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The $pH$ of $0.1\, M$ monobasic acid is $4.50$ Calculate the concentration of species $H ^{+},$ $A^{-}$ and $HA$ at equilibrium. Also, determine the value of $K_{a}$ and $pK _{a}$ of the monobasic acid.
Solution
$pH =-\log \left[ H ^{+}\right]$
Therefore, $\left[ H ^{+}\right]=10^{- pH } =10^{-4.50} $
$=3.16 \times 10^{-5} $
$\left[ H ^{+}\right]=\left[ A ^{-}\right]=3.16 \times 10^{-5}$
Thus, $K_{ a }=\left[ H ^{+}\right]\left[ A ^{-}\right] /[ HA ]$
${[HA]_{eqlbm}} = 0.1 – \left( {3.16 \times {{10}^{ – 5}}} \right) \simeq 0.1$
$K_{ a }=\left(3.16 \times 10^{-5}\right)^{2} / 0.1=1.0 \times 10^{-8}$
$p K_{ a }=-\log \left(10^{-8}\right)=8$
Alternatively, "Percent dissociation" is another useful method for measure of strength of a weak acid and is given as:
Percent dissociation
$ = {[HA]_{{\rm{dissociated }}}}/{[HA]_{{\rm{initial }}}} \times 100\% \,\,\,\,\,\,\left( {7.32} \right)$
