Hydroxylapatite, Ca10(PO4)6(OH)2, has a solubility constant of Ksp = 2.34 × 10-59, and dissociates according to

Ca10(PO4)6(OH)2(s) --> 10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq)

Solid hydroxylapatite is dissolved in water to form a saturated solution. What is the concentration of Ca2 in this solution if [OH–] is somehow fixed at 8.40 × 10-6 M?

Does this require the quadratic formula or is there a simpler way to solve for x?

I would assume 2x+8.4E-6 = 8.4E-6 and see what happens.

Is my step here correct mathematically?

2.34e-59 = (10x)^10(6x)^6(8.4e-6+2x)

and then I solve for x?

or is it

2.34e-59 = (10x)(6x)(8.4e-6+2x)

and then I solve for x?

how did you solve it???

To determine the concentration of Ca2+ in the saturated solution, we need to use the solubility product constant (Ksp) and the molar ratio of ions in the dissociation equation.

Given:
Ksp = 2.34 × 10^-59
[OH-] = 8.40 × 10^-6 M

From the dissociation equation:
Ca10(PO4)6(OH)2(s) --> 10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq)

The molar ratio of Ca2+ to OH- is 10:2, which means for every 10 moles of Ca2+ produced, 2 moles of OH- are consumed. Therefore, the concentration of OH- does not affect the concentration of Ca2+.

Since the solubility of hydroxylapatite determines the concentration of Ca2+, we can use the solubility product constant to calculate it.

The Ksp expression for the dissociation is:
Ksp = [Ca2+]^10[PO43-]^6[OH-]^2

Given that [OH-] is 8.40 × 10^-6 M, we substitute its value into the expression:

Ksp = [Ca2+]^10[PO43-]^6 (8.40 × 10^-6 M)^2

We can simplify the expression further by substituting the Ksp value:

2.34 × 10^-59 = [Ca2+]^10[PO43-]^6 (8.40 × 10^-6 M)^2

To solve for [Ca2+], we need the concentration of PO43-. However, it hasn't been given in the question. Without that information, we cannot determine the concentration of Ca2+ in the solution.

Therefore, the concentration of Ca2+ in the solution cannot be determined with the given information.

Ca10(PO4)6(OH)2 >10Ca^2+ + 6PO4^3-+2OH^2

solid.............10x......6x......2x

Ksp = 2.34E-59 = (Ca^2+)^10(PO4^3-)^6*(OH^-)^2.

(Ca^2+)= 10x
(PO4^3-) = 6x
(OH^-) = 8.4E-6 + 2x

Substitute into Ksp expression and solve for x, then Ca^2+ = 10x.

nevermind i got it thanks for the help!