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.
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?
6 answers
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?
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?
nevermind i got it thanks for the help!
how did you solve it???
2 answers