NTA Abhyas JEE Main2020ChemistryIonic EquilibriumPractice
Minimum moles of NH 3 required to be added to 1 L solution so as to dissolve 0 .1 mol of AgCl ( K sp = 1 .0 × 10 − 10 ) by the reaction is: AgCl(s) + 2 NH 3 ⇌ Ag NH 3 2 + + Cl - Given K f of Ag(NH 3 ) 2 + =10 8
Options
- A0 .5 mol
- B1 .0 mol
- C1 .1 mol
- D1 .2 mol
Correct answer
D. 1 .2 mol
Step-by-step solution
AgCl s + aq ⇌ Ag aq + + Cl aq - ; K sp Ag aq + + 2 NH 3 aq ⇌ Ag NH 3 2 aq + ; K f --------------------------------------------------------------------------------------------------- AgCl s + 2 NH 3 ⇌ Ag NH 3 2 aq + + Cl aq - K = K sp × K f C - 0 .2 M 0 .1 M 0 .1 M Equilibrium concentration K = 1 0 - 1 0 × 1 0 8 = 1 0 - 2 = Ag [ NH 3 ) 2 + Cl - NH 3 2 1 0 - 2 = 1 0 - 1 × 1 0 - 1 M - 0 . 2 2 ∴ M − 0 .2 = 1 ∴ M = 1 .2 M