], we need the K_b of Ac^−, which we obtain from the K_a of its conjugate acid HAc by applying Equation 18.7. We set up a reaction table to find [OH^−] and make the usual assumption that K_b is small, so [Ac^−]_{\text{init}}  ≈  [Ac^−]. Notice that Na^+ is a spectator ion and is not included in the base-dissociation equation.

              K_a  (\text{of }HA)  ×  K_b  (\text{of }A^−)  =  K_w         (18.7)

Solution Writing the base-dissociation equation and K_b expression:

Setting up the reaction table (Table 1), with  x  =  [Ac^−]_{\text{reacting}}  =  [HAc]  =  [OH^−]:

Solving for K_b of Ac^−:

Making the assumption: Because K_b is small, 0.25 M − x ≈ 0.25 M.
Substituting into the expression for K_b and solving for x:

Checking the assumption:
             \frac{1.2×10^{−5}  M}{0.25  M}  ×  100  =  4.8×10^{−3}\% (<5%; assumption is justified.)
Also note that             \frac{0.25}{5.6×10^{−10}}  =  4.5×10^8  >  400
Solving for pH:

            [H_3O^+]  =  \frac{K_w}{[OH^−]}  =  \frac{1.0×10^{−14}}{1.2×10^{−5}}  =  8.3×10^{−10}  M
            pH  =  −\log  (8.3×10^{−10})  =  9.08

Check The K_b calculation seems reasonable: ∼10×10^{−15}/2×10^{−5}  =  5×10^{−10}. Because Ac^− is a weak base, [OH^−]  >  [H_3O^+]; thus, pH > 7, which makes sense.