what is the solubility in mol/L of insoluble Ca3(PO4)2 (Ksp=1.00x10^-26)in a solution containing .00100 M Na3PO4?
a. 2.15x10^-7
b. 7.18x10^-8
c. 7.18x10^-9
d. 1.00x10^-23
e. answer not given
x = solubility Ca3(PO4)2.
Ca3(PO4)2 ==> 3Ca^2+ + 2PO4^3-
.....x..........3x.......2x
............Na3PO4 --> 3Na^+ + PO4^3-
initial......001M.........0......0
change......-0.001.....0.001....0.001
equil........0.........0.001....0.001
Ksp = (Ca^2+)^3(PO4)^2
Substitute and solve for x.
For Ca^2+ you substitute 2x
For PO4^3- you substitute 3x+0.001 (that's 3x from the Ca3(PO4)2 and 0.001 from the Na3PO4.
2.15*10^-8
answer
To find the solubility of an insoluble substance, you need to calculate the concentration of the soluble ions in the solution. In this case, we have Ca3(PO4)2, which dissociates into Ca2+ and 3 PO43- ions.
First, let's determine the balanced equation for the dissociation of Ca3(PO4)2:
Ca3(PO4)2 ⇌ 3 Ca2+ + 2 PO43-
The solubility product constant (Ksp) for this reaction is given as 1.00 x 10^-26. The Ksp expression is written using the concentrations of the ions:
Ksp = [Ca2+]^3 * [PO43-]^2
We are given that the solution contains 0.00100 M Na3PO4. Na3PO4 dissociates into 3 Na+ ions and 1 PO43- ion.
To determine the concentration of the PO43- ions, we need to calculate the concentration of Na3PO4 after dissociation. Given that the initial concentration of Na3PO4 is 0.00100 M, and it dissociates into 1 PO43- ion:
[PO43-] = [Na3PO4] = 0.00100 M
Substituting the concentration of PO43- into the Ksp expression:
1.00 x 10^-26 = [Ca2+]^3 * (0.00100 M)^2
Rearranging the equation:
[Ca2+]^3 = (1.00 x 10^-26) / (0.00100 M)^2
Calculating the concentration of Ca2+:
[Ca2+]^3 = 1.00 x 10^-26 / 1.00 x 10^-6
[Ca2+]^3 = 1.00 x 10^-20
Taking the cube root of both sides:
[Ca2+] = (1.00 x 10^-20)^ (1/3)
[Ca2+] = 1.00 x 10^-6.67
[Ca2+] ≈ 7.18 x 10^-7 M
Therefore, the solubility of Ca3(PO4)2 in the given solution is approximately 7.18 x 10^-7 M.
The correct answer is (a) 2.15 x 10^-7.