The \(\mathrm{pH}\) of a \(0.063-M\) solution of hypobromous acid \((\mathrm{HOBr}\) but usually written \(\mathrm{HBrO}\) ) is 4.95. Calculate \(K_{\mathrm{a}}\).

HBrO is a weak acid, so it partially dissociates into its ions in water as follows: \[ \mathrm{HBrO} \rightleftharpoons \mathrm{H^+} + \mathrm{BrO^-} \]

For this dissociation reaction, we can write the Ka expression as: \[K_\mathrm{a} = \frac{[\mathrm{H^+}][\mathrm{BrO^-}]}{[\mathrm{HBrO}]} \]

The pH of the solution is 4.95. We can use the relationship between pH and hydrogen ion concentration to find [H+]: \[ \mathrm{pH} = -\log [\mathrm{H^+}] \] Solving for [H+]: \[ [\mathrm{H^+}] = 10^{-\mathrm{pH}} = 10^{- 4.95} \] \[[\mathrm{H^+}] = 1.122 \times 10^{-5} \, M \]

Since HBrO is a weak acid, we'll assume that its initial concentration doesn't change significantly during the dissociation process. Here's the ICE table: Initial: \[ [\mathrm{HBrO}]_0 = 0.063 \, M \quad [\mathrm{H^+}]_0 = 0 \, M \quad [\mathrm{BrO^-}]_0 = 0 \, M \] Change: \[ -x \quad +x \quad +x \] Equilibrium: \[ [\mathrm{HBrO}] = 0.063 - x \, M \quad [\mathrm{H^+}] = 1.122 \times 10^{-5} + x \, M \quad [\mathrm{BrO^-}] = x \, M \]

Now we can substitute the expressions from the ICE table into the Ka expression: \[K_\mathrm{a} = \frac{(1.122 \times 10^{-5} + x)(x)}{(0.063 - x)} \] Since HBrO is a weak acid, we can assume x is much smaller than the initial concentration of HBrO: \[K_\mathrm{a} \approx \frac{(1.122 \times 10^{-5})(x)}{0.063} \] Now, we can solve for x, which is the equilibrium concentration of BrO-: \[ x \approx \frac{[\mathrm{H^+}] K_\mathrm{a}}{K_\mathrm{a} + 1.12 \times 10^{-5}} \] Since the equilibrium concentration of H+ is known, we can solve for Ka: \[ K_\mathrm{a} \approx \frac{[\mathrm{H^+}]^2}{0.063} - [\mathrm{H^+}] \] \[ K_\mathrm{a} \approx \frac{(1.122 \times 10^{-5})^2}{0.063} - 1.122 \times 10^{-5} \] \[ K_\mathrm{a} \approx 2.0 \times 10^{-9} \] Therefore, the acid dissociation constant (Ka) of hypobromous acid is approximately \(2.0 \times 10^{-9}\).