QuestionAmmonia, $\mathrm{NH}_{3}$ is used as a fertilizer in agriculture due to its impact on soil chemistry. If a solution of ammonia has a pH of 8.5 , calculate the concentration of $\mathrm{NH}_{3}$ at equilibrium. [Given the base dissociation constant, $\mathrm{K}_{\mathrm{b}}$ for $\mathrm{NH}_{3}$ is $1.8 \times 10^{-5}$ ] [5 marks]

Studdy Solution
Solve for the concentration of $\mathrm{NH}_3$.
Rearranging the equilibrium expression to solve for $[\mathrm{NH}_3]$:
$$[\mathrm{NH}_3] = \frac{[\mathrm{OH}^-]^2}{K_b}$$
Substitute the known values:
$$[\mathrm{NH}_3] = \frac{(3.16 \times 10^{-6})^2}{1.8 \times 10^{-5}}$$
Calculating this gives:
$$[\mathrm{NH}_3] = \frac{9.9856 \times 10^{-12}}{1.8 \times 10^{-5}}$$
$$[\mathrm{NH}_3] = 5.55 \times 10^{-7} \, \text{M}$$