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.
Low End
High End
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
etc.
i figured it out but how do i get the Low end and High end?
To determine the molarity of Ag+ ions required to initiate precipitation and to bring precipitation to 99.99% completion for each anion, we need to use the solubility product expressions for AgBr and Ag3PO4.
The solubility product expression for AgBr is given as:
Ksp(AgBr) = [Ag+][Br-]
Similarly, the solubility product expression for Ag3PO4 is:
Ksp(Ag3PO4) = [Ag+]^3[PO43-]
Let's calculate the molarity of Ag+ ions required for each case:
(a) [Ag+] at which AgBr precipitation begins:
To initiate precipitation, the concentration of Ag+ ions must exceed the solubility product of AgBr.
Ksp(AgBr) = [Ag+][Br-]
[Ag+] = Ksp(AgBr)/[Br-] = (5.0 * 10^-13)/(0.31 mol)
(b) [Ag+] at which AgBr precipitation is 99.99% complete:
To bring the precipitation of AgBr to 99.99% completion, we need to find a concentration of Ag+ ions where only 0.01% of AgBr remains dissolved. This means that 99.99% of AgBr has precipitated.
Thus, the concentration of AgBr ions remaining in solution will be (0.01/100) * initial concentration of Br- ions. We can then use this concentration to calculate [Ag+].
[Ag+] = ((5.0 * 10^-13)/(0.31 mol))* (0.01/100)
(c) [Ag+] at which Ag3PO4 precipitation begins:
To initiate precipitation of Ag3PO4, the concentration of Ag+ ions must exceed the solubility product of Ag3PO4.
Ksp(Ag3PO4) = [Ag+]^3[PO43-]
[Ag+] = (Ksp(Ag3PO4)/[PO43-])^(1/3) = ((1.8 * 10^-18)/(0.31 mol))^(1/3)
(d) [Ag+] at which Ag3PO4 precipitation is 99.99% complete:
Similar to part (b), we need to find the concentration of Ag+ ions that results in 99.99% completion of Ag3PO4 precipitation.
[Ag+] = (((1.8 * 10^-18)/(0.31 mol))^(1/3)) * (0.01/100)
(e) To achieve complete separation by precipitation, we need to find the range of [Ag+] values that result in the precipitation of both AgBr and Ag3PO4. This can be done by considering the higher and lower end values of [Ag+] required for precipitation of each compound.
Low end [Ag+] = [Ag+] at which Ag3PO4 precipitation begins (part c)
High end [Ag+] = [Ag+] at which AgBr precipitation is 99.99% complete (part b)
So, the [Ag+] range for complete separation by precipitation is between the low end and high end values calculated above.