Determining Ka from the Solution pH Nicotinic acid (niacin) is a monoprotic acid with the formula HC6H4NO2. A solution that is 0.012 M in nicotinic acid has a pH of 3.39 at 25°C. What is the acidionization constant, Ka, for this acid at 25°C? What is the degree of ionization of nicotinic acid in

Question

Determining Ka from the Solution pH

Nicotinic acid (niacin) is a monoprotic acid with the formula [latex]\mathrm{HC}_{6} \mathrm{H}_{4} \mathrm{NO}_{2}[/latex]. A solution that is [latex]0.012 \mathrm{M}[/latex] in nicotinic acid has a [latex]\mathrm{pH}[/latex] of 3.39 at [latex]25^{\circ} \mathrm{C}[/latex]. What is the acidionization constant, [latex]K_{a}[/latex], for this acid at [latex]25^{\circ} \mathrm{C}[/latex] ? What is the degree of ionization of nicotinic acid in this solution?

PROBLEM STRATEGY

It is important to realize that when we say the solution is [latex]0.012 \mathrm{M}[/latex], this refers to how the solution is prepared. The solution is made up by adding [latex]0.012 \mathrm{~mol}[/latex] of substance to enough water to give a liter of solution. Once the solution is prepared, some molecules ionize, so the actual concentration is somewhat less than [latex]0.012 \mathrm{M}[/latex]. To solve for [latex]K_{a}[/latex], you follow the three steps outlined in Chapter 15 for solving equilibrium problems.

Accepted Answer

STEP 1 Abbreviate the formula for nicotinic acid as HNic. Then [latex]1 \mathrm{~L}[/latex] of solution contains [latex]0.012 \mathrm{~mol} \mathrm{HNic}[/latex] and [latex]0 \mathrm{~mol} \mathrm{Nic}{ }^{-}[/latex], the acid anion, before ionization. The [latex]\mathrm{H}_{3} \mathrm{O}^{+}[/latex]concentration at the start is that from the self-ionization of water. It is usually much smaller than that obtained from the acid (unless the solution is extremely dilute or [latex]K_{a}[/latex] is quite small), so you can write [latex]\left[\mathrm{H}_{3} \mathrm{O}^{+} ight]=\sim 0[/latex] (meaning approximately zero). If [latex]x[/latex] mol HNic ionizes, [latex]x[/latex] mol each of [latex]\mathrm{H}_{3} \mathrm{O}^{+}[/latex]and [latex]\mathrm{Nic}^{-}[/latex]is formed, leaving [latex](0.012-x)[/latex] mol HNic in solution. You can summarize the situation as follows:

[latex]\begin{array}{lccc} ext { Concentration }(M) & \mathbf{H N i c}(\boldsymbol{a q})+\mathbf{H}_{2} \mathbf{O}(l)& ightleftharpoons & \mathbf{H}_{3} \mathbf{O}^{+}(\mathbf{a q})&+&\mathbf{N i c}^{-}(\boldsymbol{a q}) \ ext { Starting } & 0.012 & &\sim 0 && 0 \ ext { Change } & -x && +x && +x \ ext { Equilibrium } & 0.012-x & &x && x\end{array}[/latex]

Thus, the molar concentrations of [latex]\mathrm{HNic}, \mathrm{H}_{3} \mathrm{O}^{+}[/latex], and [latex]\mathrm{Nic}^{-}[/latex]at equilibrium are [latex](0.012-x), x[/latex], and [latex]x[/latex], respectively.

STEP 2 The equilibrium-constant equation is

[latex]K_{a}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+} ight]\left[\mathrm{Nic}^{-} ight]}{[\mathrm{HNic}]}[/latex]

When you substitute the expressions for the equilibrium concentrations, you get

[latex]K_{a}=\frac{x^{2}}{(0.012-x)}[/latex]

STEP 3 The value of [latex]x[/latex] equals the numerical value of the molar hydronium-ion concentration, and can be obtained from the [latex]\mathrm{pH}[/latex] of the solution.

[latex]x=\left[\mathrm{H}_{3} \mathrm{O}^{+} ight]=\operatorname{antilog}(-\mathrm{pH})=\operatorname{antilog}(-3.39)=4.1 imes 10^{-4}=0.00041[/latex]

You can substitute this value of [latex]x[/latex] into the equation obtained in Step 2. Note first, however, that

[latex]0.012-x=0.012-0.00041=0.01159 \simeq 0.012 \quad ext { (to two significant figures) }[/latex]

This means that the concentration of undissolved acid is equal to the original concentration of the acid within the precision of the data. (We will make use of this type of observation in later problem solving.) Therefore,

[latex]K_{a}=\frac{x^{2}}{(0.012-x)} \simeq \frac{x^{2}}{0.012} \simeq \frac{(0.00041)^{2}}{0.012}=\mathbf{1 . 4} imes \mathbf{1 0}^{-\mathbf{5}}[/latex]

To obtain the degree of ionization, note that [latex]x[/latex] mol out of 0.012 mol of nicotinic acid ionizes. Hence,

[latex] ext { Degree of ionization }=\frac{x}{0.012}=\frac{0.00041}{0.012}=\mathbf{0 . 0 3 4}[/latex]

The percent ionization is obtained by multiplying this by 100 , which gives [latex]3.4 \%[/latex].

Question 17.1: Determining Ka from the Solution pH Nicotinic acid (niacin) ...

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