A weak acid HA has a degree of dissociation x | vedclass

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(D) For a weak acid $HA$,the dissociation equilibrium is: $HA \rightleftharpoons H^{+} + A^{-}$.At $t=0$,concentration is $C$ (or $a$).At equilibrium,

Vedclass · Accepted Answer

$\log\left(\frac{x}{1-x}\right)$

Vedclass · Answer

(D) For a weak acid $HA$,the dissociation equilibrium is: $HA \rightleftharpoons H^{+} + A^{-}$.At $t=0$,concentration is $C$ (or $a$).At equilibrium,concentrations are: $[HA] = C(1-x)$,$[H^{+}] = Cx$,$[A^{-}] = Cx$.The dissociation constant $K_{a}$ is given by: $K_{a} = \frac{[H^{+}][A^{-}]}{[HA]} = \frac{(Cx)(Cx)}{C(1-x)} = \frac{Cx^{2}}{1-x}$.Taking negative logarithm on both sides: $-\log K_{a} = -\log\left(\frac{Cx^{2}}{1-x}\right)$.$pK_{a} = -\log(Cx) - \log\left(\frac{x}{1-x}\right)$.Since $pH = -\log[H^{+}] = -\log(Cx)$,we have $pK_{a} = pH - \log\left(\frac{x}{1-x}\right)$.Therefore,$pH - pK_{a} = \log\left(\frac{x}{1-x}\right)$.

$A$ weak acid $HA$ has a degree of dissociation $x$. Which option gives the correct expression for $pH - pK_{a}$?

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