posted by Tanner .
The solubility products of AgBr and Ag3PO4 are 5.0 10-13 and 1.8 10-18, respectively. If Ag+ is added (without changing the volume) to 1.00 L of a solution containing 0.31 mol Br ‾ and 0.31 mol PO43-, calculate the molarity of Ag+ ions required to initiate precipitation and to bring the precipitation to 99.99% completion for each anion.
(a) [Ag+] at which AgBr precipitation begins
(b) [Ag+] at which AgBr precipitation is 99.99% complete
(c) [Ag+] at which Ag3PO4 precipitation begins
(d) [Ag+] at which Ag3PO4 precipitation is 99.99% complete
e) Give the [Ag+] range in which a complete separation by precipitation can be achieved.
I can get you started but it's too long to work out in excruciating detail.
AgBr ==> Ag^+ + Br^-
Ksp = (Ag^+)(Br^-) = 5.0E-13
When Ag^+ is added ion by ion, AgBr will start to ppt when (Ag^+) = 5.0E-13/0.31 = 1.6E-12.
When 99.99% has been pptd that will leave (Br^-) of 0.31 x (0.01/100) = 3.1E-5 and
(Ag^+) = 5.0E-13/3.1E-5 = 1.6E-8. Ag3PO4 will not have begun to ppt then
for (Ag^+) must be at least
(Ag^+) = cube root(Ksp/PO4) = 1.8E-6
i figured it out but how do i get the Low end and High end?