(C) The dissociation equilibrium is $HA \rightleftharpoons H^{+} + A^{-}$.For a weak acid, the concentration of $H^{+}$ is given by $[H^{+}] = \sqrt{K

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(C) The dissociation equilibrium is $HA \rightleftharpoons H^{+} + A^{-}$.For a weak acid, the concentration of $H^{+}$ is given by $[H^{+}] = \sqrt{K_{a} \times c}$.Taking negative logarithm on both sides, we get $pH = \frac{1}{2} (pK_{a} - \log c)$.Given $pK_{a} = 4$ and concentration $c = 10 \ mM = 10 \times 10^{-3} \ M = 10^{-2} \ M$.Substituting the values: $pH = \frac{1}{2} [4 - \log(10^{-2})]$.$pH = \frac{1}{2} [4 - (-2)] = \frac{1}{2} [6] = 3$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)pH of weak Acids and weak Bases

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Consider the dissociation equilibrium of the following weak acid $HA \rightleftharpoons H^{+}_{(aq)} + A^{-}_{(aq)}$. If the $pK_{a}$ of the acid is $4$, then the $pH$ of $10 \ mM$ $HA$ solution is . . . . . . . (Nearest integer)
[Given : The degree of dissociation can be neglected with respect to unity]

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pH of weak Acids and weak Bases

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