.......Ca(OH)2 ==> Ca^2+ + 2OH^-
I......solid........0.......0
C......solid........x.......2x
E......solid........x.......2x
Ksp = (Ca^2)(OH^-)^2
Substitute the E line into the Ksp expression and solve for (Ca^2+). 2x that will be (OH^-) if you want to know that. The solubility of Ca(OH)2 in the saturated solution will be the same as (Ca^2+) if want to know that.
Calculate the concentration of Ca+2 in a saturated solution of Ca(OH)2 if the Ksp = 6.5x10-6 at 25oC.
can someone explain the steps to do this please? :)
2 answers
Solubility in pure water of all 1:2 or 2:1 ionizations with Ksp values is => Cube-Rt(Ksp/4).
For Ca(OH)2 ; Ksp=6.5E-5
Solubility = Cube-Rt(6.5E-5/4) = 0.0118M in Ca^+2 and 2(0.0118M)in OH^-.
In pure water (no common ion effect) the following can be used to calculate amount of salt delivering ions into solution ...
for 1:3 Izns => S = 4th-Rt(Ksp/27)
for 1:4 Izns => S = 5th-Rt(Ksp/256)
for 2:3 Izns => S = 5th-Rt(Ksp/108)
For Ca(OH)2 ; Ksp=6.5E-5
Solubility = Cube-Rt(6.5E-5/4) = 0.0118M in Ca^+2 and 2(0.0118M)in OH^-.
In pure water (no common ion effect) the following can be used to calculate amount of salt delivering ions into solution ...
for 1:3 Izns => S = 4th-Rt(Ksp/27)
for 1:4 Izns => S = 5th-Rt(Ksp/256)
for 2:3 Izns => S = 5th-Rt(Ksp/108)
3 answers