# chemistry

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If the concentration of Pb2+ is found to be 2.3 x 10-9 mol/L in a saturated solution of Pb3(PO4)2, what is the Ksp of Pb3(PO4)2 ?

• chemistry -

Write the ionization (solubility) equation for Pb3(PO4)2.
Write the Ksp expresion.
Plug in concns of Pb and PO4 and solve for Ksp. Post your work if you get stuck.

• chemistry -

Ionization equation for Pb3(PO4)2 is
Pb3(PO4)2 --> Pb^2+ + 2PO4^3-

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

Ksp = [2.3 x 10-9 mol][4.6 x 10-9 mol]
= 1.1 x 10-16 mol

I am stuck

• chemistry -

Ionization equation isn't what you have. And you didn't raise the concns to powers in the Ksp.
Pb3(PO4)2 <==> 3Pb^+2 + 2PO4^-3

Ksp = (Pb^+2)^3*(PO4^-3)^2

You know [Pb3(PO4)2] = 2.3 x 10^-9
Therefore (Pb^+2) = 3 times that.
And (PO4^-3) is 2 times that.
Plug those into the Ksp expression I wrote and solve for Ksp.

• chemistry -

Pb3(PO4)2 <==> 3Pb^+2 + 2PO4^-3

Since [Pb3(PO4)2] = 2.3 x 10^-9

(Pb^+2) = 6.9 x 10^-9

(PO4^-3)= 4.6 x 10^-9

Ksp = (Pb^+2)^3*(PO4^-3)^2

= (6.9 x 10^-9)^3*(4.6 X 10^-9)^2

=(3.2850 X 10^-25)(2.116 X 10^-17)

=6.951 X 10^-42

It seems................quite small

• chemistry -

That's the right answer. But think about the small number (and it IS small), but when you cube and square negative numbers they get even smaller. For example, if we just use 1 x 10^-9 as the solubility, then cube one and square the other (that's to the 5th power), then (10^-9)^5 = 10^-45. The difference between that and the correct answer is that the numbers were a little larger than 1 x 10^-9 and that makes up for the 1000 or so difference (when you raise one to the third and the other to the second power).

• chemistry -

THANK YOU!!!!!

• chemistry -

Ksp = [Pb2+]^3 [PO4]^2
= (2.3E-9)^3 (2/3 * 2.3E-9)^2
= 4/9 * (2.3E-9)^5

• chemistry -

You have given in the problem that the solubility of Pb3(PO4)2 is 2.3E-9. Given that, then concn of Pb MUST be 3 x that (from the ionization equation) and concn PO4^-3 MUST be 2 x that (brom the ionization equation). IF you had 1 Pb ion produced for every 1 molecule of Pb3(PO4)2 that ionized, and 2/3 PO4^- ion produced for every molecule fo Pb3(PO4)2 that ionized, then what you propose would be ok. But that isn't the ionization equation.
There is another way of doing it; in fact, I always use it because it's faster.
Pb3(PO4)2 <==> 3Pb^+2 + 2PO4^-3

I call solubility of Pb3(PO4)2 = y (I don't use x because it gets confused with the times sign).
If (Pb3(PO4)2 = y, then (Pb^+2) = 3y and (PO4^-3) = 2y
Substituting into the Ksp expression, I have Ksp = (Pb^+2)^3(PO4^-3)^2 =
(3y)^3(2y)^2 = 108y^5 and since y = 2.3E-9, then Ksp = 108(2.3E-9)^5 = 6.95E-42. The time saving advantage here is that I do all the calculator stuff at one time and get the answer. The other way we must multiply 3x solubility, then 2x solubility, then key in the two numbers, cube one, square the other one, and key them in and multiply them. The latter method you key in one number, raise it to the 5th power, multiply the existing number already in the calculator by 108 and voila!

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