Question 17.8: Calculating Concentrations of Species in a Salt Solution Wha...

Sodium nicotinate gives \mathrm{Na}^{+}and nicotinate ions in solution. Only the nicotinate ion hydrolyzes. Write HNic for nicotinic acid and \mathrm{Nic}^{-}for the nicotinate ion. The hydrolysis of nicotinate ion is

\operatorname{Nic}^{-}(a q)+H_{2}O(l)\Longrightarrow\operatorname{HNic}(a q)+\operatorname{OH}^{-}(a q)

Nicotinate ion acts as a base, and you can calculate the concentration of species in solution as in Example 17.5. First, however, you need K_{b} for the nicotinate ion. This is related to K_{a} for nicotinic acid by the equation K_{a} K_{b}=K_{w}. Substituting, you get

K_{b}=\frac{K_{w}}{K_{a}}=\frac{1.0 \times 10^{-14}}{1.4 \times 10^{-5}}=7.1 \times 10^{-10}

Now you can proceed with the equilibrium calculation. We will only sketch this calculation, because it is similar to that in Example 17.5. You let x=[\mathrm{HNic}]= \left[\mathrm{OH}^{-}\right], then substitute into the equilibrium-constant equation.

\frac{[\mathrm{HNic}]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{Nic}^{-}\right]}=K_{b}

This gives

\frac{x^{2}}{0.10-x}=7.1 \times 10^{-10}

Solving this equation, you find that x=\left[\mathrm{OH}^{-}\right]=8.4 \times 10^{-6}. Hence,

\mathrm{pH}=14.00-\mathrm{pOH}=14.00+\log \left(8.4 \times 10^{-6}\right)=\mathbf{8 . 9 2}

As expected, the solution has a pH greater than 7.00.