Calculating Concentrations of Species in a Solution of a Diprotic Acid Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6 (Figure 17.5). What is the pH of a 0.10 M solution? What is the concentration of ascorbate ion, C6H6O6^2-? The acid ionization constants are Ka1 = 7.9 × 10^-5 and

Question

Calculating Concentrations of Species in a Solution of a Diprotic Acid

Ascorbic acid (vitamin C) is a diprotic acid, [latex]\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}[/latex] (Figure 17.5). What is the [latex]\mathrm{pH}[/latex] of a [latex]0.10 \mathrm{M}[/latex] solution? What is the concentration of ascorbate ion, [latex]\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}{ }^{2-}[/latex] ? The acid ionization constants are [latex]K_{a 1}=7.9 imes 10^{-5}[/latex] and [latex]K_{a 2}=1.6 imes 10^{-12}[/latex].

PROBLEM STRATEGY

Diprotic acids have two ionization constants, one for the loss of each proton. To be exact, you should account for both reactions. However, [latex]K_{a 2}[/latex] is so much smaller than [latex]K_{a 1}[/latex] that the smaller amount of hydronium ion produced in the second reaction can be neglected. Follow the three steps for solving equilibrium problems.

Considering only the first ionization, set up a table of concentrations (starting, change, and equilibrium). Substitute these values into the equilibrium equation for [latex]K_{a 1}[/latex], solve for [latex]x=\left[\mathrm{H}_{3} \mathrm{O}^{+} ight][/latex], and then solve for [latex]\mathrm{pH}[/latex] (as in Example 17.2). Ascorbate ion is produced in the second ionization; its concentration, as we show, equals [latex]K_{a 2}[/latex].

Accepted Answer

CALCULATION of pH Abbreviate the formula for ascorbic acid as [latex]\mathrm{H}_{2}[/latex] Asc. Hydronium ions are produced by two successive acid ionizations.

[latex] \begin{aligned} & \mathrm{H}_{2} \mathrm{Asc}(a q)+\mathrm{H}_{2} \mathrm{O}(l) ightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\operatorname{HAsc}^{-}(a q) ; K_{a 1}=7.9 imes 10^{-5} \ & \mathrm{HAsc}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) ightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\operatorname{Asc}^{2-}(a q) ; K_{a 2}=1.6 imes 10^{-12} \end{aligned} [/latex]

If you let [latex]x[/latex] be the amount of [latex]\mathrm{H}_{3} \mathrm{O}^{+}[/latex]formed in the first ionization, you get the following results:

[latex] \begin{array}{cccc} ext { Concentration }(M) & \mathbf{H}_{2} \mathbf{A s c}(\boldsymbol{a q})+\mathbf{H}_{2} \mathbf{O}(\boldsymbol{l}) & ightleftharpoons \mathbf{H}_{3} \mathbf{O}^{+}(\boldsymbol{a q})&+\mathbf{H A s c}^{-}(\boldsymbol{a q}) \ ext { Starting } & 0.10 & \sim 0 & 0 \ ext { Change } & -x & +x & +x \ ext { Equilibrium } & 0.10-x & x & x \end{array} [/latex]

Now substitute into the equilibrium-constant equation for the first ionization.

[latex] \begin{aligned} \frac{\left[\mathrm{H}_{3} \mathrm{O}^{+} ight]\left[\mathrm{HAsc}^{-} ight]}{\left[\mathrm{H}_{2} \mathrm{Asc} ight]} & =K_{a 1} \ \frac{x^{2}}{0.10-x} & =7.9 imes 10^{-5} \end{aligned} [/latex]

Assuming [latex]x[/latex] to be much smaller than 0.10 , you get

[latex]\frac{x^{2}}{0.10} \simeq 7.9 imes 10^{-5}[/latex]

or

[latex] \begin{gathered} x^{2} \simeq 7.9 imes 10^{-5} imes 0.10 \ x \simeq 2.8 imes 10^{-3}=0.0028 \end{gathered} [/latex]

(Note that [latex]0.10-x=0.10-0.0028=0.10[/latex], correct to two significant figures. Therefore, the assumption that [latex]0.10-x \simeq 0.10[/latex] is correct.)

The hydronium-ion concentration is [latex]0.0028 M[/latex], so

[latex]\mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+} ight]=-\log (0.0028)=\mathbf{2 . 5 5}[/latex]

ASCORBATE-ION CONCENTRATION Ascorbate ion, [latex]\mathrm{Asc}^{2-}[/latex], which we will call [latex]y[/latex], is produced only in the second reaction. Assume the starting concentrations of [latex]\mathrm{H}_{3} \mathrm{O}^{+}[/latex] and [latex]\mathrm{HAsc}^{-}[/latex]for this reaction to be those from the first equilibrium. The amounts of each species in [latex]1 \mathrm{~L}[/latex] of solution are as follows:

[latex] \begin{array}{cccc} ext { Concentration }(M) & \mathbf{H A s c}^{-}(\boldsymbol{a q})+\mathbf{H}_{2} \mathbf{O}(l) & ightleftharpoons \mathbf{H}_{3} \mathbf{O}^{+}(\boldsymbol{a q}) & +\mathbf{A s c}^{2-}(\boldsymbol{a q}) \ ext { Starting } & 0.0028 & 0.0028 & 0 \ ext { Change } & -y & +y & +y \ ext { Equilibrium } & 0.0028-y & 0.0028+y & y \end{array} [/latex]

Now substitute into the equilibrium-constant equation for the second ionization.

[latex] \begin{aligned} & \frac{\left[\mathrm{H}_{3} \mathrm{O}^{+} ight]\left[\mathrm{Asc}^{2-} ight]}{\left[\mathrm{HAsc}^{-} ight]}=K_{a 2} \ & \frac{(0.0028+y) y}{0.0028-y}=1.6 imes 10^{-12} \end{aligned} [/latex]

This equation can be simplified if you assume that [latex]y[/latex] is much smaller than 0.0028 . (Again, this assumes that the reaction occurs to only a small extent, as you expect from the magnitude of the equilibrium constant.) That is,

[latex] \begin{aligned} & 0.0028+y \simeq 0.0028 \ & 0.0028-y \simeq 0.0028 \end{aligned} [/latex]

Then the equilibrium equation reads

[latex]\frac{(0.0028) y}{0.0028} \simeq 1.6 imes 10^{-12}[/latex]

Hence, [latex]y \simeq 1.6 imes 10^{-12}[/latex]

(Note that [latex]1.6 imes 10^{-12}[/latex] is indeed much smaller than 0.0028 , as you assumed.) The concentration of ascorbate ion equals [latex]K_{a 2}[/latex], or [latex]1.6 imes 10^{-12} M[/latex].

Question 17.4: Calculating Concentrations of Species in a Solution of a Dip...

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