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
Explain a general step-wise approach to evaluate the $pH$ of the weak electrolyte.
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The degree of ionization of a $0.1 \,M$ bromoacetic acid solution is $0.132$ Calculate the $pH$ of the solution and the $p K_{ a }$ of bromoacetic acid.
The degree of ionization of a $0.1 \,M$ bromoacetic acid solution is $0.132$ Calculate the $pH$ of the solution and the $p K_{ a }$ of bromoacetic acid.
Given
$(i)$ $\begin{gathered}
HCN\left( {aq} \right) + {H_2}O\left( l \right) \rightleftharpoons {H_3}{O^ + }\left( {aq} \right) + C{N^ - }\left( {aq} \right) \hfill \\
{K_a} = 6.2 \times {10^{ - 10}} \hfill \\
\end{gathered} $
$(ii)$ $\begin{gathered}
C{N^ - }\left( {aq} \right) + {H_2}O\left( l \right) \rightleftharpoons HCN\left( {aq} \right) + O{H^ - }\left( {aq} \right) \hfill \\
{K_b} = 1.6 \times {10^{ - 5}} \hfill \\
\end{gathered} $
These equilibria show the following order of the relative base strength
Given
$(i)$ $\begin{gathered}
HCN\left( {aq} \right) + {H_2}O\left( l \right) \rightleftharpoons {H_3}{O^ + }\left( {aq} \right) + C{N^ - }\left( {aq} \right) \hfill \\
{K_a} = 6.2 \times {10^{ - 10}} \hfill \\
\end{gathered} $
$(ii)$ $\begin{gathered}
C{N^ - }\left( {aq} \right) + {H_2}O\left( l \right) \rightleftharpoons HCN\left( {aq} \right) + O{H^ - }\left( {aq} \right) \hfill \\
{K_b} = 1.6 \times {10^{ - 5}} \hfill \\
\end{gathered} $
These equilibria show the following order of the relative base strength
- [AIEEE 2012]
The ionization constant of $HF$, $HCOOH$ and $HCN$ at $298\, K$ are $6.8 \times 10^{-4}, 1.8 \times 10^{-4}$ and $4.8 \times 10^{-9}$ respectively. Calculate the ionization constants of the corresponding conjugate base.
The ionization constant of $HF$, $HCOOH$ and $HCN$ at $298\, K$ are $6.8 \times 10^{-4}, 1.8 \times 10^{-4}$ and $4.8 \times 10^{-9}$ respectively. Calculate the ionization constants of the corresponding conjugate base.
Given the two concentration of $HCN (K_a = 10^{-9})$ are $0.1\,M$ and $0.001\,M$ respectively. What will be the ratio of degree of dissociation ?
Given the two concentration of $HCN (K_a = 10^{-9})$ are $0.1\,M$ and $0.001\,M$ respectively. What will be the ratio of degree of dissociation ?