Question

Calculate the molar solubility of calcium hydroxide in a solution buffered at each ph.

Answer 1

Calcium hydroxide dissociates and dissolves according to 
Ca(OH)₂(s)? Ca₂?(aq) + 2OH?(aq)
 The ions molarities in a saturated solution which satisfy the equilibrium equation
Ksp = [ Ca²?] . [OH?]₂
The solubility is constant at 25°c is (1)
Ksp = 5.02 × 10∧ - 6
s is the molar solubility of Ca(OH)₂. 
If s moles is dissolved per liter then all salt dissociates and dissolves, forming one calcium ion per salt molecule.
The calcium ion molarities is [Ca²?] = s. The hydroxide ion concentration does not change due to buffering
pH: [OH?] = 10∧ (pH-14)
∴ Ksp = s. (10∧(pH - 14))² = s. 10∧(2. pH-28)
s = Ksp/10∧ (2.PH -28) = Ksp. 10∧(28-2.PH)
When PH is at 4
s= 5.02 × 10∧(20) = 5.02 × 10∧14 mol/L
Whe PH is at 7
 s= 5.02 × 10∧-6. 10∧(14) = 5.02 × 10∧8 mol/L
When PH is at 9
s= 5.02 × 10∧-6. 10∧(10) = 5.02 × 10∧4 mol/L

Answer 2

Dissociation reaction of calcium hydroxide can be represented as follows:

Ca(OH)2 (s)      ⇄         Ca2+(aq) +         2OH-(aq)

The solubility product of Ca(OH)2 is mathematically expressed as 
                    Ksp = [Ca2+] [OH-]^2

Given: Ksp = 4.68 x 10-6

Now, we know that for aqueous system, pH + pOH = 14
and pOH = -log(OH-)
Thus, [OH-] = 10^(pH - 14)

∴ Ksp = [Ca2+] [10^(pH - 14)]^2 =  [Ca2+] [10^(2.pH - 28)]
∴ [Ca2+]    =    Ksp/10^(2.pH - 28)
                  =   Ksp 10^(28 - 2.pH)

Now, at pH = 4
[Ca2+] = Ksp 10^(28 - 2.pH)
            = (4.68 x 10-6) 10^(28 - 2X4)
            = 4.68 X 10^14 mol/dm3

At pH = 7,
[Ca2+] = Ksp 10^(28 - 2.pH)
            = (4.68 x 10-6) 10^(28 - 2X7)
            = 4.68 X 10^8 mol/dm3
At pH = 9
[Ca2+] = Ksp 10^(28 - 2.pH)
            = (4.68 x 10-6) 10^(28 - 2X9)
            = 4.68 X 10^4 mol/dm3

Answer:
Following is molar solubility of calcium hydroxide at different pH
at pH 4 = 4.68 X 10^14 mol/dm3
at pH 7 = 4.68 X 10^8 mol/dm3
at pH 9 = 4.68 X 10^4 mol/dm3