NTA Abhyas JEE Main2020ChemistryIonic EquilibriumPractice
For Ag 2 CO 3 , K sp = 6.2 × 10 − 12 . For AgCl , K sp = 2.8 × 10 − 10 . Solid Ag 2 CO 3 and solid AgCl are added to a beaker containing Na 2 CO 3 ( aq ) . Under these conditions the [ CO 3 2 − ] = 1.00   M . Calculate the [ Cl − ] in solution when equilibrium is established.
Options
- A1 . 1 × 1 0 - 4
- B1 . 2 6 × 1 0 - 8
- C0.15
- D2 . 8 × 1 0 - 6
Correct answer
A. 1 . 1 × 1 0 - 4
Step-by-step solution
[Ag + ] to be calculated from K sp (Ag 2 CO 3 ) with lower K sp but with greater solubility Ag + = K sp Ag 2 CO 3 CO 3 2 - = 6 . 2 × 1 0 - 1 2 1 = 2.49 x 10 - 6 M Cl - = K sp Ag Cl Ag + = 2 . 8 × 1 0 - 1 0 2 . 4 9 × 1 0 - 6 = 1.12 x 10 - 4 M