If the $\mathrm{pH}$ of $0.10 \mathrm{M}$ monobasic acid at $298 \mathrm{~K}$ is 5.0, the value of $\mathrm{p} K_a$ at the same temperature is

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Aicharya · Accepted Answer

Given: $\mathrm{pH}$ for monobasic acid $=5.0$ The reaction for dissociation of monobasic acid is given by

Since, $\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$ $ \therefore \quad\left[\mathrm{H}^{+}\right]=10^{-5} \mathrm{M} $ Thus, $\quad x=10^{-5} \mathrm{M}$ Now, $K_a=\frac{\left[\mathrm{H}^{+}\right]\left[A^{-}\right]}{[\mathrm{H} A]}$ $ \begin{array}{rlrl} \therefore & K_a & =\frac{\left[10^{-5}\right]\left[10^{-5}\right]}{[0.10]} \\ & & K_a & =10^{-9} \\ \Rightarrow & \mathrm{p} K_a & =-\log \left[K_a\right]=-\log \left[10^{-9}\right] \\ \therefore & \mathrm{p} K_a & =9 \end{array} $

If the $\mathrm{pH}$ of $0.10 \mathrm{M}$ monobasic acid at $298 \mathrm{~K}$ is 5.0, the value of…

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