NTA Abhyas JEE Main2020ChemistryIonic EquilibriumPractice
K b for CH 2 ClCOO - is 6.4 × 10 - 12 . The pH of 0.1 M CH 2 ClCOONa in water is :
Options
- A7.9
- B6.9
- C1.9
- D12.1
Correct answer
A. 7.9
Step-by-step solution
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