Question

The equilibrium constant at \(25^{\circ} \mathrm{C}\) for the reaction $$ \mathrm{Br}_{2}(l)+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{Br}^{-}(a q)+\mathrm{HBrO}(a q) $$ is \(1.2 \times 10^{-9} .\) This is the system present in a bottle of "bromine water." Assuming that HBrO does not ionize appreciably, what is the \(\mathrm{pH}\) of the bromine water?

Short answer

Answer: The pH of the bromine water is approximately 2.974.

Step-by-step solution

Write the balanced chemical equation

Write down the given balanced chemical equation to understand the reaction: $$ \mathrm{Br}_{2}(l) + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}^{+}(a q) + \mathrm{Br}^{-}(a q) + \mathrm{HBrO}(a q) $$ The equilibrium constant expression for this reaction is: $$ K = \frac{[\mathrm{H}^+][\mathrm{Br}^-][\mathrm{HBrO}]}{[\mathrm{Br}_{2}(l)][\mathrm{H}_{2}\mathrm{O}]} $$ Since the concentrations of pure liquids like \(\mathrm{Br}_{2}(l)\) and \(\mathrm{H}_{2}\mathrm{O}\) do not change appreciably, they can be considered constant, so: $$ K' = \frac{[\mathrm{H}^+][\mathrm{Br}^-][\mathrm{HBrO}]}{constant} $$ where \(K' = K * constant\)

Calculate K'

Use the given value of \(K = 1.2 \times 10^{-9}\) at \(25^{\circ}C\) to calculate K'. Since \(K'\) is equal to the product of the constant and K, and the constant is given as a dimensionless quantity: $$ K' = K = 1.2 \times 10^{-9} $$

Use the stoichiometry of the reaction to find H+ concentration

Let's assume that x moles of \(\mathrm{Br}_{2}(l)\) react. Every mole of \(\mathrm{Br}_{2}(l)\) being consumed produces 1 mole of \(\mathrm{H}^+\), 1 mole of \(\mathrm{Br}^-\), and 1 mole of \(\mathrm{HBrO}\). Based on this, we can write the following relationships: $$ [\mathrm{H}^+] = x $$ $$ [\mathrm{Br}^-] = x $$ $$ [\mathrm{HBrO}] = x $$

Input the concentrations into the K' equation

Using the above relations, the K' equation can be written as: $$ K' = \frac{x^3}{constant} $$ Substituting the value of K': $$ 1.2 \times 10^{-9} = x^3 $$

Solve for x (H+ ion concentration)

Find the concentration of H+ ions by solving the equation above: $$ x = (\sqrt[3]{1.2 \times 10^{-9}}) $$ Calculate the value of x: $$ x \approx 1.061 \times 10^{-3} $$

Calculate the pH of the solution

Use the definition of pH to find the pH of the solution: $$ \mathrm{pH} = -\log_{10} [\mathrm{H}^+] $$ Substitute the value of x: $$ \mathrm{pH} = -\log_{10} (1.061 \times 10^{-3}) \approx 2.974 $$ So the pH of the bromine water is approximately 2.974.