SOMEBODY PLEASE HELP ME! This is my last question for the chemistry and thank you. :D
A BS Medical Technology student was asked to determine the molecular weight of a diprotic acid using acid-base titration. This student performed four trials. Firstly, the student used potassium hydrogen phthalate (KHP, 204.22 g/mol) as standard to determine the exact concentration of the NaOH titrant. The results of the student are summarized in the table below:
Note: The stoichiometric relationship of KHP to NaOH is 1:1 (KHP, 204.22 g/mol)
Standardization: Trial 1 Trial 2 Trial 3 Trial 4
KHP Mass: 0.5033g 0.5066g 0.6989g 0.6843g
Volume NaOH used: 24.32mL 25.61mL 24.67mL 24.56mL
After the standardization, the student weighed different amounts of the unknown acid. In a similar fashion, the student performed acid-base titration using phenolphthalein as indicator. The results of the molecular weight determination are summarized below:
Note: The stoichiometric relationship of Unknown acid to NaOH is 1:2
Molecular Weight Determination: Trial 1 Trial 2 Trial 3 Trial 4
Unknown Mass: 0.1234g 0.1034g 0.1178g 0.1322g
Volume NaOH used: 21.75mL 20.56mL 24.39mL 25.08mL
The potential unknown given by the professor to the student are as follows:
-Fumaric acid: 116.07 g/mol
-Oxalic acid: 90.03 g/mol
-Succinic acid: 118.09 g/mol
-Malonic acid: 104.06 g/mol
-Tartaric acid: 150.09 g/mol
-Citric acid: 192.12 g/mol
Now the question are:
1. What is the molarity of NaOH solution at the standardization procedure in Trial 1?
2. What is the molarity of NaOH solution at the standardization procedure in Trial 2?
3. What is the average molarity of NaOH solution at the standardization procedure?
4. How many moles of diprotic acid have reacted in the molecular weight determination step for Trial 3?
5. Considering all the trials in molecular weight determination, what is the average molecular weight of the acid as analyzed by the student?
6. If the percent error of the student in his analysis was around 2.37%, what is the identity of the diprotic acid given to him?
It seems that u guys are tam students taking up canvas exams lol
1. What is the molarity of NaOH solution at the standardization procedure in Trial 1?
KHP + NaOH ==> NaKHP + H2O
mols KHP = g/molar mass = 0.5033/204.22 = 0.002464
M NaOH = mols/L = 0.002464/0.02432 = 0.1013 M
2. What is the molarity of NaOH solution at the standardization procedure in Trial 2?
Do this the same way as trial 1
3. What is the average molarity of NaOH solution at the standardization procedure?
Do trial 3 and trial 4 as above, then take the average of the 4 trials.
4. How many moles of diprotic acid have reacted in the molecular weight determination step for Trial 3?
I will use the data for trial 1 but you use trial 3 the same way. Let's call the diprotic acid H2A, then H2A + 2NaOH ==> Na2A + 2H2O
mols NaOH used = M x L = 0.1013 x 0.02175 = 0.002203
mols H2A will be 1/2 that or 0.001102
Then since mols = g/molar mass then molar mass = g/mols and
molar mass = 0.1234/0.001102 = 112
5. Considering all the trials in molecular weight determination, what is the average molecular weight of the acid as analyzed by the student?
Do all of the trials for molar mass using the AVERAGE molarity of the NaOH then take the average of the molar masses.
6. If the percent error of the student in his analysis was around 2.37%, what is the identity of the diprotic acid given to him?
I worked only the first trials but with 2.37% error that gives a lot of leeway. I would use the AVERAGE for the molarity of the NaOH from the first part, determine the molar mass of each of the trials, take the average of that and make the decision as to the identity of the unknown. Frankly, 2.37% is NOT good. A good analytical chemist should be able to do this with way less than 1% (say +/- 0.02%) so it isn't a guessing game as to the identity of the unknown.