Calculating the Common-Ion Effect on Acid Ionization (Effect of a Strong Acid) The degree of ionization of acetic acid, HC2H3O2, in a 0.10 M aqueous solution at 25°C is 0.013. Ka at this temperature is 1.7 × 10^-5. Calculate the degree of ionization of HC2H3O2 in a 0.10 M solution at 25°C to which

Question

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

The degree of ionization of acetic acid, [latex]\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}[/latex], in a [latex]0.10 \mathrm{M}[/latex] aqueous solution at [latex]25^{\circ} \mathrm{C}[/latex] is 0.013. [latex]K_{a}[/latex] at this temperature is [latex]1.7 imes 10^{-5}[/latex]. Calculate the degree of ionization of [latex]\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}[/latex] in a [latex]0.10 \mathrm{M}[/latex] solution at [latex]25^{\circ} \mathrm{C}[/latex] to which sufficient [latex]\mathrm{HCl}[/latex] is added to make it [latex]0.010 \mathrm{M} \mathrm{HCl}[/latex]. How is the degree of ionization affected?

PROBLEM STRATEGY

This is an acid-ionization problem, but it differs from the simple ionization illustrated by Example 17.2. Here you have a starting concentration of [latex]\mathrm{H}_{3} \mathrm{O}^{+}[/latex] [latex](=0.010 M)[/latex] from the addition of a strong acid [latex](\mathrm{HCl})[/latex]. This gives a different type of equation in Step 2, but you solve it in Step 3 by using a similar approximation method. You assume that [latex]x[/latex] is small compared with starting concentrations of acid and [latex]\mathrm{H}_{3} \mathrm{O}^{+}[/latex], so that the resulting equation is linear, rather than quadratic. The calculation of degree of ionization follows that in Example 17.2.

Accepted Answer

STEP 1 Starting concentrations are [latex]\left[\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} ight]=0.10 \mathrm{M},\left[\mathrm{H}_{3} \mathrm{O}^{+} ight]=0.010 \mathrm{M}[/latex] (from [latex]\mathrm{HCl}[/latex] ), and [latex]\left[\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}{ }^{-} ight]=0[/latex]. The acetic acid ionizes to give an additional [latex]x[/latex] [latex]\mathrm{mol} / \mathrm{L}[/latex] of [latex]\mathrm{H}_{3} \mathrm{O}^{+}[/latex]and [latex]x \mathrm{~mol} / \mathrm{L}[/latex] of [latex]\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}{ }^{-}[/latex]. The table is

[latex]\begin{array}{cccc} ext { Concentration }(M) & \mathrm{HC}_{2} \mathbf{H}_{3} \mathbf{O}_{2}(a q)+\mathbf{H}_{2} \mathbf{O}(l)& ightleftharpoons \mathbf{H}_{3} \mathbf{O}^{+}(a q)&+\mathrm{C}_{2} \mathbf{H}_{3} \mathbf{O}_{2}{ }^{-}( ext {aq }) \ ext { Starting } & 0.10 & 0.010 & 0 \ ext { Change } & -x & +x & +x \ ext { Equilibrium } & 0.10-x & 0.010+x & x\end{array}[/latex]

STEP 2 You substitute into the equilibrium-constant equation

[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]

obtaining

[latex]\frac{(0.010+x) x}{0.10-x}=1.7 imes 10^{-5}[/latex]

STEP 3 To solve this equation, assume that [latex]x[/latex] is small compared with 0.010 . Then

[latex] \begin{gathered} 0.010+x \simeq 0.010 \ 0.10-x \simeq 0.10 \end{gathered} [/latex]

The equation becomes

[latex]\frac{0.010 x}{0.10} \simeq 1.7 imes 10^{-5}[/latex]

Solving for [latex]x[/latex], you get

[latex]x=1.7 imes 10^{-5} imes \frac{0.10}{0.010}=1.7 imes 10^{-4}[/latex]

(Check that [latex]x[/latex] can indeed be neglected in [latex]0.010+x[/latex] and [latex]0.10-x[/latex].) The degree of ionization of [latex]\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}[/latex] is [latex]x / 0.10=\mathbf{0 . 0 0 1 7}[/latex]. This is much smaller than the value for [latex]0.10 M \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(0.013)[/latex], because the addition of [latex]\mathrm{HCl}[/latex] represses the ionization of [latex]\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}[/latex].

Question 17.9: Calculating the Common-Ion Effect on Acid Ionization (Effect...

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