Calculate the $pH$ of the solution in which $0.2 \,M\, NH _{4} Cl$ and $0.1 \,M\, NH _{3}$ are present. The  $pK _{ b }$ of ammonia solution is $4.75$

Vedclass pdf generator app on play store
Vedclass iOS app on app store

$NH _{3}+ H _{2} O \rightleftharpoons NH _{4}^{+}+ OH ^{-}$

The ionization constant of $NH _{3}$

$K_{ b }=$ antilog $\left(- pK _{ b }\right)$ i.e.

$K_{b}=10^{-4.75}=1.77 \times 10^{-5} \,M$

$NH _{3}+ H _{2} O \rightleftharpoons NH _{4}^{+}+ OH ^{-}$

Initial concentration $(M)$

$0.10$                              $0.20$         $0$

Change to reach equilibrium $(M)$

$-x$                                  $+x$            $+x$

At equilibrium $(M)$

$0.10-x$                       $0.20+x$           $x$

$K_{ b }=\left[ NH _{4}^{+}\right]\left[ OH ^{-}\right] /\left[ NH _{3}\right]$

$=(0.20+x)(x) /(0.1-x)=1.77 \times 10^{-5}$

As $K_{ b }$ is small, we can neglect $x$ in comparison to $0.1 \,M$ and $0.2\, M$. Thus,

$\left[ OH ^{-}\right]= x =0.88 \times 10^{-5}$

Therefore, $\left[ H ^{+}\right]=1.12 \times 10^{-9}$

$pH =-\log \left[ H ^{+}\right]=8.95$

Similar Questions

Heat of neutralisation of weak acid and strong base is less than the heat of neutralisation of strong acid and strong base due to

What concentration of $Ac^-$ ions will reduce $H_3O^+$ ion to $2 × 10^{-4}\ M$ in $0.40\ M$ solution of $HAc$ ? $K_a (HAc) = 1.8 × 10^{-5}$ ?

The dissociation constant of a substituted benzoic acid at $25^{\circ} \mathrm{C}$ is $1.0 \times 10^{-4}$. The $\mathrm{pH}$ of a $0.01 \ \mathrm{M}$ solution of its sodium salt is

  • [IIT 2009]

${K_a} = 1.4 \times {10^{ - 5}}$ of propanoic acid. Calculate its $pH$ of $0.1$ $M$ solution.

Calculate $\left[ {{S^{ - 2}}} \right]$ and $\left[ {H{S^{ - 2}}} \right]$ of the solution which contain$0.1$ $M$ ${H_2}S$ and $0.3$ $M$ $HCl$.

[ ${H_2}S$ of ${K_a}\left( 1 \right) = 1.0 \times {10^{ - 7}}$ and ${K_a}\left( 2 \right) = 1.3 \times {10^{ - 13}}$ ]