Following four solutions are prepared by mixing different volumes of $NaoH$ and $HCl$ of different concentrations, $pH$ of which one of them will be equal to 1 ?

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$60 mL \frac{M}{10} HCl + 40 mL \frac{M}{10} NaOH$

$55 mL \frac{M}{10} HCl + 45 mL \frac{M}{10} NaOH$

$100 mL \frac{M}{10} HCl + 100 mL \frac{M}{10} NaOH$

$75 mL \frac{M}{5} HCl + 25 mL \frac{M}{5} NaOH$

answer is B.

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Detailed Solution

$pH$ of $75 mL \frac{M}{5} HCl + 25 mL \frac{M}{5} NaOH$ will be equal to 1 .

Total volume $= 75 mL + 25 mL = 100 mL$

Number of $mmol$ of $HCl = 75 mL \times \frac{M}{5} = 15 mmol$

Number of $mmol$ of $NaOH = 25 mL \times \frac{M}{5} = 5 mmol$

$5 mmol$ of \ $NaOH$ will neutralise $5 mmol$ of $HCl$ .

$15 - 5 = 10 mmol$ of $HCl$ will remain.

$[H^{+}] = [HCl] = \frac{10 mmol}{100 mL} = 0.1 M$

$pH = - \log_{10} [H^{+}]$

$pH = - \log_{10} 0.1 M = 1$

Therefore, the correct option is (B).

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