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).<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the molar solubility of ( \\text{Mg(OH)}2 ) in different solutions, we need to use the solubility product constant (( K<\/em>{\\text{sp}} )) and account for the common ion effect when applicable.<\/p>\n\n\n\n The dissociation equation for ( \\text{Mg(OH)}_2 ) in water is:<\/p>\n\n\n\n [ The solubility product expression is:<\/p>\n\n\n\n [ 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:<\/p>\n\n\n\n Thus, the ( K_{\\text{sp}} ) expression becomes:<\/p>\n\n\n\n [ In pure water, there are no ions present initially, so we can directly use the equation:<\/p>\n\n\n\n [ Substitute the value of ( K_{\\text{sp}} ):<\/p>\n\n\n\n [ Solve for ( s ):<\/p>\n\n\n\n [ [ Thus, the molar solubility of ( \\text{Mg(OH)}_2 ) in pure water is approximately ( 1.65 \\times 10^{-4} \\, \\text{M} )<\/strong>.<\/p>\n\n\n\n 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:<\/p>\n\n\n\n [ Since ( [\\text{Mg}^{2+}] ) is initially 0.0862 M, let the additional solubility of ( \\text{Mg(OH)}_2 ) be ( s ), then:<\/p>\n\n\n\n [ For small ( s ), we approximate ( 0.0862 + s \\approx 0.0862 ), so:<\/p>\n\n\n\n [ Substitute the value of ( K_{\\text{sp}} ):<\/p>\n\n\n\n [ Solve for ( s^2 ):<\/p>\n\n\n\n [ [ 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} )<\/strong>.<\/p>\n\n\n\n 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:<\/p>\n\n\n\n [ 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:<\/p>\n\n\n\n [ For small ( s ), ( 0.0355 + 2s \\approx 0.0355 ), so:<\/p>\n\n\n\nGiven:<\/h3>\n\n\n\n
\n
\\text{Mg(OH)}_2 (s) \\rightleftharpoons \\text{Mg}^{2+} (aq) + 2 \\text{OH}^-(aq)
]<\/p>\n\n\n\n
K_{\\text{sp}} = [\\text{Mg}^{2+}][\\text{OH}^-]^2
]<\/p>\n\n\n\n\n
K_{\\text{sp}} = s(2s)^2 = 4s^3
]<\/p>\n\n\n\n(a) Molar Solubility in Pure Water<\/strong>:<\/h3>\n\n\n\n
K_{\\text{sp}} = 4s^3
]<\/p>\n\n\n\n
1.8 \\times 10^{-11} = 4s^3
]<\/p>\n\n\n\n
s^3 = \\frac{1.8 \\times 10^{-11}}{4} = 4.5 \\times 10^{-12}
]<\/p>\n\n\n\n
s = \\sqrt[3]{4.5 \\times 10^{-12}} \\approx 1.65 \\times 10^{-4} \\, \\text{M}
]<\/p>\n\n\n\n(b) Molar Solubility in 0.0862 M MgCl2<\/strong>:<\/h3>\n\n\n\n
K_{\\text{sp}} = [\\text{Mg}^{2+}][\\text{OH}^-]^2
]<\/p>\n\n\n\n
K_{\\text{sp}} = (0.0862 + s)(2s)^2
]<\/p>\n\n\n\n
K_{\\text{sp}} = (0.0862)(4s^2)
]<\/p>\n\n\n\n
1.8 \\times 10^{-11} = 0.0862 \\times 4s^2
]<\/p>\n\n\n\n
s^2 = \\frac{1.8 \\times 10^{-11}}{0.3448} \\approx 5.22 \\times 10^{-11}
]<\/p>\n\n\n\n
s = \\sqrt{5.22 \\times 10^{-11}} \\approx 7.22 \\times 10^{-6} \\, \\text{M}
]<\/p>\n\n\n\n(c) Molar Solubility in 0.0355 M KOH(aq)<\/strong>:<\/h3>\n\n\n\n
K_{\\text{sp}} = [\\text{Mg}^{2+}][\\text{OH}^-]^2
]<\/p>\n\n\n\n
K_{\\text{sp}} = [\\text{Mg}^{2+}](0.0355 + 2s)^2
]<\/p>\n\n\n\n