Question

Calculate molar solubility, S, of calcium phosphate at 298.15 K. The Solubility product, Ksp for calcium phosphate at 298.15 K is 2.07*10^-33

a. in pure water

b. in 0.050 M Ca(NO3)2

c. in 20.0 mM Na3PO4

Answer 1

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

                                                        3s                     2s

              Ksp      = [Ca^2+]^3[PO4^2-]^2

                          = (3s)^3(2s)^2

     2.07*10^-33     = 108s^5

      s^5                      = 2.07*10^-33/108

    s^5                         = 1.9*10^-35

       s                            = 1.14*10^-7 M

The molar solubility in water = 1.14*10^-7 M >>>>answer

b. in 0.050 M Ca(NO3)2

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

                                                     3s+0.05           2s

              Ksp      = [Ca^2+]^3[PO4^2-]^2

                          = (3s+0.05)^3(2s)^2

                          = (0.05)^3(2s)^2                 [ 3s+0.05 = 0.05 because of 3s<<<<<<0.05   ]

     2.07*10^-33     = 0.0005s^2

      s^2                     = 2.07*10^-33/0.0005

    s^2                         = 4.14*10^-30

       s                            = 2.035*10^-15 M

The molar solubility in 0.050 M Ca(NO3)2 is 2.035*10^-15 M >>>>answer

c. in 20.0 mM Na3PO4    = 20*10^-3 M Na3Po4   = 2*10^-2 M Na3PO4

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

                                                     3s            2s+0.02

              Ksp      = [Ca^2+]^3[PO4^2-]^2

                          = (3s)^3(2s+0.02)^2

                          = (3s)^3(0.02)^2                 [ 2s+0.02 = 0.02 because of 2s<<<<<<0.02 ]

     2.07*10^-33     = 0.0108s^3

      s^3                    = 2.07*10^-33/0.0108

    s^3                         = 1.92*10^-31

       s                            = 5.8*10^-11 M

The molar solubility in 20mMNa3PO4 is 5.8*10^-11 M >>>>answer