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??
it will be low
1. In this case, if the impurity in the KHP used by the student is neither acidic nor basic, the percent by mass of acetic acid in the vinegar solution determined by the student will not be affected. This is because the impurity does not react with acetic acid or NaOH, so it would not affect the titration process. Therefore, the percent by mass of acetic acid in the vinegar solution would remain the same.
2. When the student did not allow the NaOH pellets to completely dissolve before standardizing the solution with KHP, there would be an insufficient amount of NaOH available for the reaction. This would result in a lower-than-expected molarity of the NaOH solution. However, once the remaining NaOH pellets had completely dissolved and the student re-filled the buret, the solution would have the correct concentration of NaOH. Hence, when titrating the acetic acid, the molarity of acetic acid in the vinegar solution determined by the student would not be affected by this initial incomplete dissolution of NaOH pellets.
3. Dissolved CO2 in distilled water can react with NaOH to form sodium carbonate (Na2CO3). Sodium carbonate is a base, and it would react with the acetic acid during the titration process, resulting in inaccurate determination of the concentration of NaOH. This is because sodium carbonate consumes a portion of the NaOH, causing the amount of NaOH used in the titration to be lower than it should be. As a result, the determined concentration of the NaOH solution would be higher than its actual concentration. To ensure accurate determination of NaOH concentration, it is important to use CO2-free distilled water, which is free from dissolved CO2 and potential sources of sodium carbonate.
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 If grams too low then mols too low.
mols KHP = mols NaOH So mols NaOH too large.
M NaOH = mols NaOH/L NaOHIf mols NaOH too large than M too large
Then you titrate the vinegar with NaOH.
CH3COOH + NaOH ==> CH3COONa + H2O
mols NaOH used = M x L M too large makes mols NaOH too large
mols CH3COOH = mols NaOHthat makes mols CH3COOH too large
g CH3COOH = mols x molar mass CH3COOH so g CH3COOH too large and % CH3COOH too large
% CH3COOH = (grams CH3COOH/g sample)*100 = ?