Can someone please help me solve this problem?

Calculate the solubility of Ba3(PO4)2 in water. For Ba3(PO4)2, Ksp = 1.3 x 10- 29

Thank you!

......Ba3(PO4)2 ==> 3Ba^2+ + 2PO4^3-

I.....solid..........0..........0
C.....solid..........3x.........2x
E.....solid..........3x.........2x

Ksp = (Ba^2+)^3(PO4^3-)^2
Substitute the E line into Ksp expression and solvle for x. That gives you the solubility of the solid.

Sure! I can definitely help you solve this problem.

To calculate the solubility of Ba3(PO4)2 in water, we need to use the solubility product constant (Ksp) of Ba3(PO4)2, which is given as 1.3 x 10^(-29).

The expression for the solubility product constant is Ksp = [Ba^2+ ]^3 [PO4^3- ]^2, where [Ba^2+ ] is the concentration of Ba^2+ ions and [PO4^3- ] is the concentration of PO4^3- ions.

Since Ba3(PO4)2 dissociates into 3 Ba^2+ ions and 2 PO4^3- ions, we can represent the concentration of Ba^2+ ions as 3x (where x is the solubility of Ba3(PO4)2) and the concentration of PO4^3- ions as 2x.

Substituting these expressions into the solubility product constant equation, we get:

Ksp = (3x)^3 (2x)^2

Now, let's solve for x:

1.3 x 10^(-29) = (3x)^3 (2x)^2

(3^3)(2^2)x^5 = 1.3 x 10^(-29)

24x^5 = 1.3 x 10^(-29)

Divide both sides by 24:

x^5 = (1.3 x 10^(-29))/(24)

x^5 = 5.42 x 10^(-31)

Take the fifth root of both sides to solve for x:

x = (5.42 x 10^(-31))^(1/5)

Using a calculator, you can evaluate this expression to find the solubility of Ba3(PO4)2 in water.

I hope this explanation helps you solve the problem! Let me know if you have any further questions.