Question

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

Ksp =

(b) It is found that 0.00178 g of Ag3PO4 dissolves per 100 mL of aqueous solution at 25 oC. Calculate the solubility-product constant for Ag3PO4.

Ksp =

(c) The Ksp of Sc(OH)3 at 25 oC is 2.22e-31. What is the molar solubility of Sc(OH)3?

solubility ________= mol/L

Answer 1

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

                                        3s           2s

Ksp = [Cd⁺²]³[PO4^3-]²

        =   (3s)³ (2s)²

          = 108s⁵

         = 108*(1.19*10^-7)⁵

          = 2.57*10^-33

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

                     = 0.00178*1000/418.58*100

                      = 4.25*10^-5

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

                                   3s           s

Ksp   = [Ag⁺]³ [PO4^3-]

          = (3s)³ *s

          = 27s⁴

          = 27*(4.25*10^-5)⁴

          = 8.8*10^-17

c.   Sc(OH)3 --------> Sc⁺³ + 3OH^-

                                  s          3s

   Ksp = [Sc⁺³][OH-]³

           = s*(3s)³

   2.22*10^-31       = 27s⁴

    0.82*10^-32    = s⁴

       s     = 9.5*10^-9 mole/L