Respuesta :
Answer : The molar solubility of this metal hydroxide is, 0.16 M
Explanation :
The solubility equilibrium reaction will be:
[tex]M(OH)_2\rightleftharpoons M^{2+}+2OH^-[/tex]
Let the molar solubility be 's'.
The expression for solubility constant for this reaction will be,
[tex]K_{sp}=[M^{2+}]^3[OH^{-}]^2[/tex]
[tex]K_{sp}=(s)^3\times (2s)^2[/tex]
[tex]K_{sp}=4s^5[/tex]
Given:
[tex]K_{sp}[/tex] = solubility constant = 0.00037
Now put all the given values in the above expression, we get:
[tex]0.00037=4s^5[/tex]
[tex]s=0.16M[/tex]
Therefore, the molar solubility of this metal hydroxide is, 0.16 M
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
