Ethanoic acid and NaOH titration

1. Jul 5, 2016

TT0

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
If 50% of the acid has been neutralised, then there is 0.05 M of acetic acid left.Using Ka
$$1.8*10^-5 = \frac {x^2}{0.05-x}$$
I get x = 0.00094
-log(0.00094) =3.02687 =pH
This is not one of the answers provided. I think I interpreted the question wrong. Can someone explain to me what I did wrong?

Cheers!

2. Jul 5, 2016

Staff: Mentor

You can't ignore acetate presence.

Hint: at half titration [CH3COOH] = [CH3COO-]

3. Jul 6, 2016

TT0

I see, then the equilibrium is:

$$1.8*10^{-5} = \frac {(0.05+x)*x} {0.05-x}$$

Therefore, pH = 4.74

Thanks again!

4. Jul 6, 2016

Staff: Mentor

Much easier to solve using Henderson-Hasselbalch equation. Not to mention the fact you should remember at half titration pH=pK.

5. Jul 8, 2016

TT0

I see, thanks a lot!

6. Jul 12, 2016

epenguin

If you had written the equation for the equilibrium constant out explicitly in terms of chemical species you would probably not gone wrong in the way you did.