[shaggybill]

Hello. Here's a question that I'm having trouble figuring out the concept of.


H2SO4 + Ca(OH)2  ->  2H2O + CaSO4

List all the species present in solution after the reaction is complete.

The answer: Ca²⁺, OH^-, (H₂O)


I don't understand that answer.

[Borek]

Don't feel alone.

Have you listed all information given?

[shaggybill]

Thank you for answering. Yes, I think I have given all the information listed. This was a question on my last quiz. Question 3a was balancing that equation. 3b was calculating the number of moles of each of the reagents, and 3c was to list all the species present in solution after the reaction is complete.

I got 3a and 3b right, no problem, but I'm not sure what 3c is asking for.

Here's a link to the quiz key if you want to look at it. I don't think I'm leaving out anything important. It's question 3c.

http://chem.uncc.edu/images/stories/courses/chem1251/quizzes/q3-530key.pdf

[nj_bartel]

You were leaving out some necessary information.  You're reacting an excess of calcium hydroxide with sulfuric acid.  That means you'll have Ca²⁺ ions and OH^- ions in your aqueous solution due to leftover unreacted calcium hydroxide.  You do not have any SO₄^2- present because calcium sulfate is essentially insoluble in water.

[Borek]

You do not have any SO₄^2- present because calcium sulfate is essentially insoluble in water.

Only on High School level, in General Chemistry its solubility is in the 0.005M range.

[nj_bartel]

Ok, I assumed this was a solubility rules question.

Could you account for the lack of SO₄^2- due to LeChatlier's?

The excess would leave you with 0.114 M Ca²⁺ to counteract a pretty comparatively small amount of CaSO₄ that would normally dissolve lightly.  Would there literally be no SO₄ then, or is it just now insignificant enough of a concentration that you can discount it?

[Borek]

Ok, I assumed this was a solubility rules question.

Could be you are right, it is just that one never knows. Question was asked in GC forum, but sounded like HS.