Determine the equilibrium constant for the reaction:
$HF (a q) + {NH}_{3} (a q) ⟶ {NH}_{4}^{+} (a q) + F^{-} (a q)$
Given the equilibrium constants for the reaction
$\begin{array}{r} HF (a q) + H_{2} O (l) ⟶ H_{3} O^{+} (a q) + F^{-} (a q) \\ K_{a} = 6.9 \times 10^{- 4} \end{array}$
$\begin{array}{r} {NH}_{3} (a q) + H_{2} O (l) ⟶ {NH}_{4}^{+} (a q) + {OH}^{-} (a q) \\ K_{b} = 1.8 \times 10^{- 5} \end{array}$
$\begin{aligned} 2 H_{2} O (l) ⟶ H_{3} O^{+} (a q) + {OH}^{-} (a q) \\ K_{w} = 1.0 \times 10^{- 14} \end{aligned}$

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$8.1 \times 10^{7}$

$1.2 \times 10^{6}$

$1.2 \times 10^{- 8}$

$3.8 \times 10^{15}$

answer is B.

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Detailed Solution

Equilibrium constant for the reaction,

$K = \frac{k_{a} \times k_{b}}{k_{w}} K = \frac{6.9 \times 10^{- 4} \times 1.7 \times 10^{- 5}}{1.0 \times 10^{- 14}} K = 1.2 \times 10^{6}$

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