CH₃)₃ B – is an electron deficient compound due to inclete octate of B, thus behave as a lewis acid.

Ammonium chloride is a salt of weak base and strong acid. In this case hydrolysis constant K_h can be calculated as

K_h = $${{{K_w}} \over {{K_b}}}$$ = $${{{{10}^{ - 14}}} \over {1.77 \times {{10}^{ - 5}}}}$$ = 5.65 $$ \times $$ 10^$$-$$10

Number of equivalents of H⁺ = 20.0 × 0.050 milliequivalents

= 1.0 milliequivalents

Number of equivalents of OH^– = 2 × 30.0 × 0.10

= 6.0 milliequivalents

$$ \therefore $$ Equivalents of OH^- left after neutralization

= 6 – 1 = 5 milliequivalents

Total volume after neutralization

= 20.0 + 30.0 mL

= 50 mL

[OH^-] = $${5 \over {50}}$$ = 0.1 M

CH₃COOH $$\rightleftharpoons$$ CH₃COO^– + H⁺,   K₁ = 1.5 $$ \times $$ 10^$$-$$5

$$ \Rightarrow $$ K₁ = $${{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H^ + }} \right]} \over {\left[ {C{H_3}COOH} \right]}}$$ = 1.5 $$ \times $$ 10^$$-$$5

HCN $$\rightleftharpoons$$ CN^– + H⁺,     K₂ = 4.5 × 10^–10

K₂ = $${{\left[ {C{N^ - }} \right]\left[ {{H^ + }} \right]} \over {\left[ {HCN} \right]}}$$ = 4.5 × 10^–10

CN^$$-$$ + CH₃COOH $$\rightleftharpoons$$ HCN + CH₃COO^$$-$$

K = $${{\left[ {HCN} \right]\left[ {C{H_3}CO{O^ - }} \right]} \over {\left[ {C{N^ - }} \right]\left[ {C{H_3}COOH} \right]}}$$

$$ \Rightarrow $$ K = $${{{K_1}} \over {{K_2}}}$$ = $${{1.5 \times {{10}^{ - 5}}} \over {4.5 \times {{10}^{ - 10}}}}$$

= 3.33 × 10⁴

$$ \simeq $$ 3.0 × 10⁴