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  • Determine the degree of dissociation of 0.005   at 298K in a solution of pH = 10<div class='qna-opt

Determine the degree of dissociation of 0.005 NH_{3}  at 298K in a solution of pH = 10

Option: 1

4


Option: 2

5


Option: 3

1


Option: 4

2


Answers (1)

best_answer

Given,

pH = 10 \therefore \therefore [H^{+}]=10^{-10}

\begin{array}{ccc}{\mathrm{NH_{4}OH}} & {\rightleftharpoons} & {\mathrm{NH_{4}}^{+}+\mathrm{OH}^{-}} \\ {1} &\ {0} & {0} \\ {1-\alpha} & {\alpha} & {\alpha}\end{array}

As we know, at 298 K, [H^{+}] [OH^{-}]=10^{-14}

\Rightarrow [OH^{-}]=\frac{10^{-14}}{10^{-10}} = 10^{-4}

c\alpha = 10^{-4}

\alpha = \frac{10^{-4}}{0.005} = 0.02

\alpha = 2\%

 

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