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?
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.
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
To answer these questions, we need to follow the steps involved in the standardization procedure and the molecular weight determination. Let's tackle each question one by one.
1. To calculate the molarity of the NaOH solution in Trial 1 at the standardization procedure, we can use the formula:
Molarity (M) = (mol NaOH) / (volume NaOH in liters)
First, we need to determine the number of moles of KHP used in Trial 1 during standardization. The molar mass of KHP is given as 204.22 g/mol. We can calculate the moles of KHP using the mass and molar mass:
moles KHP = mass KHP / molar mass KHP
Substituting the given values, we get:
moles KHP = 0.5033g / 204.22 g/mol
Next, we use the stoichiometric relationship between KHP and NaOH, which is 1:1. This means that the number of moles of NaOH used in the titration will be the same as the moles of KHP:
moles NaOH = moles KHP
Finally, we convert the volume of NaOH used from milliliters to liters and plug in the values to calculate the molarity:
Molarity = moles NaOH / (volume NaOH in Liters)
2. Following the same steps as in Question 1, we calculate the molarity of the NaOH solution in Trial 2 at the standardization procedure. Use the given mass of KHP and volume of NaOH to determine the molarity.
3. To find the average molarity of NaOH solution at the standardization procedure, we add the molarities from all the trials (from Questions 1 and 2) and divide by the number of trials (2 in this case).
Average Molarity = (Molarity in Trial 1 + Molarity in Trial 2) / 2
4. To determine how many moles of diprotic acid reacted in the molecular weight determination step for Trial 3, we use the balanced stoichiometric relationship of the unknown acid and NaOH, which is 1:2. This means that for every mole of unknown acid, 2 moles of NaOH are consumed. We can calculate the moles of diprotic acid reacted using the volume of NaOH used in Trial 3 and the average molarity obtained in the previous step.
moles diprotic acid = (Molarity from Question 3) × (volume NaOH in Trial 3 in liters) × 2
5. To find the average molecular weight of the acid as analyzed by the student, we need to calculate the moles of the unknown acid used in each trial and then calculate the average molecular weight. We already have the masses of the unknown acid used in each trial. Divide each mass by the respective molecular weight of the unknown acid, and sum up the moles. Finally, divide by the number of trials (4 in this case) to get the average moles of the unknown acid. The average molecular weight can be obtained by dividing the sum of the masses of the unknown acid used in all trials by the average moles of the unknown acid.
Average Moles of Unknown Acid = (moles of Unknown Acid in Trial 1 + moles of Unknown Acid in Trial 2 + moles of Unknown Acid in Trial 3 + moles of Unknown Acid in Trial 4) / 4
Average Molecular Weight = (mass of Unknown Acid in Trial 1 + mass of Unknown Acid in Trial 2 + mass of Unknown Acid in Trial 3 + mass of Unknown Acid in Trial 4) / Average Moles of Unknown Acid
6. To determine the identity of the diprotic acid given to the student, we can calculate the percent error for each potential unknown acid using the experimental average molecular weight (from Question 5) and the known molecular weights. The percent error can be calculated using the formula:
Percent Error = (|Experimental Value - Known Value| / Known Value) × 100
For each potential unknown acid, substitute the experimental and known molecular weights into the formula to calculate the percent error. The diprotic acid with a percent error closest to 2.37% would likely be the correct identified acid.