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 $4.75$
$NH _{3}+ H _{2} O \rightleftharpoons NH _{4}^{+}+ OH ^{-}$
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, $\left[ H ^{+}\right]=1.12 \times 10^{-9}$
$pH =-\log \left[ H ^{+}\right]=8.95$
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
${K_a} = 1.4 \times {10^{ - 5}}$ 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 ${K_a}\left( 2 \right) = 1.3 \times {10^{ - 13}}$ ]