What is the #[OH^-]# of a #4.0 times 10^-4# #M# solution of #Ca(OH_2)#?

1 Answer
anor277
Apr 7, 2017

#[HO^-]=8.0xx10^-4*mol*L^-1#............

Explanation:

Calcium hydroxide speciates in aqueous solution according to the following reaction:

#Ca(OH)_2(s) stackrel(H_2O)rarrCa^(2+) + 2HO^-#

If the the concentration is #4.0xx10^-4*mol*L^-1# with respect to #Ca^(2+)#, stoichiometry, i.e. the composition of the salt, demands the solution is TWICE this concentrated with respect to #""^(-)OH#.

And thus #[HO^-]=2xx[Ca^(2+)]=2xx4.0xx10^-4*mol*L^-1=??*mol*L^-1.#

What are #pOH# and #pH# of this solution?

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