NTA Abhyas JEE Main2020ChemistryIonic EquilibriumPractice
An aqueous solution contains an unknown concentration of B a 2 + . When 50 mL of a 1 M solution of N a 2 S O 4 is added B a S O 4 just begins to precipitate. The final volume is 500 mL. The solubility product of B a S O 4 is 1 × 1 0 - 10 . What is the original concentration of B a 2 + ?
Options
- A1.0 × 1 0 - 10 M
- B5 × 1 0 - 9 M
- C2 × 1 0 - 9 M
- D1.1 × 1 0 - 9 M
Correct answer
D. 1.1 × 1 0 - 9 M
Step-by-step solution
B a 2 + = 1 0 - 9 M (in 500 mL solution) S O 4 2 - in 500 mL solution will be 50 × 1 = M × 500 M = 0.1 B a 2 + 450 mL + N a 2 S O 4 50 mL, 1M → B a S O 4 + 2 N a + K s p B a S O 4 = B a 2 + S O 4 2 - 1 0 - 10 = B a 2 + × 0.1 Now, we have to calculate B a 2 + in original solution (450 mL) 1 0 - 9 × 500 = 450 × M M = 1 0 - 9 × 500 450 = 10 9 × 1 0 - 9 = 1.11 × 1 0 - 9 M