Ncert Exercises & Solutions

7.64 The ionization constant of chloroacetic acid is 1.35 × 10–3. What will be the pH of 0.1M acid and its 0.1M sodium salt solution?

It is given that K_{a} for {ClCH}_{2} COOH is 1.35 \times 10^{- 3} .

\Rightarrow K_{α} = {cα}^{2}

∴ α = \sqrt{\frac{K_{α}}{c}}

= \sqrt{\frac{1.35 \times 10^{- 3}}{0.1}}

α = \sqrt{1.35 \times 10^{- 2}}

= 0.116

∴ [H^{+}] = cα = 0.1 \times 0.116

= . 0116

\Rightarrow pH = - \log [H^{+}] = 1.94

{ClCH}_{2} COONa is the salt of a weak acid i . e ., {ClCH}_{2} COOH and a strong base i . e ., NaOH .

{CICH}_{2} {COO}^{-} + H_{2} O \leftrightarrow {CICH}_{2} COOH + {OH}^{-}

K_{b} = \frac{[{CICH}_{2} COOH] [{OH}^{-}]}{[{CICH}_{2} {COO}^{-}]}

K_{b} = \frac{K_{w}}{K_{α}}

K_{b} = \frac{10^{- 14}}{1.35 \times 10^{- 3}}

= 0.740 \times 10^{- 11}

Also, K_{b} = \frac{x^{2}}{0.1}

0.740 \times 10^{- 11} = \frac{x^{2}}{0.1} (where x is the concentration of {OH}^{-} and {CICH}_{2} COOH)

0.740 \times 10^{- 11} = \frac{x^{2}}{0.1}

0.074 \times 10^{- 11} = x^{2}

\Rightarrow x^{2} = 0.74 \times 10^{- 12}

x = 0.86 \times 10^{- 6}

[{OH}^{-}] = 0.86 \times 10^{- 6}

∴ [H^{+}] = \frac{K_{w}}{o . 86 \times 10^{- 6}}

= \frac{10^{- 14}}{0.86 \times 10^{- 6}}

[H^{+}] = 1.162 \times 10^{- 8}

pH = - \log [H^{+}]

= 7.94