The common-ion effect and solubility The solubility of a slightly soluble compou
ID: 508721 • Letter: T
Question
The common-ion effect and solubility The solubility of a slightly soluble compound can be greatly affected by the addition of a soluble compound with a common ion, that is, with one of the ions in the added soluble compound being identical to one of the ions of the slightly soluble compound. The general result of the addition of the common ion is to greatly reduce the solubility of the slightly soluble compound. In other words, the addition of the common ion results in a shift in the equilibrium ofthe slightly soluble compound Part B Calculate the molar solubility in NaOH Based on the given value of the Ksp, what is the molar solubility of Mg(OH)2 in 0.190 MNaoH? Express your answer with the appropriate units. molar solubility Value M Submit Hints My Answers Give Up Review PartExplanation / Answer
Mg(OH)2 -------------> Mg+2 (aq) + 2OH-
s 2s
s 2s +0.19
Ksp = [Mg+2][OH-]2
4.85*10^-12 = s*(0.19)^2 2s is neglisible
s= 1.34*10^-10M is molar solubility
part-A
Mg(OH)2 -------------> Mg+2 (aq) + 2OH-
s 2s
Ksp = [Mg+2][OH-]2
4.85*10^-12 = s*(2s)^2
4.85*10^-12 = 4s^3
s= 0.000106M
part-B
Mg(OH)2 -------------> Mg+2 (aq) + 2OH-
s 2s
s 2s +0.202
Ksp = [Mg+2][OH-]2
4.85*10^-12 = s*(0.202)^2 2s is neglisible
s= 1.2*10^-10M
SrF2 ---------------> Sr^+2 + 2F^-
s 2s
s+0.014 2s
Ksp = [Sr^+2][F^-]^2
4.3*10^-9 = (0.014)*s^2
s =0.000554M is molar solubility
SrF2 ---------------> Sr^+2 + 2F^-
s 2s
s 2s +0.02
Ksp = [Sr+2][F^-]^2
4.3*10^-9 = s*(0.02)^2
s = 1.075*10-6 M is molar solubility
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