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The following solutions were prepared by mixing different volumes of NaOH of HCl 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

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

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

pH of which one of them will be equal to 1?

(a) (iv)

(b) (i)

(c) (ii)

(d) (iii)

1 Answer

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Best answer

(d) (iii)

No of moles of HCl = 0.2 × 75 × 10-3 = 15 × 10-3

No of moles of NaOH = 0.2 × 25 × 10-3 = 5 × 10-3 

No of moles of HCl after mixing = 15 × 10-3 – 5 × 10-3

= 10 × 10-3

for (iii) solution, PH of 0.1MHCl = -log10 (0.1) = 1.

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