# 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.