What is the pH of a 0.000460 M solution of #Ca(OH)_2#?

1 Answer
Jun 21, 2016

Answer:

Approx. #11#

Explanation:

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

Thus #pOH=-log_10{0.009200}# #=# #-(-3.04)=3.04#

But it is a FACT that #pH +pOH=14#

Thus #pH~=11#

Why is the concentration of #HO^-# double that of #Ca(OH)_2#?

PS If you have trouble with the #log# function, voice your problem . Someone here will help you.