Question

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

Answer 1

`[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?