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

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SUMMARY

The forum discussion centers on the calculation of pH during the titration of ethanoic acid (acetic acid) with sodium hydroxide (NaOH). The user initially calculated a pH of 3.03 using the dissociation constant (Ka) of acetic acid, but this was incorrect due to misinterpretation of the equilibrium conditions. The correct pH at the half-titration point, where [CH3COOH] equals [CH3COO-], is 4.74, as derived using the Henderson-Hasselbalch equation. The discussion emphasizes the importance of recognizing acetate presence and correctly applying equilibrium constants in titration calculations.

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Homework Statement


Screen Shot 2016-07-05 at 5.12.39 PM.png


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!
 
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You can't ignore acetate presence.

Hint: at half titration [CH3COOH] = [CH3COO-]
 
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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!
 
Much easier to solve using Henderson-Hasselbalch equation. Not to mention the fact you should remember at half titration pH=pK.
 
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I see, thanks a lot!
 
TT0 said:
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.
 
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