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.

$pH =-\log \left[ H ^{+}\right]$

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)$