Question 14.16: Calculating the pH of a Basic Salt Solution Calculate the pH......
Calculating the pH of a Basic Salt Solution
Calculate the pH of a 0.10 M solution of NaCN; K_a for HCN is 4.9 × 10^{-10} .
STRATEGY
Use the procedure summarized in Figure 14.7.
Step 1 The species present initially are Na^+ (inert), CN^- (base), and H_2O (acid or base).
Step 2 There are two possible proton-transfer reactions:
\begin{array}{ll}\mathrm{CN}^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{OH}^{-}(a q) & K_{\mathrm{b}} \\\mathrm{H}_2 \mathrm{O}(l)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) & K_{\mathrm{w}}\end{array}Step 3 As shown in Worked Example 14.14b, K_b = K_w/(K_a \ for \ HCN) = 2.0 × 10^{-5} . Because K_{\mathrm{b}} \gg K_{\mathrm{w}}, CN^- is a stronger base than H_2O and the principal reaction is proton transfer from H_2O to CN^-.
Step 4
Step 5 The value of x is obtained from the equilibrium equation:
Step 7 \left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\frac{K_{\mathrm{w}}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.0 \times 10^{-14}}{1.4 \times 10^{-3}}=7.1 \times 10^{-12}
Step 8 \mathrm{pH}=-\log \left(7.1 \times 10^{-12}\right)=11.15
The solution is basic, which agrees with the blue color of the indicator in Figure 14.9.
