Lab 12 of the Concentration of Acetic Acid in Vinegar Post-Laboratory Questions

Question

Lab 12 of the Concentration of Acetic Acid in Vinegar Post-Laboratory Questions 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. before standardizing the solution with KHP. However, by the time the student refilled NaO a NaOH solution, a student did not allow the NaOH pellets to completely dissolve the buret with H 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. custom page 142

Explanation / Answer

If we go by normal method

1)KHP standardization with NaOH.
mols KHP = grams/molar mass KHP
mols KHP = mols NaOH
M NaOH = mols NaOH/L NaOH
2)Titrate the vinegar with NaOH.
CH3COOH + NaOH ==> CH3COONa + H2O
mols NaOH used = M x L (M = Molarity of Naoh)
mols CH3COOH = mols NaOH
Mass in Grams of CH3COOH = mols x molar mass CH3COOH
% CH3COOH = (grams CH3COOH/grams of sample)*100 = Final Percentage

A)The answer to 1st part:

Adding impure KHP (just weighed wrong--no addition acid or base)
mols KHP = grams/molar mass KHP -  If wieght is too low then mols also will be too low.
mols KHP = mols NaOH Than mols of NaOH will be too large.
M NaOH = mols NaOH/L NaOH If mols NaOH too large than Molarity will be too large

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

mols NaOH used = M x L If M too large makes mols of NaOH also too large
mols CH3COOH = mols NaOH That makes mols CH3COOH also too large
g CH3COOH = mols x molar mass CH3COOH so wiegth CH3COOH too large and % CH3COOH also too large
% CH3COOH = (grams CH3COOH/g sample)*100 = So final percentage also will be too high.

B)Answer to Part 2

Second Part is same as the first part here Initially the dissolved Naoh is very less so Molartiy of acetic acid will be less but when titrated with the more Naoh the Molarity will increase which will automatically increase the molarity of Acetic acid in Vinegar as:

When you titrate the vinegar with NaOH all previous Naoh also dissolves which increases the Molarity of NaoH
CH3COOH + NaOH ==> CH3COONa + H2O

mols NaOH used = M x L If M too large makes mols of NaOH also too large
mols CH3COOH = mols NaOH That makes mols CH3COOH also too large
g CH3COOH = mols x molar mass CH3COOH so wiegth CH3COOH too large and % CH3COOH also too large
% CH3COOH = (grams CH3COOH/g sample)*100 = So final percentage also will be too high.

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