Kb for CH2ClCOO- is 6.4×10^(-12) . The pH of 0.1 M CH2ClCOONa in water is :

Correct answer: (A) 7.9. CH_2 ClCOO^(-) + H_2 O ⇌ CH_2 ClCOOH + OH^(-) OH^(−) = ch = C⋅√(K_H c) = √(K_H ·c) = √(K_w × c K_a) = √(K_b × c) = √(6.4 × 1 0^(- 1 2) × 0.1) = 8 × 1 0^(- 7) ∴ pOH = 6.1 pH = 7.9

Q49 / 102 · NTA Abhyas (JEE Main) 2020 · Chemistry · Ionic Equilibrium

K_b for CH₂ClCOO^- is $6.4 \times 10^{- 12}$ . The pH of 0.1 M CH₂ClCOONa in water is :

Correct answer
A
7.9
B
6.9
C
1.9
D
12.1

Solution

{CH}_{2} {ClCOO}^{-} + H_{2} O ⇌ {CH}_{2} ClCOOH + {OH}^{-}

[{OH}^{-}] = ch = C \cdot \sqrt{\frac{K_{H}}{c}} = \sqrt{K_{H} ·c} = \sqrt{\frac{K_{w} × c}{K_{a}}}

= \sqrt{K_{b} \times c} = \sqrt{6.4 \times 1 0^{- 12} \times 0.1}

= 8 \times 1 0^{- 7}

∴ pOH = 6.1

pH = 7.9

Hence, the correct answer is option: 'A'.

Kb for CH2ClCOO- is 6.4×10-12 . The pH of 0.1 M CH2ClCOONa in water is :

Solution

More Ionic Equilibrium Questions

K_b for CH₂ClCOO^- is $6.4 \times 10^{- 12}$ . The pH of 0.1 M CH₂ClCOONa in water is :