Calculating Concentrations of Species in a Salt Solution What is the pH of 0.10 M sodium nicotinate at 25°C (a problem posed in the chapter opening)? The Ka for nicotinic acid was determined in Example 17.1 to be 1.4 × 10^-5 at 25°C.

Question

Calculating Concentrations of Species in a Salt Solution

What is the [latex]\mathrm{pH}[/latex] of [latex]0.10 \mathrm{M}[/latex] sodium nicotinate at [latex]25^{\circ} \mathrm{C}[/latex] (a problem posed in the chapter opening)? The [latex]K_{a}[/latex] for nicotinic acid was determined in Example 17.1 to be 1.4 [latex] imes 10^{-5}[/latex] at [latex]25^{\circ} \mathrm{C}[/latex].

Accepted Answer

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

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

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 [latex]K_{b}[/latex] for the nicotinate ion. This is related to [latex]K_{a}[/latex] for nicotinic acid by the equation [latex]K_{a} K_{b}=K_{w}[/latex]. Substituting, you get

[latex] K_{b}=\frac{K_{w}}{K_{a}}=\frac{1.0 imes 10^{-14}}{1.4 imes 10^{-5}}=7.1 imes 10^{-10} [/latex]

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 [latex]x=[\mathrm{HNic}]=[/latex] [latex]\left[\mathrm{OH}^{-} ight][/latex], then substitute into the equilibrium-constant equation.

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

This gives

[latex]\frac{x^{2}}{0.10-x}=7.1 imes 10^{-10}[/latex]

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

[latex]\mathrm{pH}=14.00-\mathrm{pOH}=14.00+\log \left(8.4 imes 10^{-6} ight)=\mathbf{8 . 9 2}[/latex]

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

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

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