Calculating the Common-Ion Effect on Acid Ionization (Effect of a Conjugate Base) A solution is prepared to be 0.10 M acetic acid, HC2H3O2, and 0.20 M sodium acetate, NaC2H3O2. What is the pH of this solution at 25°C? Ka for acetic acid at 25°C is 1.7 × 10^-5.

Question

Calculating the Common-Ion Effect on Acid Ionization (Effect of a Conjugate Base)

A solution is prepared to be [latex]0.10 \mathrm{M}[/latex] acetic acid, [latex]\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}[/latex], and [latex]0.20 \mathrm{M}[/latex] sodium acetate, [latex]\mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}[/latex]. What is the [latex]\mathrm{pH}[/latex] of this solution at [latex]25^{\circ} \mathrm{C}[/latex] ? [latex]K_{a}[/latex] for acetic acid at [latex]25^{\circ} \mathrm{C}[/latex] is [latex]1.7 imes 10^{-5}[/latex].

Accepted Answer

STEP 1 Consider the equilibrium

[latex]\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) ightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}{ }^{-}(a q)[/latex]

Initially, [latex]1 \mathrm{~L}[/latex] of solution contains [latex]0.10 \mathrm{~mol}[/latex] of acetic acid. Sodium acetate is a strong electrolyte, so [latex]1 \mathrm{~L}[/latex] of solution contains [latex]0.20 \mathrm{~mol}[/latex] of acetate ion. When the acetic acid ionizes, it gives [latex]x[/latex] mol of hydronium ion and [latex]x[/latex] mol of acetate ion. This is summarized in the following table:

[latex]\begin{array}{cccc} ext { Concentration }(M) & \mathrm{HC}_{2} \mathrm{H}_{3} \mathbf{O}_{2}( ext { aq })+\mathrm{H}_{2} \mathbf{O}(l) & ightleftharpoons \mathrm{H}_{3} \mathbf{O}^{+}( ext {aq })&+\mathrm{C}_{2} \mathbf{H}_{3} \mathbf{O}_{2}{ }^{-}( ext {aq }) \ ext { Starting } & 0.10 & \sim 0 & 0.20 \ ext { Change } & -x & +x & +x \ ext { Equilibrium } & 0.10-x & x & 0.20+x\end{array}[/latex]

STEP 2 The equilibrium-constant equation is

[latex]\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+} ight]\left[\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-} ight]}{\left[\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} ight]}=K_{a}[/latex]

Substituting into this equation gives

[latex]\frac{x(0.20+x)}{0.10-x}=1.7 imes 10^{-5}[/latex]

STEP 3 To solve the equation, assume that [latex]x[/latex] is small compared with 0.10 and 0.20 . Then

[latex] \begin{aligned} & 0.20+x \simeq 0.20 \ & 0.10-x \simeq 0.10 \end{aligned} [/latex]

The equilibrium equation becomes

[latex]\frac{x(0.20)}{0.10} \simeq 1.7 imes 10^{-5}[/latex]

Hence,

[latex]x \simeq 1.7 imes 10^{-5} imes \frac{0.10}{0.20}=8.5 imes 10^{-6}[/latex]

(Note that [latex]x[/latex] is indeed much smaller than 0.10 or 0.20 .) The hydronium-ion concentration is [latex]8.5 imes 10^{-6} \mathrm{M}[/latex], and

[latex]\mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+} ight]=-\log \left(8.5 imes 10^{-6} ight)=\mathbf{5 . 0 7}[/latex]

For comparison, the [latex]\mathrm{pH}[/latex] of [latex]0.10 \mathrm{M}[/latex] acetic acid is 2.88 .

Question 17.10: Calculating the Common-Ion Effect on Acid Ionization (Effect...

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