HA ⇌ H^{+} + A^{–}

pH = -log[H^{+}]

log[H^{+}] = -4.15

\( [H^{+}] = 7.08\times 10^{-5}M\, [A^{–}] \)

\( = 7.08\times 10^{-5}K_a=\dfrac{[H^{+}][A^{-}]}{[HA]}\)

\([H^{+}] =\dfrac{(7.08\times 10^{-5})(7.08\times 10^{-5})}{10^{-2}}\)

\( =5.0\times 10^{-7}\\ \\ \)

\( p_{K_{a}}=-log\, K_{a} \)

\( = -log\, (5.0\times 10^{-7})\)

\( =7-0.699=6.301\)

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