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.

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

Ksp = (Ca^+2)^3*(PO4^-3)^2
If x = solubility of Ca3(PO4)2, then 3x is the concn of Ca^+2 and 2x is the concn of PO4^-3
Substitute th 3x and 2x into the Ksp expression, solve for x, then 3x will be the (Ca^+2)

should it be 0?

is the answer 0?

To calculate the concentration of calcium ion (Ca2+) in a saturated solution of Ca3(PO4)2, you need to use the solubility product constant (Ksp) and the stoichiometry of the compound.

The balanced equation for the dissolution of Ca3(PO4)2 is:
Ca3(PO4)2(s) ⇌ 3Ca2+(aq) + 2PO4^3-(aq)

The Ksp expression for this equation is:
Ksp = [Ca2+]^3 [PO4^3-]^2

The given Ksp value is 1.3 × 10^(-32) at 25°C.

Since Ca3(PO4)2 dissociates into three Ca2+ ions and two PO4^3- ions, we can assume that the concentration of Ca2+ ions will be three times that of the concentration of Ca3(PO4)2.

Let's assume the concentration of Ca3(PO4)2 in the saturated solution is 'x'. Then, the concentration of Ca2+ ions ([Ca2+]) will be 3x.

Substituting these values into the Ksp expression, we get:
Ksp = (3x)^3 (2x)^2 = 1.3 × 10^(-32)

Expanding and rearranging the equation gives:
(27x^3)(4x^2) = 1.3 × 10^(-32)

Simplifying further:
108x^5 = 1.3 × 10^(-32)

To solve for 'x', the concentration of Ca3(PO4)2, divide both sides by 108:
x^5 = (1.3 × 10^(-32))/(108)

Taking the fifth root of both sides:
x = ((1.3 × 10^(-32))/(108))^(1/5)

Evaluating this expression using a calculator, we find that x is approximately 2.77 × 10^(-7) M.

Therefore, the concentration of calcium ions (Ca2+) in a saturated solution of Ca3(PO4)2 is approximately 3 × 2.77 × 10^(-7) M, which is approximately 8.31 × 10^(-7) M.

it should be 25°C not 25°C

3.3x10^-7