NTA Abhyas JEE Main2020ChemistryIonic EquilibriumPractice
Calculate H + and % dissociation of 0.1 M solution of N H 4 O H solution. The ionisation constant for N H 4 O H is k b = 2.0 × 1 0 - 5
Options
- A7.09 × 1 0 - 12 M , 3 %
- B7.09 × 1 0 - 12 M , 1.4 %
- C9.02 × 1 0 - 12 M , 2.4 %
- D9.02 × 1 0 - 12 M , 3 %
Correct answer
B. 7.09 × 1 0 - 12 M , 1.4 %
Step-by-step solution
k b = ( C α ) 2 C 1 - α = C α 2 1 - α negligible 1 - α = 1 k b = C α 2 α = k b C = 2 × 1 0 - 5 0.1 = 2 × 1 0 - 4 α = 1.41 × 1 0 - 2 α = 1.4 % O H - = C α = 0.1 × 1.4 × 1 0 - 2 = 1.4 × 1 0 - 3 M H - = K w O H - = 1 0 - 14 1.4 × 1 0 - 3 = 7.09 × 1 0 - 12 M