If 35.9 grams of #Ca(OH)_2# are dissolved in 535 mL of solution, what is the molarity of the solution?

1 Answer
May 15, 2016

#(35.9*g)/(74.10*g*mol^-1)xx1/(0.535*L)# #=# #0.906*mol*L^-1#

Explanation:

This value, #0.906*mol*L^-1# with respect to #"calcium hydroxide"# makes no sense at all, as calcium hydroxide has fairly limited water solubility.

This site list a solubility product for calcium hydroxide, #K_(sp)=5.5xx10^-6#.