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.
Didn't I do this for you yesterday?
no???
I was sick all week and i just got this worksheet today
http://www.jiskha.com/display.cgi?id=1463494088
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
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.
@DrBob222 what do you mean by ppt?
precipitate = ppt
precipitated = pptd
To find the final concentration of the calcium ion in the solution, we need to determine the amount of calcium hydroxide that dissolves when sodium hydroxide is added. This can be determined by calculating the precipitation reaction using the solubility product constant.
The solubility product constant (Ksp) for calcium hydroxide (Ca(OH)2) is given as 6.5 × 10^-6. The equation for this reaction is:
Ca(OH)2 ⇌ Ca2+ + 2OH-
Let's assume x represents the amount of calcium hydroxide that dissolves, and the resulting concentration of Ca2+ ions is also x. The concentration of OH- ions will then be 2x since the reaction stoichiometry is 1:2 for Ca2+:OH-.
Using the given molarity of the original Ca(OH)2 solution (0.0001 M) and assuming x is small compared to the initial concentration, we can write the following equation for equilibrium:
Ksp = [Ca2+][OH-]^2
Substituting the values:
6.5 × 10^-6 = x(2x)^2
6.5 × 10^-6 = 4x^3
Solving for x:
x^3 = (6.5 × 10^-6) / 4
x^3 = 1.625 × 10^-6
Taking the cube root of both sides:
x ≈ 1.06 × 10^-2
Therefore, the amount of calcium hydroxide that dissolves is approximately 1.06 × 10^-2 moles.
Since the initial concentration of Ca(OH)2 was 0.0001 M, and we added 0.10 moles of sodium hydroxide to the solution, the final volume will be 1 L + 0.10 L = 1.10 L.
To find the final concentration of the calcium ion, divide the amount of calcium hydroxide dissolved by the final volume of the solution:
Final concentration of calcium ion = (1.06 × 10^-2 moles) / 1.10 L
Final concentration of calcium ion ≈ 9.64 × 10^-3 M
Therefore, the final concentration of the calcium ion in the solution is approximately 9.64 × 10^-3 M.