Lab 8 Discussion

Lab 8 Discussion - We dissolved TUMS in 0.3M HCl and the...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
In order to determine the percentage of acetic acid in vinegar in part two, we used titration. We used Sodium Hydroxide, which is a strong base, to do this. We did two trials to ensure accuracy. We titrated 20mL of vinegar with 0.3M NaOH. We then found the moles of NaOH and used this number to determine the moles of CH3COOH and the percent content in vinegar. Unfortunately, we may have caused minor errors in titration since it is such a sensitive test and it depends on our perception. CH 3 COOH(aq)+OH - (aq) -> CH 3 COO - (aq)+H 2 0(l) K=[CH 3 COO - ]/[CH 3 COOH][OH - ] In the third part of the experiment we wanted to determine the mount of calcium carbonate (the active ingredient) in TUMS. To do this we used back titration.
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: We dissolved TUMS in 0.3M HCl and the resulting mixture was titrated with 0.3M NaOH. We determined the amount of unrelated HCl by analyzing the amount of NaOH used, and once we found the unreacted amount we subtracted it from the total to determine the amount of calcium carbonate. The percent content of the active ingredient was 81%. Once, again, this part relied on titration which is a tricky procedure and is reliant on personal perception, thus minor error may have occurred. Step 1: CaCO 3 (aq)+H + (aq)->Ca 2 + (aq)+HCO 3-(aq) K=[[Ca 2 + ][HCO 3-]]/[[CaCO 3 ][H + ]] Step 2: HCO 3-(aq)+OH-(aq)->CO 3 2-(aq)+H 2 O(l) K=[CO 3 2-]/[[HCO 3-][OH-]]...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online