What Went Wrong in My Ethanoic Acid and NaOH Titration Calculation?

[TT0]

Homework Statement

Homework Equations

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!

 

[Borek]

You can't ignore acetate presence.

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

 

[TT0]

I see, then the equilibrium is:

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

Therefore, pH = 4.74

Answer is B.

Thanks again!

 

[Borek]

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

 

[TT0]

I see, thanks a lot!

 

[epenguin]

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!

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.