If #Ca(OH)_2(aq)=0.01molL^-1#, what is #pH# of the solution?

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Answer 1 · 2016-07-10T09:25:18

#pH~=12#

We know (or should know) that in aqueous solution #pH+pOH=14#.

Thus, if we calculate #pOH#, #pH# is directly available.

#pOH=-log_10([HO^-])# #=-log_10(0.02)# (why #0.02# here and not #0.01#?) #=# #-(-1.70)# #=# #1.70#

And from the above, #pH=14-1.70# #=# #??#

If #Ca(OH)_2(aq)=0.01*mol*L^-1#, what is #pH# of the solution?