Question

1. Using the K_sp values given, find the solubility in moles per litre of the following.

a) Calcium phosphate in water

b) Calcium phosphate in 0.010 M CaCl₂ solution

PLEASE SHOW ALL WORK, THANK YOU

Answer 1

Ksp of Ca3(PO4)2 2.07*10^-33

Ca3(PO4)2 ------------------------> 3Ca^2+ (aq) + 2PO4^3-

                                                           3s                2s

Ksp   = [Ca2^+]^3[PO4^3-]^2

            = (3s)^3*(2s)^2

1.3*10^-33          = 108s^5

s^5                    = 1.3*10^-33/108   

s^5                   = 1.2*10^-35

     s                    = 1.037*10^-7

solubility of Ca3(PO4)2 in water = 1.037*10^-7 mole/L

b.   CaCl2 --------------------> Ca^2+ (aq) + 2Cl^-

    0.01M                                0.01M

     Ca3(PO4)2 ------------------------> 3Ca^2+ (aq) + 2PO4^3-

                                                           3s+0.01            2s

   [Ca^2+]    = 3s+0.01   = 0.01M                    [3s<<<<<<0.01]

Ksp   = [Ca2^+]^3[PO4^3-]^2

1.3*10^-33   = (0.01)^3 * (2s)^2

4s^2            = 1.3*10^-33/(0.01)^3

4s^2             = 1.3*10^-27

s^2             = 1.3*10^-27/4

s^2                = 3.25*10^-28

s                  = 1.8 *10^-14

solubility of Ca3(PO4)2 in 0.01M CaF2 is 1.8*10^-14 mole/L