Determine the equilibrium constant for the reaction HF(aq) + NH₃(aq) ? F?(aq) + NH₄⁺(aq)
Given the equilibrium constants for the following reactions.
$\begin{array}{ll}\mathrm{HF}(a q) +\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{F}-(a q) +\mathrm{H}_{3} \mathrm{O}^{-}(a q) & K_{\mathrm{a}}=6.9 \times 10^{-4} \\\mathrm{NH}_{3}(a q) +\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_{4}^{+}(a q) +\mathrm{OH}-(a q) & K_{\mathrm{b}}=1.8 \times 10^{-5} \\2 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{OH}-(a q) +\mathrm{H}_{3} \mathrm{O}^{+}(a q) & K_{\mathrm{W}}=1.0 \times 10^{-14}\end{array}$

Question

Determine the equilibrium constant for the reaction HF(aq) + NH₃(aq) ? F?(aq) + NH₄⁺(aq)
Given the equilibrium constants for the following reactions.

\begin{array}{ll}\mathrm{HF}(a q) +\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{F}-(a q) +\mathrm{H}_{3} \mathrm{O}^{-}(a q) & K_{\mathrm{a}}=6.9 \times 10^{-4} \\\mathrm{NH}_{3}(a q) +\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_{4}^{+}(a q) +\mathrm{OH}-(a q) & K_{\mathrm{b}}=1.8 \times 10^{-5} \\2 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{OH}-(a q) +\mathrm{H}_{3} \mathrm{O}^{+}(a q) & K_{\mathrm{W}}=1.0 \times 10^{-14}\end{array}

Quizplus · Accepted Answer

The equilibrium constant for the given reaction can be calculated using the relation $ K = \frac{K_a \cdot K_b}{K_w} $. Substituting the given values: $ K = \frac{(6.9 \times 10^{-4}) \cdot (1.8 \times 10^{-5})}{1.0 \times 10^{-14}} = 1.2 \times 10^6 $.

Determine the Equilibrium Constant for the Reaction HF(aq)+ NH3(aq) A) 12 × 10?8 B) 1

Determine the Equilibrium Constant for the Reaction HF(aq)+ NH₃(aq) A) 12 × 10?⁸
B) 1