Question
The ionization constant of NH4+ in water is 5.6 × 10−10 mol L–1 at 25oC. The rate constant for the reaction of NH4+ and OH− to form NH3 and H2O at 25oC is 3.4 × 1010 L mol−1 s−1. The rate constant for proton transfer from water to NH3 at 25o C is
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Solution
$NH_{4}^{+} + OH^{-}\overset{\quad K_{f} = 3.4 \times 10^{10}M^{- 1}S^{- 1}K_{b} = ?\quad}{\rightarrow}NH_{3} + H_{2}O$
Given: $NH_{4}^{+} \rightleftharpoons NH_{3} + H^{+};K_{1} = 5.6 \times 10^{- 10}\text{ M}$
and $H_{2}O \rightleftharpoons H^{+} + OH^{-};K_{2} = 1.0 \times 10^{- 14}\text{ }\text{M}^{2}$
$\therefore NH_{4}^{+} + OH^{-} \rightleftharpoons NH_{3} + H_{2}O;K_{eq} = \frac{K_{1}}{K_{2}}$
Now, $\frac{3.4 \times 10^{10}}{K_{b}} = \frac{5.6 \times 10^{- 10}}{10^{- 14}} \Rightarrow K_{b} = 6.07 \times 10^{5}\text{ }\text{S}^{- 1}$
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