Question

What is the solubility of m(oh)2 in a 0.202 m solution of m(no3)2? ksp = 9.05×10−18?

Answer 1

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.

Answer 2

M(NO3)2 ==> [M2+] + 2 [NO3-] 0.202 M ==> 0.202 M M(OH)2 ==> [M2+] + 2[OH-] 5.05*10^-18 ===> s + [2s]^2 5.05*10^-18 ===> 0.202 + [2s]^2 5.05*10^-18 = 0.202 * 4s^2 4s^2 = 25*10^-18 s^2 = 6.25*10^-18 s = 2.5*10^-9 So, the solubility is 2.5*10^-9