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)$
For a concentrated solution of a weak electrolyte ( $K _{ eq }=$ equilibrium constant) $A _2 B _3$ of concentration ' $c$ ', the degree of dissociation " $\alpha$ ' is
Calculate $\left[ {{S^{ - 2}}} \right]$ and $\left[ {H{S^{ - 2}}} \right]$ of the solution which contain$0.1$ $M$ ${H_2}S$ and $0.3$ $M$ $HCl$.
[ ${H_2}S$ of ${K_a}\left( 1 \right) = 1.0 \times {10^{ - 7}}$ and ${K_a}\left( 2 \right) = 1.3 \times {10^{ - 13}}$ ]
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.
Derive ${K_a} \times {K_b} = {K_w}$ equation.
${K_a} = 1.4 \times {10^{ - 5}}$ of propanoic acid. Calculate its $pH$ of $0.1$ $M$ solution.