Question

A metal hydroxide with the general formula M(OH)2 has a Ksp value of 0.00037. Calculate the molar solubility of this metal hydroxide.

Answer 1

Answer : The molar solubility of this metal hydroxide is, 0.16 M

Explanation :

The solubility equilibrium reaction will be:

$M(OH)_2\rightleftharpoons M^{2+}+2OH^-$

Let the molar solubility be 's'.

The expression for solubility constant for this reaction will be,

$K_{sp}=[M^{2+}]^3[OH^{-}]^2$

$K_{sp}=(s)^3\times (2s)^2$

$K_{sp}=4s^5$

Given:

$K_{sp}$ = solubility constant = 0.00037

Now put all the given values in the above expression, we get:

$0.00037=4s^5$

$s=0.16M$

Therefore, the molar solubility of this metal hydroxide is, 0.16 M

Answer 2

Answer:

The molar solubility of this metal hydroxide is 0.0452 mol/L

Explanation:

Step 1: Data given

General formula = M(OH)2

Ksp = 0.00037

Step 2: The balanced equation

M(OH)2 → M^2+ + 2OH-

Step 3: Calculate the molar solubility of this metal hydroxide.

If we have X of M(OH)2, it will completely react. So there will react X

For 1 mol M(OH)2 we have 1 mol M^2+ and 2 moles OH-

There will be formed X of M^2+ and 2X of OH-

Ksp = [M^2+]*[OH-]²

Ksp = 0.00037 =  X*(2X)² = 4X³

X = 0.0452

The molar solubility of this metal hydroxide is 0.0452 mol/L