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If the ionization constant of hypochlorous acid (HOCl) is $2.5 \times 10^{-5}$, the pH of $1.0 \mathrm{M}$ of its solution is $(\log 5=0.7)$
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Verified Answer
The correct answer is:
2.3
$$
\begin{aligned}
& \text { Given, ionisation constant }\left(K_a\right)=2.5 \times 10^{-5} \\
& \text { and molarity }(c)=1.0 \mathrm{M} \\
& \because K_a=C \alpha^2 \\
& \qquad \begin{aligned}
\therefore & =\sqrt{\frac{K_a}{C}}=\sqrt{\frac{2.5 \times 10^{-5}}{1}} \\
& =0.5 \times 10^{-3} \\
\text { or } \quad \alpha & =5 \times 10^{-3}
\end{aligned}
\end{aligned}
$$
or
Also, for dissociation of $\mathrm{HOCl}$, the required relation is as follows
$$
(\because C=1 \mathrm{~m})
$$
$$
\begin{aligned}
& \therefore \quad\left[\mathrm{H}^{+}\right]=\alpha=5 \times 10^{-4} \\
& \text { and } \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\
& =-\log \left[5 \times 10^{-3}\right] \\
& =-\log 5+3 \log 10 \\
& =-0.7+3 \\
& \mathrm{pH}=2.3 \\
&
\end{aligned}
$$
\begin{aligned}
& \text { Given, ionisation constant }\left(K_a\right)=2.5 \times 10^{-5} \\
& \text { and molarity }(c)=1.0 \mathrm{M} \\
& \because K_a=C \alpha^2 \\
& \qquad \begin{aligned}
\therefore & =\sqrt{\frac{K_a}{C}}=\sqrt{\frac{2.5 \times 10^{-5}}{1}} \\
& =0.5 \times 10^{-3} \\
\text { or } \quad \alpha & =5 \times 10^{-3}
\end{aligned}
\end{aligned}
$$
or
Also, for dissociation of $\mathrm{HOCl}$, the required relation is as follows
$$
(\because C=1 \mathrm{~m})
$$
$$
\begin{aligned}
& \therefore \quad\left[\mathrm{H}^{+}\right]=\alpha=5 \times 10^{-4} \\
& \text { and } \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\
& =-\log \left[5 \times 10^{-3}\right] \\
& =-\log 5+3 \log 10 \\
& =-0.7+3 \\
& \mathrm{pH}=2.3 \\
&
\end{aligned}
$$
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