What is the pH of a #5.00 x 10^-2# M aqueous solution of #Ba(OH)_2#?

1 Answer
anor277
Apr 28, 2017

#pH=13#

Explanation:

We know that in water under standard conditions, #pH+pOH=14#.

Now for #5.00xx10^-2*mol*L^-1# #Ba(OH)_2#, #[HO^-]=0.100*mol*L^-1#; and thus #pOH=-log_10[HO^-]# #=-log_10(10^-1)=1#. And thus #pH=13#.

For another example see this linky.

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