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
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?
1 answer