# Titration help

osker246

## Homework Statement

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.

a. 0.00mL
b. 5.00mL
c. 10.00mL
d. 30.00mL

## Homework Equations

pH=pKa +log [base]/[acid]

## The Attempt at a Solution

Im having trouble with C when the pH is at its equivalance point.

so,

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-

[BrO-]=2.00mmol/30.00ml=.0667 M

I set up my ice chart:

BrO-<=====>OH-+HBrO
I .0667 0 0
C -x +x +x
E .0667-x x x

Where

Kb=[OH-][HBrO]/[BrO-]=x2/.0667-x=4.0x10-6

Assume x is small to find x=[OH-]=5.2x10-4

pOH=3.28
pH= 14-pOH=10.72

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 shouldn't 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.