NTA Abhyas NEET2020ChemistryIonic EquilibriumPractice
The solubility of B a S O 4 in water is 2.42 × 1 0 - 3 g L - 1 at 298 K. The value of its solubility product ( K s p ) will be (Given molar mass of B a S O 4 = 233 g m o l - 1 )
Options
- A1.08 × 1 0 - 14 m o l 2 L - 2
- B1.08 × 1 0 - 12 m o l 2 L - 2
- C1.08 × 1 0 - 10 m o l 2 L - 2
- D1.08 × 1 0 - 8 m o l 2 L - 2
Correct answer
C. 1.08 × 1 0 - 10 m o l 2 L - 2
Step-by-step solution
Solubility of BaSO 4 , s = 2.42 × 10 − 3 233 molL − 1 = 1.04 × 1 0 - 5 ( m o l L - 1 ) BaSO 4 s ⇌ Ba 2 + s aq + SO 4 2 − s sq K s p = [ B a 2 + ] [ S O 4 2 - ] = s 2 = ( 1.04 × 1 0 - 5 ) 2 = 1.08 × 1 0 - 10 m o l 2 L - 2