Question

What is the solubility of m(oh)2 in a 0.202 m solution of m(no3)2? ksp 4.45*10^-12?

Answer 1

To find the solubility of M(OH)2 in a 0.202 M solution of M(NO3)2, we can use the solubility product constant (Ksp). From the balanced equation, we can determine that the concentration of M²+ ions in the solution is 0.202 M. By calculating the concentration of OH- ions and using the given Ksp value, we can determine that the molar solubility of M(OH)2 is approximately 7.37 x 10^-14 M.

Step-by-step explanation:

To find 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).

The balanced equation for the dissociation of M(NO3)2 is:

M(NO3)2 → M²+ + 2NO3-

From the equation, we can see that 1 mole of M(NO3)2 yields 1 mole of M²+ ions, so the concentration of M²+ ions in the solution is also 0.202 M.

The Ksp expression for M(OH)2 is:

Ksp = [M²+][OH-]²

Since the stoichiometric coefficient for OH- is 2 in the balanced equation, the concentration of OH- ions will be twice that of M²+ ions.

Therefore, [OH-] = 2 * 0.202 M = 0.404 M

Now, using the Ksp expression and the given Ksp value of 4.45 x 10^-12, we can solve for the molar solubility of M(OH)2:

4.45 x 10^-12 = (0.202)(0.404)²

Solving this equation, we find that the molar solubility of M(OH)2 is approximately 7.37 x 10^-14 M.

Answer 2

M(NO₃)₂ fully separates into M²⁺ and NO₃ ²⁻ and M(OH)₂ partially separates as M²⁺ and 2OH⁻

M(NO₃)₂ → M²⁺ + 2NO₃²⁻

0.202 M 0.202 M

M(OH)₂(s) ↔ M²⁺ (aq) + 2OH⁻(aq)

I - -

C -X +X +2X

E X 2X

Ksp = [M²⁺ (aq)] [OH⁻(aq)]²

4.45 * 10∧-12 = (0.202 + X ) (2X)²

Since X is very small, (0.202 + X ) = 0.202

4.45 * 10-12 = 0.202 * 4X²

X = 2.347 × 10∧-6 M

Hence the solubility of M(OH)2 is 2.347 × 10∧-6 M