Ncert Exercises & Solutions

7.66 Calculate the pH of the resultant mixtures:

a) 10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl

b) 10 mL of 0.01M H2SO4 + 10 mL of 0.01M Ca(OH)2

c) 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH

(a) Moles of H_{3} O^{+} = \frac{25 \times 0.1}{1000} = . 0025 mol

moles of {OH}^{-} = \frac{10 \times 0.2 \times 2}{1000} = . 0040 mol

Thus, excess of {CH}^{-} = . 0015 mol

[{OH}^{-}] = \frac{. 0015}{35 \times 10^{- 3}} mol / L = . 0428

pOH = - \log [OH]

= 1.36

pH = 14 - 1.36

= 12.63

(b) Moles of H_{3} O^{+} = \frac{2 \times 10 \times . 01}{1000} = . 0002 mol

Moles of {OH}^{-} = \frac{2 \times 10 \times . 01}{1000} = . 0002 mol

Since there is neither an excess of H_{3} O^{+} or {OH}^{-}

(c) Moles of H_{2} O^{+} = \frac{2 \times 10 \times 0.1}{1000} = . 002 mol

Moles of {OH}^{-} = \frac{10 \times 0.1}{1000} = 0.001 mol

Excess of H_{3} O^{+} = . 001 mol

Thus, [H_{3} O^{+}] = \frac{. 001}{20 \times 10^{- 3}} = \frac{10^{- 3}}{20 \times 10^{- 3}} = 0.5

∴ pH = - \log (0.05)

= 1.30

The solution is neutral . Hence, pH = 7 .