Teeth and bones are composed, to a first approximation, of calcium phosphate, Ca3(PO4)2(s). The Ksp for this salt is 1.3 10-32 at 25°C. Calculate the concentration of calcium ion in a saturated solution of Ca3(PO4)2.
Please, help!
Let S = solubility of Ca3(PO4)2
.........Ca3(PO4)2 ==> 3Ca^2+ + 2PO4^3-
initial....solid........0.........0
change......S..........+3S.......2S
equil.......S...........3S.......2S
Ksp = (Ca^3+)^3(PO4^3-)^2
Ksp = (3S)^3(2S)^2
Solve for S.
To determine the concentration of calcium ion in a saturated solution of Ca3(PO4)2, we need to use the solubility product constant (Ksp).
The solubility product constant is given by the expression:
Ksp = [Ca2+][PO42-]^3
We know the value of Ksp, which is 1.3 × 10^-32.
Let's assume that 'x' represents the molar solubility of Ca3(PO4)2. Since the compound dissociates into one calcium ion and two phosphate ions, the molar solubility (x) will be equal to the concentration of calcium ion ([Ca2+]) in the saturated solution.
Now we can set up the equation using the values we have:
Ksp = [Ca2+][PO42-]^3
1.3 × 10^-32 = (x)(x^2)^3
1.3 × 10^-32 = x^7
Now we can solve for 'x' by taking the seventh root of both sides of the equation:
x = (1.3 × 10^-32)^(1/7)
Using a calculator to perform the calculation, we find that x ≈ 5.16 × 10^-6 mol/L.
Therefore, the concentration of calcium ion in a saturated solution of Ca3(PO4)2 is approximately 5.16 × 10^-6 mol/L.
To calculate the concentration of calcium ions in a saturated solution of Ca3(PO4)2, we need to use the solubility product constant (Ksp) and the stoichiometry of the compound.
The balanced equation for the dissociation of Ca3(PO4)2 is:
Ca3(PO4)2(s) ↔ 3 Ca2+(aq) + 2 PO43-(aq)
From the equation, we can see that for every one mole of Ca3(PO4)2 that dissolves, we will get three moles of Ca2+ ions.
Since Ksp is given as 1.3 × 10^-32, we can write the expression for Ksp as:
Ksp = [Ca2+]^3 × [PO43-]^2
Since Ca3(PO4)2 is an ionic compound, it will completely dissociate in water. This means that the concentration of Ca2+ ions will be equal to the solubility (S) of Ca3(PO4)2.
Therefore, we can write the equation for Ksp as:
Ksp = (S)^3 × [PO43-]^2
Given that Ksp = 1.3 × 10^-32, we need to solve for S (the solubility of Ca3(PO4)2).
Taking the cube root of both sides, we get:
(S)^3 = Ksp^(1/3)
S = (Ksp^(1/3))^1
Now, we can substitute the value of Ksp into the equation:
S = (1.3 × 10^-32)^(1/3)
Using a calculator, we find:
S ≈ 7.943 × 10^-11
Therefore, the solubility of Ca3(PO4)2 is approximately 7.943 × 10^-11 M.
Since the concentration of calcium ions (Ca2+) in the solution is equal to the solubility (S) of Ca3(PO4)2, the concentration of calcium ions is also approximately 7.943 × 10^-11 M.