Question

What will be the pOH of a 0.02 M of Ca(OH)2 ?

Answer 1

`pOH=1.40`

Explanation:

`pOH=-log_10[HO^-]`...

....for calcium hydroxide we got `[HO^-]=0.04*mol*L^-1`...

...and so `pOH=-log_10(0.04)=-(-1.40)=1.40`...