Final answer:
The molar solubility of Pb(OH)2 in a PbCl2 solution can be calculated using the solubility equilibrium constant, Ksp. The presence of a common ion, Pb2+, from the PbCl2 solution affects the solubility of Pb(OH)2 by reducing it. Real-world scenarios where the solubility of Pb(OH)2 is significant include the production of pigments and the treatment of lead-contaminated water.
Step-by-step explanation:
To calculate the molar solubility of Pb(OH)2 in the given PbCl2 solution, we need to use the solubility equilibrium constant, Ksp. We can set up the dissolution reaction as follows: Pb(OH)2(s) ↔ Pb2+(aq) + 2OH-(aq). The expression for Ksp is [Pb2+][OH-]2. We can write the balanced chemical equation for PbCl2 as PbCl2(s) ↔ Pb2+(aq) + 2Cl-(aq). Since there is a common ion (Pb2+) in both the solubility product and the dissolution reaction, we need to consider the common ion effect.
The common ion effect occurs when the addition of a common ion, in this case, Pb2+, suppresses the solubility of an ionic compound (Pb(OH)2) that contains the same ion. The presence of Pb2+ ions in solution from the PbCl2 reduces the solubility of Pb(OH)2. This is because the addition of Pb2+ ions shifts the equilibrium of the dissolution reaction to the left, causing more Pb(OH)2 to precipitate and decreasing its molar solubility.
Real-world scenarios where the solubility of Pb(OH)2 might be significant include the production of pigments and the treatment of lead-contaminated water. Pb(OH)2 is commonly used as a white pigment in paints, ceramics, and cosmetics. Its solubility is an important factor in determining its effectiveness as a pigment and its potential for leaching into the environment. In water treatment, the solubility of Pb(OH)2 affects the removal of lead from contaminated water sources through precipitation and filtration processes.