Question 14.16: Calculating the pH of a Basic Salt Solution Calculate the pH......

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:

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.