What is the $\mathrm{pH}$ of acetic acid at equilibrium, given that acetic acid concentration is $0.1\mathrm{M}$ and it is $30\%$ dissociated at equilibrium. (log 3=0.47)

${\mathrm{CH}}_3\mathrm{COOH}\rightleftharpoons {\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{H}}^{+}$

because Dissociation occurs $=30\%$, means 100 moles of

${\mathrm{CH}}_3\mathrm{COOH}$, gives =30 moles of ${\mathrm{H}}^{+}$-ions.

Thus, 0.1 mole of${\mathrm{CH}}_3\mathrm{COOH}$ gives

$=\frac{30\times 0.1}{100}=0.03$ mole of $H^{+}$ions

$\left(\because \right.$ cc$\mathrm{COOH}$=0.1 $\mathrm{M}\backslash \mathrm{right})$

Also, $\because pH=-\log$ [Concentration]

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

p H=$-\log [0.03]$

p H=$-\log \left[\left(3\times {10}^{-2}\right)\right]$

$p H=-0.47+2$

$pH=1.53$