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

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answer is D.

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Detailed Solution

$pH = 1, so [H^{+}] = 10^{- 1}$

$For acid base mixture : N_{1} V_{1} - N_{2} V_{2} = N_{3} V_{3} (For NaOH and HCl, Normality = Molarity)$

$A. M_{1} (H^{+}) = \frac{60 \times \frac{1}{10} - 40 \times \frac{1}{10}}{100} = 2 \times 10^{- 3} M i.e. pH = 2.7$

$B. M_{2} (H^{+}) = \frac{55 \times \frac{1}{10} - 45 \times \frac{1}{10}}{100} = \frac{1}{100} = 10^{- 2} M i.e. pH = 2$

$C. M_{3} (H^{+}) = \frac{75 \times \frac{1}{5} - 25 \times \frac{1}{5}}{100} = 10^{- 1} M i.e. pH = 1$

$D. M_{4} (H^{+}) = \frac{100 \times \frac{1}{10} - 100 \times \frac{1}{10}}{200} = 0 i.e. pH = 7$

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