Question

Calculate the molar concentration of \(\mathrm{OH}^{-}\) ions in a \(0.550 \mathrm{M}\) solution of hypobromite ion \(\left(\mathrm{BrO}^{-} ;\right.\) \(K_{b}=4.0 \times 10^{-6}\) ). What is the \(\mathrm{pH}\) of this solution?

Short answer

The equilibrium concentration of \(\mathrm{OH}^{-}\) ions in the \(0.550\mathrm{M}\) solution of hypobromite ion is approximately \(2.98 \times 10^{-3} \mathrm{M}\), and the corresponding \(\mathrm{pH}\) of the solution is approximately \(11.47\).

Step-by-step solution

Write the expression for the base dissociation reaction

For the hypobromite ion, the base dissociation reaction is: \[\mathrm{BrO}^{-} + \mathrm{H_2O} \rightleftharpoons \mathrm{OH}^{-} + \mathrm{HBrO}\]

Write the expression for the base dissociation constant \(K_b\)

The expression for the base dissociation constant \(K_b\) is: \[K_b = \frac{[\mathrm{OH}^{-}][\mathrm{HBrO}]}{[\mathrm{BrO}^{-}]}\]

Set up an ICE table for the reaction

An ICE table will help us determine the equilibrium concentrations of the species involved in the reaction. The initial concentration of \(\mathrm{BrO}^{-}\) is \(0.550 \mathrm{M}\), while the initial concentrations of \(\mathrm{OH}^{-}\) and \(\mathrm{HBrO}\) are both \(0\). ``` [BrO⁻] [OH⁻] [HBrO] I 0.550 M 0 0 C -x +x +x E 0.550-x M x x ```

Solve for the equilibrium concentration of \(\mathrm{OH}^{-}\) ions

At equilibrium, we can plug the ICE table values into the \(K_b\) expression: \[4.0 \times 10^{-6} = \frac{x \cdot x}{0.550 - x}\] Assuming that x is small compared to \(0.550\), we can approximate the denominator to be \(0.550\): \[4.0 \times 10^{-6} = \frac{x^2}{0.550}\] Now, solve for x (the equilibrium concentration of \(\mathrm{OH}^{-}\) ions): \[x^2 = 4.0 \times 10^{-6} \cdot 0.550\] \[x = \sqrt{4.0 \times 10^{-6} \cdot 0.550} \approx 2.98 \times 10^{-3} \mathrm{M}\]

Calculate \(\mathrm{pOH}\) and \(\mathrm{pH}\) of the solution

Now that we have the equilibrium concentration of \(\mathrm{OH}^{-}\) ions, we can calculate the \(\mathrm{pOH}\) and \(\mathrm{pH}\) of the solution: \[\mathrm{pOH} = -\log{[\mathrm{OH}^{-}]} = -\log{(2.98 \times 10^{-3})} \approx 2.53\] Recall that \(\mathrm{pH} + \mathrm{pOH} = 14\), so we can find the \(\mathrm{pH}\) of the solution: \[\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 2.53 \approx 11.47\] Thus, the \(\mathrm{pH}\) of the \(0.550 \mathrm{M}\) solution of hypobromite ion is approximately \(11.47\).