Ksp a constant?: EK (7th ed). Chemistry pg. 156 #118

[Benwa]

Question 118 on page 156 of the EK chemistry book (7th ed.) asks what the Ksp is for Reaction 1 in the presence of NaOH. The correct answer is B, but why isn't A correct? Isn't Ksp always the same unless the temperature changes (pg. 75)? The solubility, not Ksp (solubility product), changes with a common ion. So shouldn't the Ksp only be concerned about the OH- produced by Ca(OH)2 dissociating?

Thanks!

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[MedHopeful20134]

In the presence of NaOH, there are already some OH- ions in solution. When you try to dissolve Ca(OH)2, it will not fully dissolve.
You are right. Ksp is not changing here. The question requires you to know that Ksp= [Ca2+] [OH-]2, and that all of OH- will not come from Ca(OH)2, as there are already some OH- in solution. Hence, answer is Ksp = [Ca2+] [OH-] total.