1. The generic metal hydroxide M(OH)2 has K sp = 7.25×10 12 . (NOTE: In this par
ID: 538783 • Letter: 1
Question
1. The generic metal hydroxide M(OH)2 has Ksp = 7.25×1012. (NOTE: In this particular problem, because of the magnitude of the Ksp and the stoichiometry of the compound, the contribution of OH from water can be ignored. However, this may not always be the case.)
a. What is the solubility of M(OH)2 in pure water?
b. What is the solubility of M(OH)2 in a 0.202 M solution of M(NO3)2?
2.) Fe2+(aq)+6CN(aq)[Fe(CN)6]4(aq) where Kf=4.21×1045.
The average human body contains 5.20 L of blood with a Fe2+ concentration of 2.70×105M . If a person ingests 5.00 mL of 25.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion? find % Fe2+
3.) For 530.0 mL of pure water, calculate the initial pH and the final pH after adding 0.020 mol of HCl.
Explanation / Answer
Q1.
M(OH)2 <-> M+2 + 2OH-
Ksp = [M+2][OH-]^2
Ksp = 7.25*10^-12
a)
solubility in pure water
Ksp = 7.25*10^-12
Ksp = (S)(2S)^2
4*S^3 = Ksp
S = (Ksp/4)^(1/3) = ( 7.25*10^-12)/4))^(1/3)
S = 121924*10^-4 M of M(OH)2
B)
in M(NO3)2 = M+2 + NO3- so [M+2] = 0.202 M
Ksp = [M+2][OH-]^2
7.25*10^-12 = 0.202 * (2S)^2
S = ((7.25*10^-12)/ (0.202) / 4 ) ^(1/2) = 0.00000299545 M = 2.995*10^-6 M of M(OH)2
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