Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations, pH of which one of them will be equal to 1? [2018]

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Solution

Q.
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 [2018]

1 B

2 A

3 D

4 C

Ans.
(4)

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.    $M_{1} (H^{+}) = \frac{60 \times \frac{1}{10} - 40 \times \frac{1}{10}}{100} = 2 \times 10^{- 2} M i.e. pH = 1.698 \approx 1.7$

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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