(a) If the molar solubility of Cd3(PO4)2 at 25 oC is 1.19e-07 mol/L, what is the...

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(a) If the molar solubility of Cd3(PO4)2 at 25 oC is 1.19e-07mol/L, what is the Ksp at this temperature?
Ksp =
(b) It is found that 0.00178 g of Ag3PO4 dissolves per 100 mL ofaqueous solution at 25 oC. Calculate the solubility-productconstant for Ag3PO4.
Ksp =
(c) The Ksp of Sc(OH)3 at 25 oC is 2.22e-31. What is the molarsolubility of Sc(OH)3?
solubility ________= mol/L

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Cd3(PO4)2 -------------> 3Cd⁺² + 2PO4^3-

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 3sÂ Â Â Â Â Â Â Â Â Â 2s

Ksp = [Cd⁺²]³[PO4^3-]²

Â Â Â Â Â Â Â =Â Â (3s)³ (2s)²

Â Â Â Â Â Â Â Â Â = 108s⁵

Â Â Â Â Â Â Â Â = 108*(1.19*10^-7)⁵

Â Â Â Â Â Â Â Â Â = 2.57*10^-33

b. solubility = W*1000/G.M.Wt * volume of solution in ml

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.00178*1000/418.58*100

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 4.25*10^-5

Â Â Ag3(PO4) ----------> 3Ag⁺Â Â + PO4^3-

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 3sÂ Â Â Â Â Â Â Â Â Â s

KspÂ Â = [Ag⁺]³ [PO4^3-]

Â Â Â Â Â Â Â Â Â = (3s)³ *s

Â Â Â Â Â Â Â Â Â = 27s⁴

Â Â Â Â Â Â Â Â Â = 27*(4.25*10^-5)⁴

Â Â Â Â Â Â Â Â Â = 8.8*10^-17

c.Â Â Sc(OH)3 --------> Sc⁺³ + 3OH^-

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â sÂ Â Â Â Â Â Â Â Â 3s

Â Â Ksp = [Sc⁺³][OH-]³

Â Â Â Â Â Â Â Â Â Â = s*(3s)³

Â Â 2.22*10^-31 Â Â Â Â Â = 27s⁴

Â Â Â 0.82*10^-32Â Â Â = s⁴

^{Â Â Â Â Â Â} sÂ Â Â Â = 9.5*10^-9 mole/L

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(a) If the molar solubility of Cd3(PO4)2 at 25 oC is 1.19e-07 mol/L, what is the...

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