Final answer:
The molar solubility of Sn(OH)2 in a buffer solution can be calculated using the common ion effect. By comparing the concentration of NH3 (which can be assumed to be equal to the concentration of OH-) with the solubility product constant (Ksp) of Sn(OH)2, the molar solubility can be determined. The molar solubility of Sn(OH)2 in the buffer solution is approximately 0.005 M. Therefore, correct option is b.
Step-by-step explanation:
The molar solubility of Sn(OH)2 in a buffer solution containing equal concentrations of NH3 and NH4+ can be calculated using the common ion effect. When NH4+ is added to the solution, it will react with OH- to form NH3 and water, effectively reducing the concentration of OH-.
This reduction in OH- concentration will shift the equilibrium of Sn(OH)2 solubility to favor the dissolution of the compound.
The molar solubility of Sn(OH)2 can be calculated by comparing the concentration of OH- after the addition of NH4+ with the solubility product constant (Ksp) of Sn(OH)2. In this case, since the concentration of NH3 and NH4+ are equal, the concentration of OH- can be assumed to be equal to the concentration of NH3.
The Ksp of Sn(OH)2 is 5.45 x 10^-27. Using the Ksp expression: Ksp = [Sn2+][OH-]^2, we can substitute the concentration of OH- as [NH3] and solve for [Sn2+].
Let x be the molar solubility of Sn(OH)2 in the solution. Since 2 moles of OH- are produced for every mole of Sn(OH)2 that dissolves, the concentration of OH- is 2x. Therefore, the Ksp equation becomes: 5.45 x 10^-27 = x * (2x)^2. Solving for x gives x ≈ 0.005 M.