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, $\left[ H ^{+}\right]=1.12 \times 10^{-9}$
$pH =-\log \left[ H ^{+}\right]=8.95$
Calculate the $pH$ of a $0.10 \,M$ ammonia solution. Calculate the pH after $50.0 \,mL$ of this solution is treated with $25.0 \,mL$ of $0.10 \,M$ $HCl$. The dissociation constant of ammonia, $K_{b}=1.77 \times 10^{-5}$
At $298$ $K$ temperature, the ${K_b}$ of ${\left( {C{H_3}} \right)_2}NH$ is $5.4 \times {10^{ - 4}}$ $0.25$ $M$ solution.
Derive the equation of ionization constant $({K_b})$ of weak base.
If degree of ionisation is $0.01$ of decimolar solution of weak acid $HA$ then $pKa$ of acid is
It has been found that the $pH$ of a $0.01$ $M$ solution of an organic acid is $4.15 .$ Calculate the concentration of the anion, the ionization constant of the acid and its $p{K_a}$