Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations. (i) 60 mL M 10 HCl + 40 mL M 10 NaOH (ii) 55 mL M 10 HCl + 45 mL M 10 NaOH (iii) 75 mL M 5 HC

Question

Following solutions were prepared by mixing different volumes of $NaOH$ and $HCl$ of different concentrations.

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

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

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

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

The $pH$ of which one of them will be equal to $1$ ?NEET 2018

TestHub · Accepted Answer

Equivalent of acid = N 1 V 1 Equivalent of base = N 2 V 2 As N 1 V 1 > N 2 V 2 , acid is left at the end of the reaction. The 'n' factor of NaOH and HCl are 1 , 1 . So, N = M for both. N final solution = H + = N 1 V 1 - N 2 V 2 V 1 + V 2 = 1 5 × 75 - 1 5 × 25 75 + 25 = 1 10 = 10 − 1

Chemistry - Ionic Equilibrium Question with Solution | TestHub

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