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$30 \operatorname{cc}$ of $\frac{\mathrm{M}}{3} \mathrm{HCl}, 20 \operatorname{cc}$ of $\frac{\mathrm{M}}{2} \mathrm{HNO}_{3}$ and $40 \mathrm{cc}$ of $\frac{\mathrm{M}}{4} \mathrm{NaOH}$ solutions are mixed and the volume was made up to $1 \mathrm{dm}^{3}$. The $\mathrm{pH}$ of the resulting solution is
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Total milliequivalents of $\mathrm{H}^{+}$
$$
=30 \times \frac{1}{3}+20 \times \frac{1}{2}=20
$$
Total milliequivalents of $\mathrm{OH}^{-}$
$$
=40 \times \frac{1}{4}=10
$$
Milli equivalence of $\mathrm{H}^{+}$left
$$
\begin{gathered}
=20-10=10 \\
\therefore \quad\left[\mathrm{H}^{+}\right]=\frac{10}{1000} \text { g ions } / \mathrm{dm}^{3}=10^{-2} \\
\therefore \quad \mathrm{pH}=2
\end{gathered}
$$
$$
=30 \times \frac{1}{3}+20 \times \frac{1}{2}=20
$$
Total milliequivalents of $\mathrm{OH}^{-}$
$$
=40 \times \frac{1}{4}=10
$$
Milli equivalence of $\mathrm{H}^{+}$left
$$
\begin{gathered}
=20-10=10 \\
\therefore \quad\left[\mathrm{H}^{+}\right]=\frac{10}{1000} \text { g ions } / \mathrm{dm}^{3}=10^{-2} \\
\therefore \quad \mathrm{pH}=2
\end{gathered}
$$
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