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The $\mathrm{pH}$ of a $10^{-8}$ molar solution of $\mathrm{HCl}$ in water is
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1454 Upvotes
Verified Answer
The correct answer is:
between 6 and 7
$\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log 10^{-8}=8$
It is not possible for acid, so it is $\left[\mathrm{H}^{+}\right]$, the $\left[\mathrm{H}^{+}\right]$ of water is also added. Total $\left[\mathrm{H}^{+}\right]$ in solution
$$
\begin{aligned}
&=\left[\mathrm{H}^{+}\right] \text {of } \mathrm{HCl}+\left[\mathrm{H}^{+}\right] \text {of water } \\
&=\left(1 \times 10^{-8}+1 \times 10^{-7}\right) \mathrm{M} \\
&=(1+10) \times 10^{-8}=11 \times 10^{-8} \mathrm{M} \\
\therefore \quad \mathrm{pH} &=-\log \left[\mathrm{H}^{+}\right] \\
&=-\log 11 \times 10^{-8} \\
&=-\log 11+8 \log 10 \\
&=-1.0414+8=6.9586
\end{aligned}
$$
It is not possible for acid, so it is $\left[\mathrm{H}^{+}\right]$, the $\left[\mathrm{H}^{+}\right]$ of water is also added. Total $\left[\mathrm{H}^{+}\right]$ in solution
$$
\begin{aligned}
&=\left[\mathrm{H}^{+}\right] \text {of } \mathrm{HCl}+\left[\mathrm{H}^{+}\right] \text {of water } \\
&=\left(1 \times 10^{-8}+1 \times 10^{-7}\right) \mathrm{M} \\
&=(1+10) \times 10^{-8}=11 \times 10^{-8} \mathrm{M} \\
\therefore \quad \mathrm{pH} &=-\log \left[\mathrm{H}^{+}\right] \\
&=-\log 11 \times 10^{-8} \\
&=-\log 11+8 \log 10 \\
&=-1.0414+8=6.9586
\end{aligned}
$$
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