Question

The pH of NH4OH solution is 10.72 in 0.015 M solution . Calculate its dissociation constant?

Answer 1

here is your answer to the question

Answer 2

Answer: The dissociation constant of the reaction is $1.79\times 10^{-5}$.

Explanation:

$pOH=14-10.72=3.28=-\log{OH^-}$

$[OH^-]=0.00052 M$

              $NH_4OH\rightleftharpoons NH_4^++OH^-$

Initial    c                   0 0

At eq'm   $c-c\alpha$          $c\alpha$    $c\alpha$

$[OH^-]=c\alpha =0.00052 M$

$\alpha =\frac{0.00052 M}{0.015 M}=0.034$

Expression of $K_d$

$K_d=\frac{[NH_4^+][OH^-]}{[NH_4OH]}=\frac{c^2\alpha ^2}{c-c\aipha}=\frac{c\alpha ^2}{(1-\alpha )}$

$K_d=\frac{(0.015 M)(0.034)^2}{1-0.034}$

$K_d=1.79\times 10^{-5}$

The dissociation constant of the reaction is $1.79\times 10^{-5}$.