Question

1. When performing this experiment, a student mistakenly used impure KHP to standardize the NaOH solution. If the impurity is neither acidic nor basic, will the percent by mass of acetic acid in the vinegar solution determined by the student be too high or too low? Justify your answer with an explanation.
~Would the percent by mass be affected by an impure solution? If the actual grams of the solution aren't affected then the % by mass won't either, right?

2. When preparing a NaOH solution, a student did not allow the NaOH pellets to completely dissolve before standardizing the solution with KHP. However, by the time the student refilled the buret with NaOH to titrate the acetic acid, the remaining NaOH pellets had completely dissolved. Will the molarity of acetic acid in the vinegar solution, determined by the student, be too high or too low? Justify your answer with an explanation
~The first one would be too low and the second trial too high because of the difference in the dissolved NaOH solution, but I'm not sure about overall..

3. Distilled water normally contains dissolved CO2. When preparing NaOH standard solutions, it is important to use CO2 free distilled water. How does dissolved CO2 in distilled water affect the accuracy of the determination of a NaOH solution’s concentration? (Hint: Use your textbook, the internet, etc. to research the term acid anhydride.)
~(CH3CO)2O will interfere with the solution and cause it to be to acidic??

Answer 1

I think the easiest way to work these "too small, too large, neither" problems is to go through the formula used to calculate the end result. I have summarized them here for problem #1. My comments are included with each in bold face.

KHP standardization with NaOH.

mols KHP = grams/molar mass KHP
mols KHP = mols NaOH
M NaOH = mols NaOH/L NaOH

Then you titrate the vinegar with NaOH.
CH3COOH + NaOH ==> CH3COONa + H2O

mols NaOH used = M x L
mols CH3COOH = mols NaOH
g CH3COOH = mols x molar mass CH3COOH >
% CH3COOH = (grams CH3COOH/g sample)*100 = ?
-------------------------
Adding impure KHP (just weighed wrong--no addition acid or base)
mols KHP = grams/molar mass KHP <b>If grams too low then mols too low.</b>
mols KHP = mols NaOH <b>So mols NaOH too large.</b>
M NaOH = mols NaOH/L NaOH<b>If mols NaOH too large than M too large</b>

Then you titrate the vinegar with NaOH.
CH3COOH + NaOH ==> CH3COONa + H2O

mols NaOH used = M x L <b>M too large makes mols NaOH too large</b>
mols CH3COOH = mols NaOH<b>that makes mols CH3COOH too large</b>
g CH3COOH = mols x molar mass CH3COOH <b>so g CH3COOH too large and % CH3COOH too large</b>
% CH3COOH = (grams CH3COOH/g sample)*100 = ?

Answer 2

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Answer 3

irjurj

Answer 4

it will be low