Question

If solid NaCl is added to a saturated water solution of PbCl2 at 20o C, a precipitate is formed. How would this affect the value of the Ksp for [Pb2+][Cl-] in solution?
a) The Ksp increases.
b)The Ksp decreases.
c) The Ksp remains the same.
d) none of the above. ...?

Answer 1

Answer:

c) The Ksp remains the same.

Explanation:

The solubility product constant, Ksp is essentially an equilibrium constant which represents the extent to which a solute dissolves in a solvent.

In the given situation, the PbCl2 equilibrium can be represented as:

$PbCl_{2}(s)\rightleftharpoons Pb^{2+}(aq)+2Cl^{-}(aq)$

$Ksp = [Pb^{2+}][Cl^{-}]^{2}$

Addition of NaCl will introduce a common ion i.e. Cl-. This would increase its concentration and solubility. This will change the reaction quotient. However, the value of Ksp which is the 'equilibrium constant' will reamin unaffected since it is a constant at a given temperature.

Answer 2

For the question above, the answer would be C. The Ksp remains the same. The common ion effect will change the concentrations of the ions in thesolution, but the value of Ksp is CONSTANT at a constant temperature thus Ksp does not change.