The equilibrium constant for : CN ^ - + CH _ 3 COOH ⇌ HCN + CH _ 3 COO ^ - is : ( Given pK _ b for CN ^ - = 4.69and pK _ b for CH _ 3 COO ^ - = 2.25 )

Correct answer: A. 3.7 × 1 0 ^ 4 K _ eq. = [HCN] [ CH _ 3 COO ^ - ] [ CN ^ - ] [ CH _ 3 COOH] ; K _ CH _ 3 COOH = [ CH _ 3 COO ^ - ] [H^ + ] [ CH _ 3 COOH] ; K _ HCN = [H^ + ] [ CN ^ - ] [HCN] pK_ b of CN^ − = 4.69 ∴ pK_ HCN = 9.31; K _ HCN = 4.9 × 10 ^ - 10 pK _ CH _ 3 COOH = 4.75 ; K _ CH _ 3 COOH = 1.77 × 1 0 ^ - 5 ∴ K _ eq. = K _ CH _ 3 COOH K _ HCN = 1.77 × 1 0 ^ - 5 4.9 × 1 0 ^ - 1 0 = 3.7 × 1 0 ^ 4

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Chemistry
Chapter: Ionic Equilibrium

NTA Abhyas 2020 · Chemistry · Ionic Equilibrium

The equilibrium constant for : CN ^ - + CH _ 3 COOH ⇌ HCN + CH _ 3...

The equilibrium constant for :

{CN}^{-} + {CH}_{3} \text{COOH} ⇌ \text{HCN} + {CH}_{3} {COO}^{-}

is :

( Given{pK}_{b}for{CN}^{-} = 4.69and{pK}_{b}for{CH}_{3}{COO}^{-} = 2.25 )

Options

A $\text{3.7} \times 1 {0}^{4}$
B $\text{2.8} × 1 {0}^{- 5}$
C $\text{1.97} × 1 {0}^{4}$
D $\text{0.5} × 1 {0}^{- 5}$

Correct answer

A. $\text{3.7} \times 1 {0}^{4}$

Step-by-step solution

{K}_{eq.} = \dfrac{\left[\text{HCN}\right] \left[{CH}_{3} {COO}^{-}\right]}{\left[{CN}^{-}\right] \left[{CH}_{3} \text{COOH}\right]} ;

{K}_{{CH}_{3} \text{COOH}} = \dfrac{\left[{CH}_{3} {COO}^{-}\right] \left[H^{+}\right]}{\left[{CH}_{3} \text{COOH}\right]} ;

{K}_{HCN} = \dfrac{\left[H^{+}\right] \left[{CN}^{-}\right]}{\left[\text{HCN}\right]}

{\text{pK}_{b} \text{of}{\text{CN}^{−} = 4.69 ∴ {\text{pK}_{\text{HCN}} = 9.31; {K}_{\text{HCN}} = 4.9 \times {10}^{- 10}

{pK}_{{CH}_{3} \text{COOH}} = \text{4.75} ; {K}_{{CH}_{3} \text{COOH}} = \text{1.77} \times 1 {0}^{- 5}

∴ {K}_{eq.} = \dfrac{{K}_{{CH}_{3} \text{COOH}}}{{K}_{HCN}} = \dfrac{\text{1.77} \times 1 {0}^{- 5}}{\text{4.9} \times 1 {0}^{- 1 0}} = \text{3.7} \times 1 {0}^{4}

Practice Ionic Equilibrium on Quantrex Academy →

More from Ionic Equilibrium

Given is a concentrated solution of a weak electrolyte A_xB_y of concentration 'c' and… JEE Main 2026 M_3 A_2 is a sparingly soluble salt of molar mass y g mol ^ -1 and solubility x g L ^ -1… JEE Main 2026 Arrange the following resultant mixtures in increasing order of their pH values A. 10 mL… JEE Main 2026 20 mL of a solution of acetic acid required 28.4 mL of 0.1 M NaOH for its neutralization.… JEE Main 2026 The pH of a solution obtained by mixing 5 mL of 0.1 M NH_4OH solution with 250 mL of 0.1… JEE Main 2026 The first and second ionization constants of a weak dibasic acid H_2 A are 8.1 × 10^ -8… JEE Main 2026 At 25°C , 20.0 mL of 0.2 M weak monoprotic acid HX is titrated against 0.2 M NaOH. The pH… JEE Main 2026 The solubility product constants of Ag_2CrO_4 and AgBr are 32x and 4y respectively at 298… JEE Main 2026 Full Ionic Equilibrium list All NTA Abhyas JEE Main PYQs