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Chemistry Question on Law Of Chemical Equilibrium And Equilibrium Constant

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?

Answer

It is given that Ka for ClCH2COOH is 1.35 × 10-3 .
⇒ Kα = cα2
∴ α = Kαc\sqrt{\frac{K_\alpha}{c}} = 1.35×10−30.1\sqrt{\frac{1.35\times10^{-3}}{0.1}} (∴ Concentration of acid = 0.1m)
α = 1.35×10−2\sqrt{1.35 × 10^{-2}} = 0.116
∴ [H+] = cα = 0.1 × 0.116 = .0116 ⇒ pH = -log [H+] = 1.94
ClCH2COONa is the salt of a weak acid i.e., ClCH2COOH and a strong base i.e., NaOH.
CICH2COO- + H2O ↔\leftrightarrow CICH2COOH + OH-
Kh = [CICH2COOH][OH−][CICH2COO−]\frac{[CICH_2COOH][OH^-] }{[CICH_2COO^-]}
Kh = KwKα\frac{K_w}{K_\alpha}
Kh = 10−141.35×10−3\frac{10^{-14}}{1.35 × 10^{-3}} = 0.740 × 10-11
Also, Kh = x20.1\frac{x^2}{0.1} (Where x is the concentration of OH- and CICH2COOH)
0.740 ×10-11 = x20.1\frac{x^2}{0.1} = 0.074 × 10-11 = x2
⇒ x2 = 0.74 × 10-12
⇒ x = 0.86 × 10-6
[OH-] = 0.86 × 10-6
∴ [H+] = Kw0.86×10−6\frac{K_w }{0.86 × 10^{-6}}= 10−140.86×10−6\frac{10^{-14} }{0.86 × 10^{-6}}
[H+] = 1.162 × 10-8
pH = -log[H+] = 7.94