Value of Kspfor PbSO4 systems at 25 °C is equal to:
Correct answer: C. 1.6 x 10^-8mol2dm-6
- A. 1.6 x 10^-5 mol2dm-6
- B. 1.6 x 10^-6 mol2dm-6
- C. 1.6 x 10^-8mol2dm-6
- D. 1.6 x 10^-7 mol2dm-6
Explanation
The question is knowledge-based and it can be solved that way (from Punjab textbook page no243) The solubility of PbSO4 at 25 degrees is 0.038 PbSO^-4 ⇌Pb^+2+SO4^2− s=mass/molar mass So S =1.254×10^−4 Here, we need s^2 So S^2 =1.573×10^-8
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About Chemical Equilibrium
Reversible reactions reach dynamic equilibrium when forward and reverse rates become equal, and Le Chatelier's principle predicts the effect of changing concentration, pressure or temperature. The chapter also covers solubility product, the common ion effect, buffer action and the conditions used in Haber's process, with equilibrium shifts distinguished from changes in the equilibrium constant.
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