Question

Drinking water is often disinfected with \(\mathrm{Cl}_{2},\) which hydrolyzes to form \(\mathrm{HClO},\) a weak acid but powerful disinfectant: $$ \mathrm{Cl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HClO}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{Cl}^{-}(a q) $$ The fraction of \(\mathrm{HClO}\) in solution is defined as $$ \frac{[\mathrm{HClO}]}{[\mathrm{HClO}]+\left[\mathrm{ClO}^{-}\right]} $$ (a) What is the fraction of \(\mathrm{HClO}\) at \(\mathrm{pH} 7.00\left(K_{\mathrm{a}}\right.\) of \(\mathrm{HClO}=\) \(\left.2.9 \times 10^{-8}\right) ?\) (b) What is the fraction at pH \(10.00 ?\)

Short answer

(a) The fraction of \(\mathrm{HClO}\) at pH 7.00 is approximately 0.775.(b) The fraction of \(\mathrm{HClO}\) at pH 10.00 is approximately 0.0034.

Step-by-step solution

- Write the balanced chemical equation and relevant expressions

Write the hydrolysis reaction: \[\mathrm{Cl}_{2}(aq) + 2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HClO}(aq) + \mathrm{H}_{3} \mathrm{O}^{+}(aq) + \mathrm{Cl}^{-}(aq)\]Identify the base dissociation of \(\mathrm{HClO}\):\[\mathrm{HClO}(aq) \leftrightarrow \mathrm{H}^{+}(aq) + \mathrm{ClO}^{-}(aq)\]

- Identify the equilibrium expression

Given the dissociation constant \(K_a\) of \(\mathrm{HClO}\), write the expression for the equilibrium constant:\[K_a = \frac{[\mathrm{H}^+][\mathrm{ClO}^-]}{[\mathrm{HClO}]}\]Given: \(K_a = 2.9 \times 10^{-8}\)

- Calculate the concentration of \(\mathrm{H}^+\) at pH 7.00

Use the pH to find the \([\mathrm{H}^+]\) concentration:\[pH = -\log[\mathrm{H}^+]\]At pH 7.00:\[[\mathrm{H}^+] = 10^{-7}\ \text{M}\]

- Calculate the ratio of \([\mathrm{ClO}^-]\) to \([\mathrm{HClO}]\) at pH 7.00

Substitute \([\mathrm{H}^+]\) into the equilibrium expression to solve for the ratio:\[2.9 \times 10^{-8} = \frac{10^{-7} [\mathrm{ClO}^-]}{[\mathrm{HClO}]}\]Solve for \(\frac{[\mathrm{ClO}^-]}{[\mathrm{HClO}]}\):\[\frac{[\mathrm{ClO}^-]}{[\mathrm{HClO}]} = \frac{2.9 \times 10^{-8}}{10^{-7}} = 0.29\]

- Calculate the fraction of \(\mathrm{HClO}\) at pH 7.00

Use the fraction formula:\[\frac{[\mathrm{HClO}]}{[\mathrm{HClO}] + [\mathrm{ClO}^-]}\]Express the relation:\[[\mathrm{HClO}] = 1\quad \text{and} \quad [\mathrm{ClO}^-] = 0.29 \frac{[\mathrm{HClO}]}{1 + 0.29} = \frac{1}{1.29} \approx 0.775\]

- Calculate the concentration of \(\mathrm{H}^+\) at pH 10.00

Use the pH to find the \([\mathrm{H}^+]\) concentration:\[pH = -\log[\mathrm{H}^+]\]At pH 10.00:\[[\mathrm{H}^+] = 10^{-10}\ \text{M}\]

- Calculate the ratio of \([\mathrm{ClO}^-]\) to \([\mathrm{HClO}]\) at pH 10.00

Substitute \([\mathrm{H}^+]\) into the equilibrium expression to solve for the ratio:\[2.9 \times 10^{-8} = \frac{10^{-10} [\mathrm{ClO}^-]}{[\mathrm{HClO}]}\]Solve for \(\frac{[\mathrm{ClO}^-]}{[\mathrm{HClO}]}\):\[\frac{[\mathrm{ClO}^-]}{[\mathrm{HClO}]} = \frac{2.9 \times 10^{-8}}{10^{-10}} = 290\]

