Titration was used to determine the molarity of acetic acid in vinegar. A primary standard solution of KHP was used to standardize the NaOH.
1. When performing this experiment, impure KHP was used to standardize the NaOH solution. If the impurity is neither acidic or basic, will the percent by mass id acetic acid in the vinegar solution determined by the student be too high or low? Justify answer with explanation.
It would be too low right due to a low number of moles of KHP to start off with, right?
I think the % acetic acid will be too high.
Standardization step:
mols KHP = grams KHP/molar mass KHP
mols NaOH = mols KHP
M NaOH = mols NaOH/L NaOH
Titration vinegar step:
mols NaOH titrant = M x L
mols acetic acid = mols NaOH
g acetic acid = mols acetic acid x molar mass
% acetic acid = (g acetic acid/mass sample) x 100
explanation
Standardization step:
mols KHP = grams KHP/molar mass KHP
You weigh out impure KHP so mass KHP is too high for the amount KHP present. That makes mols KHP too high. Now just follow each step down to the end. Or you can think of it as there is less KHP there; therefore, since less KHP is present that means it will take less NaOH to reach the equivalence point. So if L NaOH is too low in the standardization step that makes M NaOH too high. Now follow the steps down. Either way you end up with too high % acetic acid at the end.
mols NaOH = mols KHP
M NaOH = mols NaOH/L NaOH
Titration vinegar step:
mols NaOH titrant = M x L
mols acetic acid = mols NaOH
g acetic acid = mols acetic acid x molar mass
% acetic acid = (g acetic acid/mass sample) x 100
No, actually the percent by mass of acetic acid in the vinegar solution would be too high if impure KHP was used to standardize the NaOH solution. Let me explain why.
In this experiment, impure KHP (potassium hydrogen phthalate) is used as a primary standard to standardize the concentration of the NaOH (sodium hydroxide) solution. The impurity in the KHP is neither acidic nor basic, which means it will not react with the NaOH during the titration process.
During the titration, a known concentration of NaOH solution is slowly added to the impure KHP solution until the equivalence point is reached. At the equivalence point, the moles of NaOH added are stoichiometrically equivalent to the moles of KHP present.
However, since the impurity in the KHP is not acidic or basic, it does not react with the NaOH. Therefore, the amount of impure KHP actually reacting with the NaOH is less than the amount indicated by the stoichiometry. As a result, the calculated molarity of the NaOH solution will be higher than its actual value.
Now, let's consider the student's determination of the percent by mass of acetic acid in the vinegar solution. Acetic acid (CH3COOH) is a weak acid that reacts with NaOH during the titration. The balanced chemical equation for the reaction is:
CH3COOH + NaOH → CH3COONa + H2O
Since the molarity of the NaOH solution is higher than its actual value due to the impurity in the KHP, the student will use a higher concentration of NaOH in the titration. As a result, the calculated number of moles of acetic acid will be lower than the actual number of moles.
To calculate the percent by mass of acetic acid, the student divides the mass of acetic acid by the total mass of the vinegar solution. If the calculated moles of acetic acid are lower than the actual moles, the resulting percent by mass of acetic acid will be erroneously higher than its true value.
Therefore, when impure KHP is used to standardize the NaOH solution, the percent by mass of acetic acid in the vinegar solution determined by the student will be too high.