Q39.The solubility product of BaSO4 is 1 × 10−10 at 298 K. The solubility of BaSO4 in 0. 1M K2 SO4(aq) solution is ------ ×10−9 g L−1 (nearest integer). Given: Molar mass of BaSO4 is 233 g mol−1
What This Question Tests
This question requires calculating the solubility of BaSO₄ in the presence of a common ion and expressing it in g L⁻¹.
Concepts Tested
Formulas Used
K_sp = [Ba²⁺][SO₄²⁻]
Molar mass of BaSO₄
📚 NCERT Sections This Tests
1.27 — If The Solubility Product Of Cus Is 6 × 10–16, Calculate The Maximum Molarity Of
Chemistry Class 11 · Chapter 1
1.27 If the solubility product of CuS is 6 × 10–16, calculate the maximum molarity of CuS in aqueous solution.
2.8 — The Conductivity Of 0.20 M Solution Of Kcl At 298 K Is 0.0248 S Cm–1. Calculate
Chemistry Class 11 · Chapter 2
2.8 The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm–1. Calculate its molar conductivity.
2.9 — The Resistance Of A Conductivity Cell Containing 0.001M Kcl Solution At 298
Chemistry Class 11 · Chapter 2
2.9 The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 W. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10–3 S cm–1. 59 Electrochemistry Reprint 2025-26
📋 Question Details
- Chapter
- Ionic Equilibrium
- Topic
- Solubility product, Common ion effect
- Year
- 2023
- Shift
- 08 Apr Shift 2
- Q Number
- Q39
- Type
- Numerical
- NCERT Ref
- Class 11 Chemistry Ch 7: Equilibrium
More from this Chapter
Q95.Species acting as both Bronsted acid and base is (1) (HSO4)−1 (2) Na2CO3 (3) NH3 (4) OH−1
Q96.Let the solubility of an aqueous solution of Mg(OH)2 be x then its ksp is (1) 4x3 (2) 108x5 (3) 27x4 (4) 9x
Q97.The solubility of Mg(OH)2 is S moles/litre. The solubility product under the same condition is (1) 4 S3 (2) 3S 4 (3) 4 S2 (4) S 3
Q98.How do we differentiate between Fe3+ and Cr3+ in group III? (1) by taking excess of NH4OH solution (2) by increasing NH4+ ion concentration (3) by decreasing OH− ion concentration (4) both (b) and (c)