1. What is the molarity of NaOH solution at the standardization procedure in Trial 1?
KHP + NaOH ==> KNaP + H2O
mols KHP = grams/molar mass = 0.5033/204.22 = ?
Since this is 1:1 you know mols NaOH = mols KHP
M NaOH = mols/L = mols NaOH/0.02432 = ?
2. What is the molarity of NaOH solution at the standardization procedure in Trial 2?
See my answer above for trial 1.
3. What is the average molarity of NaOH solution at the standardization procedure?
Repeat for all the steps and calculate the average.
4. How many moles of diprotic acid have reacted in the molecular weight determination step for Trial 3?
2NaOH + H2A ==> Na2A + 2H2O
mols NaOH = avg M x L = avg M x 0.02439 L = ?
mols H2A = 1/2 * mols NaOH
5. Considering all the trials in molecular weight determination, what is the average molecular weight of the acid as analyzed by the student?
mols H2A = grams H2A/mol wt H2A. You know mols H2A and grams H2A, solve for mol wt H2A. They way the questions are arranged is a little disjointed, I think. I would do trial 1 all the way through, repeat for trials 2, 3 and 4. then average all 4 sets, and use that average number to answer the identity of the acid.
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?
IN my opinion, 2.37% error is terrible. He wouldn't get a job in my factory. Take your average molecular weight, add and subtract 1.64%, use those numbers to identify the unknown.
If you run into trouble post your work and ask your question about what you don't understand.
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?
5 answers
can i ask where did you get 0.02432?
That's trial 1 volume of NaOH of 24.32 mL. Since M = mols/L then 24.32 mL = 0.02432
thank you so much!!!!!
WHAT WAS IT
2 answers