Calculate the molar solubility of Mg(OH)2 (Ksp = 1.8 * 10-11 ) in
(a) pure water;
(b) 0.0862 M MgCl2.
(c) 0.0355 M KOH(aq).
The correct answer and explanation is :
To calculate the molar solubility of ( \text{Mg(OH)}2 ) in different solutions, we need to use the solubility product constant (( K{\text{sp}} )) and account for the common ion effect when applicable.
Given:
- ( K_{\text{sp}} ) for ( \text{Mg(OH)}_2 ) = ( 1.8 \times 10^{-11} )
The dissociation equation for ( \text{Mg(OH)}_2 ) in water is:
[
\text{Mg(OH)}_2 (s) \rightleftharpoons \text{Mg}^{2+} (aq) + 2 \text{OH}^-(aq)
]
The solubility product expression is:
[
K_{\text{sp}} = [\text{Mg}^{2+}][\text{OH}^-]^2
]
Let ( s ) represent the molar solubility of ( \text{Mg(OH)}_2 ) in moles per liter. From the dissociation equation, the concentrations of the ions in solution will be:
- ( [\text{Mg}^{2+}] = s )
- ( [\text{OH}^-] = 2s )
Thus, the ( K_{\text{sp}} ) expression becomes:
[
K_{\text{sp}} = s(2s)^2 = 4s^3
]
(a) Molar Solubility in Pure Water:
In pure water, there are no ions present initially, so we can directly use the equation:
[
K_{\text{sp}} = 4s^3
]
Substitute the value of ( K_{\text{sp}} ):
[
1.8 \times 10^{-11} = 4s^3
]
Solve for ( s ):
[
s^3 = \frac{1.8 \times 10^{-11}}{4} = 4.5 \times 10^{-12}
]
[
s = \sqrt[3]{4.5 \times 10^{-12}} \approx 1.65 \times 10^{-4} \, \text{M}
]
Thus, the molar solubility of ( \text{Mg(OH)}_2 ) in pure water is approximately ( 1.65 \times 10^{-4} \, \text{M} ).
(b) Molar Solubility in 0.0862 M MgCl2:
In the presence of ( \text{MgCl}_2 ), the concentration of ( \text{Mg}^{2+} ) is already 0.0862 M, so the solubility of ( \text{Mg(OH)}_2 ) will be suppressed due to the common ion effect. The equilibrium expression now becomes:
[
K_{\text{sp}} = [\text{Mg}^{2+}][\text{OH}^-]^2
]
Since ( [\text{Mg}^{2+}] ) is initially 0.0862 M, let the additional solubility of ( \text{Mg(OH)}_2 ) be ( s ), then:
[
K_{\text{sp}} = (0.0862 + s)(2s)^2
]
For small ( s ), we approximate ( 0.0862 + s \approx 0.0862 ), so:
[
K_{\text{sp}} = (0.0862)(4s^2)
]
Substitute the value of ( K_{\text{sp}} ):
[
1.8 \times 10^{-11} = 0.0862 \times 4s^2
]
Solve for ( s^2 ):
[
s^2 = \frac{1.8 \times 10^{-11}}{0.3448} \approx 5.22 \times 10^{-11}
]
[
s = \sqrt{5.22 \times 10^{-11}} \approx 7.22 \times 10^{-6} \, \text{M}
]
Thus, the molar solubility of ( \text{Mg(OH)}_2 ) in 0.0862 M ( \text{MgCl}_2 ) is approximately ( 7.22 \times 10^{-6} \, \text{M} ).
(c) Molar Solubility in 0.0355 M KOH(aq):
In the presence of ( \text{KOH} ), the concentration of ( \text{OH}^- ) is 0.0355 M. This will suppress the solubility of ( \text{Mg(OH)}_2 ) due to the common ion effect. The equilibrium expression is:
[
K_{\text{sp}} = [\text{Mg}^{2+}][\text{OH}^-]^2
]
Since the concentration of ( \text{OH}^- ) is already 0.0355 M, the additional concentration of ( \text{OH}^- ) from ( \text{Mg(OH)}_2 ) is negligible. Thus:
[
K_{\text{sp}} = [\text{Mg}^{2+}](0.0355 + 2s)^2
]
For small ( s ), ( 0.0355 + 2s \approx 0.0355 ), so:
[
K_{\text{sp}} = \text{Mg}^{2+}^2
]
Substitute the value of ( K_{\text{sp}} ):
[
1.8 \times 10^{-11} = [\text{Mg}^{2+}] \times (0.0355)^2
]
Solve for ( [\text{Mg}^{2+}] ):
[
[\text{Mg}^{2+}] = \frac{1.8 \times 10^{-11}}{(0.0355)^2} \approx 1.43 \times 10^{-7} \, \text{M}
]
Thus, the molar solubility of ( \text{Mg(OH)}_2 ) in 0.0355 M ( \text{KOH} ) is approximately ( 1.43 \times 10^{-7} \, \text{M} ).
Summary of Results:
- (a) In pure water: ( 1.65 \times 10^{-4} \, \text{M} )
- (b) In 0.0862 M MgCl2: ( 7.22 \times 10^{-6} \, \text{M} )
- (c) In 0.0355 M KOH: ( 1.43 \times 10^{-7} \, \text{M} )
The presence of common ions in each case suppresses the solubility of ( \text{Mg(OH)}_2 ), with the highest solubility in pure water and the lowest in the presence of both ( \text{Mg}^{2+} ) and ( \text{OH}^- ).