Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations. (i) 60 mLM/10 HCl+40mLM/10NaOH (ii) 55…

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

?

MARKS Content Team · Accepted Answer

Correct answer: (D) (iii). Equivalent of acid =N_1V_1 Equivalent of base =N_2V_2 As N_1V_1>N_2V_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_1V_1-N_2V_2)/(V_1+V_2) =1/5×75-1/5×2575+25 =1/10= 10^(−1)

Solution

Relevant Questions

Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations.(i) 60 mLM10 HCl+40mLM10NaOH(ii) 55 mLM10HCl+45mLM10NaOH(iii) 75 mLM5HCl+25mLM5NaOH(iv) 100 mLM10HCl+100 mLM10NaOHThe pH of which one of them will be equal to 1?

Solution

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