Calculate the molar solubility of calcium phosphate [Ca3(PO4)2 in a solution which is 0.10M in sodium phosphate (Na3PO4). Ksp for calcium phosphate is 1.1x10^-26

................Ca3(PO4)2 ==> 3Ca^2+ + 2[PO4]^3-

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

Then Na3PO4 is completely (100%) ionized
.............Na3PO4==> 3Na^+ + [PO4]^3-
I............0.1 M..............0...............0
C...........-0.1...............0.3............0.1
E.............0.0............0.3...............0.1
Ksp = 1.1E-26 = (Ca^2+)^3[(PO4)^3-]^2
Ca^2+ is 3x from the Ca3(PO4)2
[PO4]^3- is 2x from the Ca3PO4)2 + 0.1 for the Na3PO4 for a total of 2x + 0.1
so Ksp = 1.1E-26 = (3x)^3(2x + 0.1)^2
Solve for x and that gives you the molar solubility of Ca3(PO4)2
Post your work if you get stuck. I think you can ignore the 2x in 2x+0.1. Try it at least and see if you can because it will simplify the math.

To calculate the molar solubility of calcium phosphate, we need to use the solubility product constant (Ksp) along with the concentration of the common ion present in the solution, which in this case is phosphate (PO4^3-).

The balanced equation for the dissolution of calcium phosphate is:
Ca3(PO4)2 (s) ↔ 3 Ca^2+ (aq) + 2 PO4^3- (aq)

From the equation, we can see that for every 1 mole of calcium phosphate that dissolves, it releases 2 moles of phosphate ions (PO4^3-).

Given that the solution is 0.10 M in sodium phosphate (Na3PO4), we can consider the concentration of phosphate ions to be the same, which is 0.10 M.

Now, let's use the expression for the solubility product constant (Ksp) to calculate the molar solubility (S) of calcium phosphate:

Ksp = [Ca^2+]^3 * [PO4^3-]^2

Since the molar solubility of calcium phosphate is represented by 'S', we can substitute the concentration of phosphate ions into the equation:

1.1x10^-26 = (S)^3 * (0.10)^2

Solving for 'S', we can rearrange the equation as follows:

(S)^3 = 1.1x10^-26 / (0.10)^2

(S)^3 = 1.1x10^-26 / 0.01

(S)^3 = 1.1x10^-24

Taking the cube root of both sides:

S = (1.1x10^-24)^(1/3)

S ≈ 1.045x10^-8

Therefore, the molar solubility of calcium phosphate in the given solution is approximately 1.045x10^-8 M.

To calculate the molar solubility of calcium phosphate (Ca3(PO4)2), we need to consider the balanced chemical equation for the dissolution of calcium phosphate in water:

Ca3(PO4)2 ⇌ 3Ca2+ + 2PO43-

From the equation, we can see that for every mole of calcium phosphate that dissolves, we get 3 moles of calcium ions (Ca2+) and 2 moles of phosphate ions (PO43-). Therefore, the equilibrium expression for the solubility product constant (Ksp) can be written as:

Ksp = [Ca2+]^3 * [PO43-]^2

We are given that the Ksp for calcium phosphate is 1.1x10^-26. However, we need to find the molar solubility of calcium phosphate in a solution that is 0.10M in sodium phosphate (Na3PO4).

To determine the molar solubility, we assume that x mol/L of Ca3(PO4)2 dissolves, resulting in a concentration of 3x mol/L for Ca2+ and 2x mol/L for PO43-. Since stoichiometry tells us that the ratio of Ca2+ to PO43- is 3:2, we can express the concentration of Ca2+ in terms of x.

[Ca2+] = 3x
[PO43-] = 2x

Now, substitute these values into the Ksp expression:

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

Simplifying:

1.1x10^-26 = 54x^5

To solve for x, rearrange the equation:

x^5 = (1.1x10^-26) / 54

x = (1.1x10^-26 / 54)^(1/5)

Using a calculator, compute the right side of the equation, which equals 6.88x10^-6. Thus, the molar solubility of calcium phosphate in the solution is 6.88x10^-6 M.