Question #d40e3

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Question #d40e3

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Answer 1 · 2015-04-17T18:35:17

Start with the balanced chemical equation for the equilibrium reaction that exists when hydroxylamine reacts with water

$N H_{2} O H_{(a q)} + H_{2} O_{(l)} ⇌ N H_{3} O H_{(a q)}^{+} + O H_{(a q)}^{-}$ $NH_2OH_((aq)) + H_2O_((l)) rightleftharpoons NH_3OH_((aq))^(+) + OH_((aq))^(-)$

Use the solution's pH to determine the pOHm which will allow you to calculate the concentration of the hydroxide ions

$p H_{sol} = 14 - p O H \Rightarrow p O H = 14 - p H_{sol}$ $pH_"sol" = 14 - pOH => pOH = 14 - pH_"sol"$

$p O H = 14 - 10.11 = 3.89$ $pOH = 14 - 10.11 = 3.89$

The concentration of hydroxide ions will be

$[O H^{-}] = 10^{- p O H} = 10^{3.89} = 1.29 \cdot 10^{- 4} M$ $[OH^(-)] = 10^(-pOH) = 10^3.89 = 1.29 * 10^(-4)"M"$

Since you have $1 : 1$ $1:1$ mole ratios between all the species of interest, the initial concentration of hydroxylamine will decrease by the same amount the concentration of hydroxide increased.

At equilibrium, you'll have

$[N H_{3} O H^{+}] = [O H^{-}] = 1.29 \cdot 10^{- 4} M$ $[NH_3OH^(+)] = [OH^(-)] = 1.29 * 10^(-4)"M"$

$[N H_{2} O H] = {[N H_{2} O H]}_{0} - [O H^{-}] = 0.15 - 1.29 \cdot 10^{- 4} = 0.14987 M$ $[NH_2OH] = [NH_2OH]_0 - [OH^(-)] = 0.15 - 1.29 * 10^(-4) = "0.14987 M"$

By definition, the base dissociation constant, $K_{b}$ $K_b$ , will be equal to

$K_{b} = \frac{[O H^{-}] \cdot [N H_{3} O H^{+}]}{[N H_{2} O H]}$ $K_b = ([OH^(-)] * [NH_3OH^(+)])/([NH_2OH])$

$K_{b} = \frac{1.29 \cdot 10^{- 4} \cdot 1.29 \cdot 10^{- 4}}{0.14987} = 1.1 \cdot 10^{- 9}$ $K_b = (1.29 * 10^(-4) * 1.29 * 10^(-4))/(0.14987) = color(green)(1.1 * 10^(-9))$

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Question #d40e3

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