(a) If the molar solubility of
Mg3(PO4)2 at 25 oC is
6.26e-06 mol/L, what is the Ksp at this
temperature?
Ksp =
(b) It is found that 6.35e-07 g of
Cu3(PO4)2 dissolves per 100 mL of
aqueous solution at 25 oC. Calculate the
solubility-product constant for
Cu3(PO4)2.
Ksp =
(c) The Ksp of PbBr2 at 25 oC is
6.60e-06. What is the molar solubility of PbBr2?
solubility = mol/L
a.
the molar solubility of Mg3(PO4)2 at 25 oC is 6.26*10^-6 mol/L
Mg3PO4)2(s) ---------> 3Mg^2+ (aq) + 2PO4^3- (aq)
3s 2s
Ksp = [Mg^2+]^3[PO4^3-]^2
= (3s)^3(2s)^2
= 108s^5
= 108*(6.26*10^-6)^5
= 1.04*10^-24
b.
molarity = W*1000/G.M.Wt* volume in ml
= 6.35*10^-7*1000/(380.58*100)
= 1.7*10^-8M
Cu3(PO4)2(s) ------------> 3Cu^2+ (aq) + 2PO4^3-(aq)
3s 2s
Ksp = [Cu^2+]^3[PO4^3-]^2
= (3s)^3(2s)^2
= 108s^5
= 108(1.7*10^-8)^5
= 1.135*10^-38
c. PbBr2(s) ------------. Pb^2+ (aq) + 2Br^- (aq)
s 2s
Ksp = [Pb^2+][Br^-]^2
6.6*10^-6 = s(2s)^2
6.6*10^-6 = 4s^3
s = 0.0118
solubility of PbBr2 = 1.18*10^2mole/L
(a) If the molar solubility of Mg3(PO4)2 at 25 oC is 6.26e-06 mol/L, what is the...
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