Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations.
A. $60mL \frac{M}{10} HCl+40mL \frac{M}{10} NaOH$
B. $55mL \frac{M}{10} HCl+45mL \frac{M}{10} NaOH$
C. $75mL \frac{M}{5} HCl+25mL \frac{M}{5} NaOH$
D. $100mL \frac{M}{10} HCl+100mL \frac{M}{10} NaOH$
The pH of which of them will be equal to 1?

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answer is D.

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Detailed Solution

$pH=-log (\frac{N_{A} V_{A} {-N}_{B} V_{B}}{V_{A} {+V}_{B}}) (∴ M=N)$
$pH=-log (\frac{75 \times 0.2 - 25 \times 0.2}{75 + 25})$
${pH=-log}_{10} (\frac{15 - 5}{100})$
${pH=-log}_{10} (\frac{10}{100})$
${pH=-log}_{10}^{10^{- 1}} = 1$
(B) ${[H}^{+}]= \frac{55× \frac{1}{10} -45× \frac{1}{10}}{55+45}$
$= \frac{5.5×4.5}{100} = 10^{- 2}$
$P^{H} =2$
(A) ${[H}^{+}]= \frac{60× \frac{1}{10} -40× \frac{1}{10}}{60+40}$
$= \frac{2}{100} {=2×10}^{-2}$
$P^{H} =2-log2=1.6990$
(D) ${[H}^{+}]=100× \frac{1}{10} -100× \frac{1}{10} =0$
Neutral solution $P^{H} =7$