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The $\mathrm{pH}$ of $0.1 \mathrm{M}$ solution of monobasic acid is 2.34 . Calculate the degree of dissociation of the acid.
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1727 Upvotes
Verified Answer
The correct answer is:
$4.5 \times 10^{-2}$
$$
\mathrm{pH}=2.34, \mathrm{c}=0.1 \mathrm{M}
$$
(i) $\mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right]$
$$
\begin{gathered}
\log _{10}\left[\mathrm{H}^{+}\right]=-\mathrm{pH}=-2.34 \\
=-2-0.34-1+1 \\
=-3+0.66=\overline{3} .66
\end{gathered}
$$
$$
\therefore\left[\mathrm{H}^{+}\right]=\text {antilog } \overline{3} .66
$$
$$
=4.571 \times 10^{-3} \mathrm{~mol} \mathrm{dm}^{-3}
$$
(ii) $\left[\mathrm{H}^{+}\right]=\mathrm{c} \alpha$
$$
\therefore \alpha=\frac{\left[\mathrm{H}^{+}\right]}{\mathrm{c}}=\frac{4.571 \times 10^{-3}}{0.1}=4.571 \times 10^{-2}
$$
\mathrm{pH}=2.34, \mathrm{c}=0.1 \mathrm{M}
$$
(i) $\mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right]$
$$
\begin{gathered}
\log _{10}\left[\mathrm{H}^{+}\right]=-\mathrm{pH}=-2.34 \\
=-2-0.34-1+1 \\
=-3+0.66=\overline{3} .66
\end{gathered}
$$
$$
\therefore\left[\mathrm{H}^{+}\right]=\text {antilog } \overline{3} .66
$$
$$
=4.571 \times 10^{-3} \mathrm{~mol} \mathrm{dm}^{-3}
$$
(ii) $\left[\mathrm{H}^{+}\right]=\mathrm{c} \alpha$
$$
\therefore \alpha=\frac{\left[\mathrm{H}^{+}\right]}{\mathrm{c}}=\frac{4.571 \times 10^{-3}}{0.1}=4.571 \times 10^{-2}
$$
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