- #1

Frankenstein19

- 56

- 0

## Homework Statement

When 100.0 mL of weak acid HA was titrated with 0.09381M NaOH, 27.63mL were required to reach the equivalence point. The pH at the equivalence point was 10.99. What is the pKa of HA

## Homework Equations

pH=pKa+log([A]/[HA]

[OH]-10^(-pOH)

## The Attempt at a Solution

14-10.99=3.01 this gives me pOH

From this I can calculate the concentration of OH: 0.000977237 M

Since we are at the equivalence point, to get the moles I multiplied the above concentration by 0.02763L and got 2.7*10^(-5)

I know that initially I had 0.0025919703 moles of HA

I subtract the moles of OH that react with HA and get 0.0025919703-2.7*10^(-5)=2.56399*10^(-3)

The moles of OH are the moles of A: 2.7*10^(-5)

Using pH=pKa+log([A]/[HA]

I get 10.99=pKa-1.97

Giving me a pH of 12.96 instead of 9.69 which is the correct answer

What am I doing wrong