Given this data: You are assigned citic acid as a known. It is triprotic with a

Question

Given this data:

You are assigned citic acid as a known. It is triprotic with a molar mass of 210.14 g/mol

Amt of this used in grams to make solution: 1.672 g

Burette Readings:

1 - initial: 0.13mL final: 13.85mL

2- initial: 0.20 mL final: 13.90mL

3- initial: 0.20 mL final: 13.92mL

Ust the titration data for the known acid to calculate the moles of NaOH added to reach the equivalence point. You will do this three times, once for each trial. Use the average NaOH concentration for these calculation

-Best answer goes to best explaination there are more steps to this assignment/lab but this seems to be the part that I am confused on. The rest I think I understand

Explanation / Answer

Citric acid is triprotic, that is Molarity * 3 = Normailty

Molar Mass: 210.14g/mol

Amount used: 1.672g

Number of moles of citric acid: 1.672/210.14 = 0.00795 moles

Assuming 1l of citric acid: Molarity of solution = .00795 M

Normailty = 3*.00795 = 0.0238 N

Now, using N1V1 = N2V2

.0238 * 1 = N2 * (Volume change in burette in liters)

For the 1st reading: Volume change = 13.85 - .13 = 13.72 ml = 0.01372 l

N2 =1.7346 N = 1.7346 M since NaOH is unipolar

For the 2nd reading: Volume change = 13.90 - 0.20 = 13.70ml = 0.01370 l

N2 = 1.7372 N = 1.7372 M

For the 3rd reading: Volume change = 13.92 - 0.20 = 13.72ml = .01372 l

N2 = 1.7346 N = 1.7372 M

The 1st & 3rd reading are the same result, so we can say that the molarity is 1.7372

no of moles of NaOH = Volume * Molarity = 0.01372 * 1.7372 = 0.0238 moles of NaOH

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