Asked by Jose
In this experiment the only source of Ca2+ ions is Ca(OH)2. Using the experimental Ksp value, calculate the slubility of Ca(OH)2 (in mol/L) in .30 M NaOH.
Experiment Ksp = 1.69x10^-5
Experiment Ksp = 1.69x10^-5
Answers
Answered by
DrBob222
Ca(OH)2 ==> Ca^2+ + 2OH^-
NaOH ==> Na^+ + OH^-
Ksp = 1.69E-5 = (Ca^2+)(OH^-)^2
Set up an ICE chart.
(Ca^2+) = x
(OH^-) from Ca(OH)2 = x
(OH^-) from NaOH = 0.3
Substitute into Ksp expression to obtain
Ksp = (x)(x+0.3)^2
Solve for x.
NaOH ==> Na^+ + OH^-
Ksp = 1.69E-5 = (Ca^2+)(OH^-)^2
Set up an ICE chart.
(Ca^2+) = x
(OH^-) from Ca(OH)2 = x
(OH^-) from NaOH = 0.3
Substitute into Ksp expression to obtain
Ksp = (x)(x+0.3)^2
Solve for x.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.