- Calculate the fraction of \(\mathrm{HClO}\) at pH 10.00

Use the fraction formula:\[\frac{[\mathrm{HClO}]}{[\mathrm{HClO}] + [\mathrm{ClO}^-]}\]Express the relation:\[[\mathrm{HClO}] = 1\quad \text{and} \quad [\mathrm{ClO}^-] = 290 \frac{[\mathrm{HClO}]}{1 + 290} = \frac{1}{291} \approx 0.0034\]

Key concepts

Chlorination

Chlorination is a common method used to disinfect drinking water. This process involves adding chlorine (\text{Cl}_2) to the water. Chlorine is effective at killing bacteria, viruses, and other pathogens that can cause illness. When \text{Cl}_{2} is added to water, it reacts with water molecules to form hypochlorous acid (\text{HClO}), which is powerful at disinfecting. The chemical reaction for this process is:\[ \text{Cl}_{2}(aq) + 2 \text{H}_{2} \text{O}(l) \rightarrow \text{HClO}(aq) + \text{H}_{3} \text{O}^{+}(aq) + \text{Cl}^{-}(aq) \]This reaction not only forms \text{HClO}, but also hydronium ions (\text{H}_3\text{O}^+) and chloride ions (\text{Cl}^-), which alter the water's acidity. Understanding this reaction and its products is crucial for effective water treatment.

Weak Acids

Weak acids, like hypochlorous acid (\text{HClO}), do not fully dissociate in water. Unlike strong acids that completely ionize, weak acids only partially release their hydrogen ions (\text{H}^+). The dissociation of \text{HClO} in water is given by:\[ \text{HClO}(aq) \rightleftharpoons \text{H}^{+}(aq) + \text{ClO}^{-}(aq) \]This reversible reaction indicates that not all \text{HClO} molecules lose their protons, leading to a state of equilibrium. The balance between the undissociated \text{HClO} and the dissociated ions (\text{H}^+ and \text{ClO}^{-}) depends on the acidity (\text{pH}) of the solution.

Equilibrium Constant

The equilibrium constant \text{K}_a is a vital concept in understanding acid strength. For weak acids like \text{HClO}, the \text{K}_a value indicates the degree of dissociation in water. The equilibrium expression for \text{HClO} is:\[ K_a = \frac{[\text{H}^+][\text{ClO}^-]}{[\text{HClO}]}\]This formula relates the concentrations of the products (\text{H}^+ and \text{ClO}^-) to the reactant (\text{HClO}). A smaller \text{K}_a value (like the 2.9 \times 10^{-8} for \text{HClO}) means the acid is weaker and less dissociation occurs. By knowing \text{K}_a and the \text{H}^+ concentration, you can calculate the concentrations of the other species in the solution at equilibrium.

pH and pKa

The acidity of a solution is expressed by its \text{pH}, which is the negative logarithm of the hydrogen ion concentration (\text{[H}^+\text{]}). For example, a \text{pH} of 7.00 means \text{[H}^+\text{]} = 10^{-7} M. The \text{pKa} of an acid is similarly the negative logarithm of the dissociation constant (\text{K}\text{_a}):\[ pK_a = -\text{log}(K_a) \]For \text{HClO}, \text{pKa} = -\text{log}(2.9 \times 10^{-8}) \text{ which is approximately 7.54}. The relationship between \text{pH} and \text{pKa} helps us determine the degree of dissociation and the form of the acid in the solution. When \text{pH} = \text{pKa}, half of the acid is dissociated. At \text{pH} values less than \text{pKa}, more \text{HClO} exists in its protonated form. At \text{pH} values higher than \text{pKa}, more \text{HClO} exists as \text{ClO}^{-}.

Hydrolysis Reactions

Hydrolysis reactions involve breaking a bond in a molecule using water. In the chlorination process, \text{Cl}_2 hydrolyzes to form \text{HClO}, as shown in the reaction:\[ \text{Cl}_2(aq) + 2 \text{H}_2 \text{O}(l) \rightarrow \text{HClO}(aq) + \text{H}_3 \text{O}^+(aq) + \text{Cl}^-(aq) \]This reaction is essential for producing the disinfectant properties of \text{HClO}. Hydrolysis can affect the chemical composition and properties of water. For instance, the formation of \text{H}_3\text{O}^+ ions increases the water's acidity, influencing pH and the equilibrium of other reactions. Understanding hydrolysis reactions helps in predicting how different substances will react and transform in aqueous environments.