The solubility of Pb(OH)2 in water is 6.75 ×10–6 M. Its solubility in a buffer solution of pH = 8 would be

To determine the solubility of Pb(OH)2 in a buffer solution of pH = 8, we need to consider the equilibrium reaction of Pb(OH)2 in water:

Pb(OH)2 (s) ⇌ Pb2+ (aq) + 2OH- (aq)

The solubility of Pb(OH)2 is given as 6.75 × 10–6 M in water. This means that at equilibrium, the concentration of Pb2+ ions and OH- ions in the solution will be 6.75 × 10–6 M.

In a buffer solution, the pH is controlled by the presence of a weak acid and its conjugate base. The pH of 8 indicates that the solution is slightly basic.

To calculate the solubility of Pb(OH)2 in the buffer solution, we need to consider the effect of the OH- ions on the pH. Since the solution is already basic, the concentration of OH- ions will be higher than in pure water.

The presence of OH- ions will shift the equilibrium of the reaction to the left, reducing the solubility of Pb(OH)2. This is because the OH- ions will react with the Pb2+ ions, forming more Pb(OH)2 and decreasing the concentration of Pb2+ ions in the solution.

Therefore, the solubility of Pb(OH)2 in a buffer solution of pH = 8 would be lower than 6.75 × 10–6 M. The exact solubility would depend on the specific composition of the buffer solution and the equilibrium constant of the reaction.