Posted by Rudy on Sunday, January 26, 2014 at 7:24pm.
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 4.30 × 10-6
I understand to set this question up as such:
2.34e-59 = (10x)^10(6x)^6(4.3e-6+2x)
But I don't understand the next step, can anyone help? you don't have to give the answer just the next few steps
2 answers
wouldn't the Ca ion cncentration be 5 times the OH conc?
First, I think you left a squared term off the OH^-.
Ca10(PO4)6(OH)2(s) --> 10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq)
(Ca^2+) = 10x
(PO4^3-) = 6x
(OH^-) = 2x
Ksp = (Ca^2+)^10(PO4^3-)^6(OH^-)^2
Substituting the x values we have
Ksp = (10x)^10(6x)^6(4.3e-6 + 2x)^2
I would make the assumption that
4.3e-6 + 2x = 4.3e-6 so we have
Ksp = (10x)^10(6x)^6(4.3e-6)^2
2.34e-59 = 10^10*x^10 (6x)^6*(1.85e-11)
2.34e-59 = 10^10*x^10*4.66e4*x^6*(1.85e-11)
2.34e-59/[(10)^10*(4.66e4)*(1.85e-11) = x^16.
Ca10(PO4)6(OH)2(s) --> 10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq)
(Ca^2+) = 10x
(PO4^3-) = 6x
(OH^-) = 2x
Ksp = (Ca^2+)^10(PO4^3-)^6(OH^-)^2
Substituting the x values we have
Ksp = (10x)^10(6x)^6(4.3e-6 + 2x)^2
I would make the assumption that
4.3e-6 + 2x = 4.3e-6 so we have
Ksp = (10x)^10(6x)^6(4.3e-6)^2
2.34e-59 = 10^10*x^10 (6x)^6*(1.85e-11)
2.34e-59 = 10^10*x^10*4.66e4*x^6*(1.85e-11)
2.34e-59/[(10)^10*(4.66e4)*(1.85e-11) = x^16.
6 answers