In: Chemistry
1 - For the insoluble salt lead (ii) phosphate Pb3(PO4)2, the solubility product, Ksp is 3.10x10-44
a- Write the chemical equation and solubility product expression for Pb3(PO4)2
b- Determine the molar solubility (M) of Pb3(PO4)2
c- Identify wheather the molar solubility of Pb3(PO4)2 will increase or decrease upon the following changes and explain why?
*** The ddition of 0.10 M Pb(NO3)2
*** The addition of 0.10 M NaCl
Given that Ksp of Pb3(PO4)2 = 3.10 x 10-44
Pb3(PO4)2 ---------------> 3 Pb2+ + 2 PO43-
3s 2s s = molar solubility of Pb3(PO4)2
Ksp = [Pb2+]3 [PO43-]2
3.10 x 10-44 = (3s)3 (2s)2
3.10 x 10-44 = 108 s5
s = [3.10 x 10-44 / 108]1/5
= 0.78 x 10-9 M
Therefore, molar solubility of Pb3(PO4)2 = 0.78 x 10-9 M
c) Molar solubility (M) of Pb3(PO4)2 after the addition of 0.10 M Pb(NO3)2
Dut to common ion Pb2+ in Pb(NO3)2 and Pb3(PO4)2 , molar solubility of Pb3(PO4)2 decreases.
Molar solubility (M) of Pb3(PO4)2 after the addition of 0.10 M NaCl
Since there is no common ion, molar solubility of Pb3(PO4)2 is not effected.