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If $20 \mathrm{~mL}$ of an acidic solution of $\mathrm{pH} 3$ is diluted to $100 \mathrm{~mL}$, the $\mathrm{H}^{+}$ ion concentration will be
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$2 \times 10^{-4} \mathrm{M}$
$\mathrm{pH}=3$ (given)
$\begin{aligned} \therefore\left[\mathrm{H}^{+}\right]=1 \times 10^{-3} \mathrm{M} &=1 \times 10^{-3} \mathrm{~N} \\ N_{1} V_{1} &=N_{2} V_{2} \\ 1 \times 10^{-3} \times 20 &=N_{2} \times 100 \\ N_{2} &=2 \times 10^{-4} \\ \therefore \quad\left[\mathrm{H}^{+}\right] &=2 \times 10^{-4} \mathrm{M} \end{aligned}$
$\begin{aligned} \therefore\left[\mathrm{H}^{+}\right]=1 \times 10^{-3} \mathrm{M} &=1 \times 10^{-3} \mathrm{~N} \\ N_{1} V_{1} &=N_{2} V_{2} \\ 1 \times 10^{-3} \times 20 &=N_{2} \times 100 \\ N_{2} &=2 \times 10^{-4} \\ \therefore \quad\left[\mathrm{H}^{+}\right] &=2 \times 10^{-4} \mathrm{M} \end{aligned}$
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