In: Chemistry
Rank the following five salts in order of decreasing solubility,
in terms of mass per unit volume.
(The most soluble gets rank 1, the least soluble gets rank 5.)
Hg2CO3
(Ksp =
9.00×10-15; Note: The cation is
Hg22+)
Hg2Cl2
(Ksp =
1.10×10-18; Note: The cation is
Hg22+)
Ca3(PO4)2
(Ksp =
1.30×10-32)
Ag3PO4
(Ksp =
1.80×10-18)
Ca5(PO4)3OH
(Ksp =
6.80×10-37)
This is the 2nd time I've posted this question because previously people just listed it in order of the KSP that's given in the question. That is INCORRECT. I need to convert it to mass/volume. I've tried it 4 times and I keep getting the answer wrong.
1 ) solubility of Hg2CO3 = Ksp^1/2 = (9*10^(-15))^(1/2) =
9.5*10^-8
solubility im mass/volume = 9.5*10^-8*molarmass of Hg2CO3
Molar mass of Hg2CO3 is 461.1889 g/mol
solubility im mass/volume = 9.5*10^-8*461.1889 = 4.38*10^-5 g/L
2) solubility of Hg2Cl2 = (Ksp^1/4)^(1/3) =
((1.1*10^(-18))/4)^(1/3)
= 6.5*10^-7
solubility im mass/volume = 6.5*10^-7*molarmass of Hg2Cl2
Molar mass of Hg2CL2 is 472.09 g/mol
solubility im mass/volume = 6.5*10^-7*472.09
= 0.0003068585 = 3.07*10^-4 g/L
3)
solubility of Ca3(PO4)2 = (Ksp^1/108)^(1/5) = ((1.3*10^(-32))/108)^(1/5)
= 1.64*10^-7
solubility im mass/volume = 1.64*10^-7 *molarmass of Ca3(PO4)2
Molar mass of Ca3(PO4)2 is 310.1767 g/mol
solubility im mass/volume = 1.64*10^-7 *310.1767
= 0.0003068585 = 5.1*10^-5 g/L
4)
solubility of Ag3PO4 = (Ksp^1/27)^(1/4) = ((1.8*10^(-18))/27)^(1/4)
= 1.6*10^-5
solubility im mass/volume = 1.6*10^-5 *molarmass of Ag3PO4
Molar mass of Ag3PO4 is 418.5760 g/mol
solubility im mass/volume = 1.64*10^-7 * 418.5760
= 0.0003068585 = 6.9*10^-5 g/L
5)
solubility of Ca5(PO4)3OH = (Ksp^1/84375)^(1/9)
= ((6.8*10^(-37))/84375)^(1/9)
= 2.72*10^-5
solubility im mass/volume = 2.72*10^-5 *molarmass of Ca3(PO4)2
Molar mass of Ca5(PO4)3OH is 502.3114 g/mol
solubility im mass/volume = 2.72*10^-5 *502.3114
= 0.0003068585 = 0.014 g/L
ORDER OF SOLUBILITY
Ca5(PO4)3OH > Hg2Cl2 > Ag3PO4 > Ca3(PO4)2 > Hg2CO3