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Chemistry · EQUILIBRIUM
हिन्दीThe \( \text{pH} \) of a \( 0.1\text{ M} \) monobasic acid \( \text{HA} \) is found to be \( 4.50 \). If \( 10^{-4.50} = 3.16 \times 10^{-5} \), what is the \( \text{p}K_a \) value of this weak acid?
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Correct answer · Option B
8.0
Explanation
\( [\text{H}^+] = 3.16 \times 10^{-5}\text{ M} \). \( K_a = \frac{[\text{H}^+]^2}{c} = \frac{(3.16 \times 10^{-5})^2}{0.1} = \frac{1.0 \times 10^{-9}}{0.1} = 1.0 \times 10^{-8} \). Thus, \( \text{p}K_a = -\log(10^{-8}) = 8.0 \).
