Question
Derive the necessary equation to calculate pH of weak acid.
✓
Answer
The ionisation of weak acid take place as :
HX $( aq )+ H _2 O (l) \rightleftharpoons H _3 O ^{+}( aq )+ X ^{-}( aq )$
Initial
concen-
$\begin{array}{lllll}\text { tration }( M ) & c & \text { Excess } & 0 & 0\end{array}$
If degree of ionisation $=\alpha$
Concentration at
$K_{a}=\frac{\left[H^{+}\right]\left[X^{-}\right]}{[HX]}$
$K_{a}=\frac{(c \alpha)(c \alpha)}{(c-c \alpha)}$
$K_{a}=\frac{c \alpha^2}{1-\alpha}$
$\alpha \ll 1$ hence $K_a=c \alpha^2$
$
\alpha=\sqrt{\frac{K_{a}}{c}}
$
$
pH=-\log \left[H^{+}\right]
$
To calculate $pH \quad\left[ H ^{+}\right]= c \alpha$.
$
\begin{array}{l}
{\left[H^{+}\right]=c \sqrt{\frac{K_{a}}{c}}} \\
{\left[H^{+}\right]=\sqrt{c \cdot K_{a}}}
\end{array}
$
or
$
pH=\frac{1}{2} p_{ka}-\frac{1}{2} \log c
$
HX $( aq )+ H _2 O (l) \rightleftharpoons H _3 O ^{+}( aq )+ X ^{-}( aq )$
Initial
concen-
$\begin{array}{lllll}\text { tration }( M ) & c & \text { Excess } & 0 & 0\end{array}$
If degree of ionisation $=\alpha$
Concentration at
$K_{a}=\frac{\left[H^{+}\right]\left[X^{-}\right]}{[HX]}$
$K_{a}=\frac{(c \alpha)(c \alpha)}{(c-c \alpha)}$
$K_{a}=\frac{c \alpha^2}{1-\alpha}$
$\alpha \ll 1$ hence $K_a=c \alpha^2$
$
\alpha=\sqrt{\frac{K_{a}}{c}}
$
$
pH=-\log \left[H^{+}\right]
$
To calculate $pH \quad\left[ H ^{+}\right]= c \alpha$.
$
\begin{array}{l}
{\left[H^{+}\right]=c \sqrt{\frac{K_{a}}{c}}} \\
{\left[H^{+}\right]=\sqrt{c \cdot K_{a}}}
\end{array}
$
or
$
pH=\frac{1}{2} p_{ka}-\frac{1}{2} \log c
$
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