What is the #"pH"# of #"0.1 N"# #"Ca"("OH")_2#? Thanks.

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Answer 1 · 2017-07-01T15:46:07

#pH=13.0#

We work out (i) the #pOH#........

We have #0.1*N# with respect to #HO^-#. This is #0.05*mol*L^-1# with respect to #Ca(OH)_2(aq)#. Agreed?

Now.................

#pOH=-log_10[HO^-]=-log_10(0.1)=-(-1)=+1#.

And we can now (ii) address the #pH#.

Because #pOH+pH=14#, and thus #pH=14-1.00=13.0#.

Is this solution acidic or basic?

See here for a similar problem.

Thanks to Truong-Son who pointed out an error on my part......