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erastova [34]
3 years ago
8

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

Chemistry
2 answers:
torisob [31]3 years ago
7 0
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

sashaice [31]3 years ago
3 0
Dissociation reaction of calcium hydroxide can be represented as follows:

<span>Ca(OH)2 (s)      ⇄         Ca2+(aq) +         2OH-(aq)
</span>
The solubility product of Ca(OH)2 is mathematically expressed as 
                    Ksp = [Ca2+] [OH-]^2

Given: Ksp = <span>4.68 x 10-6

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

</span>∴ 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


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