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Consider the compounds shown below along with their solubility product constants

ID: 1055412 • Letter: C

Question

Consider the compounds shown below along with their solubility product constants MnS Ksp = 3.00e-l4 ZO_3(PO_4)_2 Ksp = 9.00e-33 Sc(OH)_3 Ksp= 8.00e-31 In order to compare relative solubility and put these compounds in rank order from least to most soluble. you first need to calculate the molar solubility of each. The molar solubility of MnS is _________ moles/L. The molar solubilityof Zn_3(PO_4)_2 is _________moles?L. The molor solubility of Sc(OH)_3 is ________ moles/L. Based on these the correct order of solubility, from least to most is SC(OH)_3, Zn_3(PO_4)_2, MnS MnS, SC(OH)_3, Zn_3(PO_4)_2 MnS, Zn_3(PO_4)_2, SC(OH)_3 Calculate the molar solubility of Nickel II hydroxide (Ni(OH)_2) = 2.0 e middot 15.

Explanation / Answer

for MnS:
MnS <-------> Mn2+ + S2-
s s
Ksp =s*s
3*10^-14 = s^2
s = 1.73*10^-7 mol/L

for Zn3(PO4)2:
Zn3(PO4)2 <-------> 3Zn2+ + 2 PO43-
3s 3s
Ksp =(3s)^3*(2s)^2
9*10^-33 = 108*S^5
S = 1.53*10^-7 mol/L

for Sc(OH)3
Sc(OH)3 <---------> Sc3+ + 3OH-
s 3s

Ksp = s (3s)^3
8*10^-31 = 27*s^4
s = 1.31*10^-8 mol/L

Answer:
Sc(OH)3 , Zn3(PO4)2 , MnS

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