Determine the [OH-] of a solution that is 0.230 M in HCO3-
Homework Answers
-Log(Molarity given) = pH
-Log(.230M)=pH
pH= .63827...
Which rounds to .638.
To find the pOH, subtract the pH from 14.
14-.638 = 13.36.
To find the concentration of [OH-], simply do 10^-pOH.
10^(-13.36) = 4.37*10^-14 (after rounding).
H2CO3 
HCO3− + H+
Ka1= 4.45×10−7
HCO3− CO32− + H+
Ka2=4.69×10−11
[H+]=√ka1*ka2
=4.57x10-9
∴pH=-log[H+]
=8.34
pOH=14-8.34=5.66
-log[OH-]=5.66
[OH-]=2.18x10-6 M
Read up more here: http://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt
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