Final answer:
The solubility of M(OH)2 in a 0.202 M solution of M(NO3)2 is approximately 0.202 M.
Step-by-step explanation:
To determine the solubility of M(OH)2 in a 0.202 M solution of M(NO3)2, we need to use the solubility product constant (Ksp) and solve for the molar solubility (M) of M(OH)2.
The equation for the dissolution of M(OH)2 is:
M(OH)2 → M²⁺ + 2OH⁻
Using the given Ksp value of 9.05 x 10-18, we can set up the expression for Ksp:
Ksp = [M²⁺] * [OH⁻]2
Since the molarity of M(NO3)2 is 0.202 M, the concentration of M²⁺ ions is also 0.202 M.
Let's assume the molar solubility of M(OH)2 is M. Therefore, the concentration of OH⁻ ions would be 2M (because of the 2:1 stoichiometric ratio).
Substituting the values into the expression for Ksp and solving for M, we get:
Ksp = (0.202)(2M)2
9.05 x 10-18 = 0.404 M²
Taking the square root of both sides:
M = 0.202 M
Therefore, the solubility of M(OH)2 in a 0.202 M solution of M(NO3)2 is approximately 0.202 M.