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 $4.75$
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, $\left[ H ^{+}\right]=1.12 \times 10^{-9}$
$pH =-\log \left[ H ^{+}\right]=8.95$
Similar Questions
The ionization constant of acetic acid is $1.74 \times 10^{-5}$. Calculate the degree of dissociation of acetic acid in its $0.05\, M$ solution. Calculate the concentration of acetate ion in the solution and its $pH$.
The ionization constant of acetic acid is $1.74 \times 10^{-5}$. Calculate the degree of dissociation of acetic acid in its $0.05\, M$ solution. Calculate the concentration of acetate ion in the solution and its $pH$.
What is the $pH$ of solution of $7$ $gm$ $N{H_4}OH$ per $500$ $mL$ ? ( ${K_b}$ of $N{H_4}OH 1.8 \times {10^{ - 5}}$, Molecular moles of $N{H_4}OH$ is $35\,g\,mo{l^{ - 1}}$ )
What is the $pH$ of solution of $7$ $gm$ $N{H_4}OH$ per $500$ $mL$ ? ( ${K_b}$ of $N{H_4}OH 1.8 \times {10^{ - 5}}$, Molecular moles of $N{H_4}OH$ is $35\,g\,mo{l^{ - 1}}$ )
When $CO_2$ dissolves in water, the following equilibrium is established
$C{O_2} + 2{H_2}O\, \rightleftharpoons {H_3}{O^ + } + HCO_3^ - $
for which the equilibrium constant is $3.8 \times 10^{-7}$ and $pH = 6.0$. The ratio of $[HCO_3^- ]$ to $[CO_2]$ would be :-
When $CO_2$ dissolves in water, the following equilibrium is established
$C{O_2} + 2{H_2}O\, \rightleftharpoons {H_3}{O^ + } + HCO_3^ - $
for which the equilibrium constant is $3.8 \times 10^{-7}$ and $pH = 6.0$. The ratio of $[HCO_3^- ]$ to $[CO_2]$ would be :-
Sulphurous acid $\left( H _{2} SO _{3}\right)$ has $Ka _{1}=1.7 \times 10^{-2}$ and $Ka _{2}=6.4 \times 10^{-8} .$ The $pH$ of $0.588 \,M\, H _{2} SO _{3}$ is ..... . (Round off to the Nearest Integer)
Sulphurous acid $\left( H _{2} SO _{3}\right)$ has $Ka _{1}=1.7 \times 10^{-2}$ and $Ka _{2}=6.4 \times 10^{-8} .$ The $pH$ of $0.588 \,M\, H _{2} SO _{3}$ is ..... . (Round off to the Nearest Integer)
- [JEE MAIN 2021]
Derive the equation of relation between weak base ionization constant ${K_b}$ and its conjugate acid ionization constant ${K_a}$
Derive the equation of relation between weak base ionization constant ${K_b}$ and its conjugate acid ionization constant ${K_a}$