The solubility product constant of Calcium hydroxide is 6.5 * 10^-6 . If 0.10 mo of sodium hydroxide is added to 1 L of 0.0001 M Ca(OH)2, what is the final concentration of the calcium ion? Show your work.

Answer 1

Didn't I do this for you yesterday?

Answer 2

no???

Answer 3

I was sick all week and i just got this worksheet today

Answer 4

http://www.jiskha.com/display.cgi?id=1463494088

Answer 5

This is a common ion problem where 0.10M NaOH is added into a solution containing 0.0001M Ca(OH)2.

Ca(OH)2 <=> Ca^+2 + 2OH^-
Ci: -------- 0M ---- 0.10M
∆Ci: ------- +x ---- +2x
Ceq: -------- x ---- 0.10+2x~0.10M

Ksp = [Ca^+2][OH^-]^2
6.5x10^-6 = [Ca^+2](0.10)^2
Solve for [Ca^+2] ...
= (6.5x10^-6)/(0.10)^2
= 6.5x10-4M

Answer 6

It may appear to be a common ion problem but it isn't. (Ca^2+) can't possibly be more than it started with. It started out being 1E-4 and there isn't enough Ca^2+ and OH^- to exceed Ksp; therefore, there is no ppt of Ca(OH)2 initially. Therefore, Ksp = (Ca^2+)(OH^-)^2 is not valid because there is no solid Ca(OH)2 initially and there won't be solid Ca(OH)2 since Ksp isn't exceeded. It started at 1E-4M and there is no Ca^2+ being added so the final (Ca^2+) = 1E-4M.

Answer 7

@DrBob222 what do you mean by ppt?

Answer 8

precipitate = ppt
precipitated = pptd