Consider the titration of 20.0mL of a 0.100 M solution of HBrO, a weak acid (Ka=2.5x10-9) with 0.200 M KOH. Calculate the pH of the following volumes of KOH.
pH=pKa +log [base]/[acid]
The Attempt at a Solution
Im having trouble with C when the pH is at its equivalance point.
10.00mL (.200 mmol KOH/mL) = 2.00 mmol OH-
So stoichiometricaly my OH- and HBrO are used to completely to give me a conjugate base of 2.00 mmol BrO-
I set up my ice chart:
I .0667 0 0
C -x +x +x
E .0667-x x x
Assume x is small to find x=[OH-]=5.2x10-4
What doesn't make sense to me is in Part A pH=4.80, part B pH=8.5, Part C=10.72 and part D pH=8.20.
Now shouldnt the pH in part D be greater than part C since I have excess OH-?
I'll type of my work for Part D in a bit. I have to leave right now. But if somebody can help my checking my answers I'd appreciate it. Thanks.