........Ca(OH)2 ==> Ca^2+ + 2OH^-
I.......solid........0.......0
C.......solid........x.......2x
E.......solid........x.......2x
Ksp = 6.5E-6 = (Ca^2+)(OH^-)^2
6.5E-6 = (x)(2x)^2 = 4x^3
Solve for x = [Ca(OH)2] in mols/L
Convert that to grams/500 mL.
g = mols x molar mass and that divided by 2 = g/500. Compare with the 0.37 g in the problem. If 0.37 g is larger than x the soln is not saturated. If 0.37 g is smaller it is saturated.
Ca(OH)2 has a Ksp of 6.5 x 10^-6. If 0.37 g of Ca(OH)2 is added to 500 mL of water and the mixture is allowed to come to equilibrium, will the solution be saturated? pls help me I stayed up all last night trying to get this one question :(
2 answers
Thank you so much!