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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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The correct answer is:
9.0
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, $K_a=\frac{\left[\mathrm{H}^{+}\right]\left[A^{-}\right]}{[\mathrm{H} A]}$
$$
\begin{array}{rlrl}
\therefore & K_a & =\frac{\left[10^{-5}\right]\left[10^{-5}\right]}{[0.10]} \\
& & K_a & =10^{-9} \\
\Rightarrow & \mathrm{p} K_a & =-\log \left[K_a\right]=-\log \left[10^{-9}\right] \\
\therefore & \mathrm{p} K_a & =9
\end{array}
$$
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, $K_a=\frac{\left[\mathrm{H}^{+}\right]\left[A^{-}\right]}{[\mathrm{H} A]}$
$$
\begin{array}{rlrl}
\therefore & K_a & =\frac{\left[10^{-5}\right]\left[10^{-5}\right]}{[0.10]} \\
& & K_a & =10^{-9} \\
\Rightarrow & \mathrm{p} K_a & =-\log \left[K_a\right]=-\log \left[10^{-9}\right] \\
\therefore & \mathrm{p} K_a & =9
\end{array}
$$
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