IChO 1973 · Theory problem 3

Ionisation of acetic and hypochlorous acids in a mixture

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Equal volumes (\(10\ \mathrm{cm^3}\)) of 0.01-molar solutions of \(\ce{CH3COOH}\) and \(\ce{HClO}\) were mixed and then diluted to a total volume of \(100\ \mathrm{cm^3}\). Ionisation constant of \(\ce{CH3COOH}\) is equal to \(1.8\times10^{-5}\) and that for \(\ce{HClO}\) is \(3{,}7\times10^{-8}\).

Problems:

Calculate:

3.1 degree of ionisation for each of the acids in the solution,

3.2 degree of ionisation of \(\ce{HClO}\) if the diluted solution would not contain \(\ce{CH3COOH}\),

3.3 pH value for the solution containing at the same time \(\ce{CH3COOH}\) and \(\ce{HClO}\).

Official English text of the problem.

Answers

  • 3.1 In the mixture:
  • Acetic acid, CC(=O)O: \(\alpha=0.125\), or \(12.5\%\).
  • Hypochlorous acid, OCl: \(\alpha=2.94\times10^{-4}\), or \(0.0294\%\).
  • 3.2 Hypochlorous acid alone: \(\alpha=6.06\times10^{-3}\), or \(0.606\%\).
  • 3.3 The mixture has \(\mathrm{pH}=3.90\).

Solution

We use concentrations in place of activities and neglect water’s self-ionisation. Besides the two acids named above, the species involved are water, O; hydronium ion, [OH3+]; acetate ion, CC(=O)[O-]; and hypochlorite ion, [O-]Cl.

3.1 Degree of ionisation of each acid in the mixture

Dilution gives the same analytical concentration for each acid:

\[ C=(0.01\ \mathrm{mol\,dm^{-3}}) \frac{10\ \mathrm{cm^3}}{100\ \mathrm{cm^3}} =1.0\times10^{-3}\ \mathrm{mol\,dm^{-3}}. \]

The ionisation equilibria are

\[ \ce{CH3COOH + H2O <=> CH3COO- + H3O+}, \]

\[ \ce{HClO + H2O <=> ClO- + H3O+}. \]

Both equilibria share the same hydronium-ion concentration,

\[ h=[\ce{H3O+}]. \]

Let \(\alpha_1\) and \(\alpha_2\) be the degrees of ionisation of acetic acid and hypochlorous acid, respectively. Their equilibrium concentrations are

\[ \begin{aligned} [\ce{CH3COO-}]&=C\alpha_1, & [\ce{CH3COOH}]&=C(1-\alpha_1),\\ [\ce{ClO-}]&=C\alpha_2, & [\ce{HClO}]&=C(1-\alpha_2). \end{aligned} \]

Consequently,

\[ K_1=\frac{h\alpha_1}{1-\alpha_1}, \qquad K_2=\frac{h\alpha_2}{1-\alpha_2}, \]

so that

\[ \alpha_1=\frac{K_1}{h+K_1}, \qquad \alpha_2=\frac{K_2}{h+K_2}. \]

Here and below, concentrations entering the equilibrium expressions are expressed in \(\mathrm{mol\,dm^{-3}}\).

Electrical neutrality requires

\[ h=[\ce{CH3COO-}]+[\ce{ClO-}] =C(\alpha_1+\alpha_2). \]

Thus \(h\) satisfies

\[ h=C\left(\frac{K_1}{h+K_1} +\frac{K_2}{h+K_2}\right). \]

Substituting \(C=1.0\times10^{-3}\), \(K_1=1.8\times10^{-5}\), and \(K_2=3.7\times10^{-8}\), the positive solution is

\[ h=1.25623\times10^{-4}\ \mathrm{mol\,dm^{-3}}. \]

Therefore,

\[ \alpha_1 =\frac{1.8\times10^{-5}} {1.25623\times10^{-4}+1.8\times10^{-5}} =0.125328 \approx\boxed{12.5\%}, \]

\[ \alpha_2 =\frac{3.7\times10^{-8}} {1.25623\times10^{-4}+3.7\times10^{-8}} =2.94446\times10^{-4} \approx\boxed{0.0294\%}. \]

Acetic acid is sufficiently ionised that retaining its depletion, through the factor \(1-\alpha_1\), improves the calculation appreciably.

3.2 Degree of ionisation of hypochlorous acid alone

Its analytical concentration remains

\[ C=1.0\times10^{-3}\ \mathrm{mol\,dm^{-3}}. \]

With no acetic acid present, let

\[ x=[\ce{H3O+}]=[\ce{ClO-}]. \]

Then \([\ce{HClO}]=C-x\), giving

\[ K_2=\frac{x^2}{C-x}. \]

Rearranging and taking the positive root,

\[ x^2+K_2x-K_2C=0, \]

\[ x=\frac{\sqrt{K_2^2+4K_2C}-K_2}{2} =6.06429\times10^{-6}\ \mathrm{mol\,dm^{-3}}. \]

Hence

\[ \alpha=\frac{x}{C} =6.06429\times10^{-3} \approx\boxed{0.606\%}. \]

The greater ionisation in this case illustrates the common-ion effect: hydronium ions supplied by acetic acid suppress the ionisation of hypochlorous acid in the mixture.

3.3 pH of the mixture

Using the hydronium-ion concentration obtained in part 3.1,

\[ \mathrm{pH} =-\log_{10}\left( \frac{[\ce{H3O+}]}{1\ \mathrm{mol\,dm^{-3}}} \right) =-\log_{10}(1.25623\times10^{-4}) =3.90093. \]

Thus,

\[ \boxed{\mathrm{pH}=3.90}. \]

The solution was checked against the official solution.