Titration Curve of unknown Acid Molarity of NaOH =.172 1. Your unknown acid is o

Question

Titration Curve of unknown Acid

Molarity of NaOH =.172

1. Your unknown acid is one of the following: oxalic acid (H2C2O4*2H20, pKa1=1.27, pKa2=4.27), citraconic acid (C5H6O4, pKa1=2.29, pKa2=6.15), or Maleic Acid (C4H4O4, pKa1=1.92, pKa2=6.09). These acids are classified as diprotic. Calculate the molar mass for your acid using the average mass from trials 1 and 2 and the average volume at the first equivalence point; then redo the calculation using the average mass from trials 1 and 2 and the average volume at the second equivalence point.

2. Using the average molar mass at the second equivalence point that you calculated in question 1, and your average values of pKa1 and pKa2, indentifiy your unknown acid. Explain.

Trial 1 Trial 2 Mass of acid used .1056g .1272 Volume @ equivalence point EP1 5.58 6.51 Volume @ equivalence point EP2 11.16 13.02 pH at half the 1st equivalence point 1.91 1.79 pH at halfway point of 1st and 2nd equiv. 4.04 3.99 Average value of pKa1 and Pka2 1.85, 4.02

Explanation / Answer

First eqv point:

general: M = m/n, n = c*V    at equiv point: n (base) = n (acid)

Trial 1: n = 0.172mole/L * 0.00558L = 9.6*10-4 mole

M = 0.1056g / 9.6*10-4 mole = 110 g/mole

Trial 2: n = 0.172mole/L * 0.00651L = 1.1*10-3 mole

M = 0.1272g/ 1.1*10-3 mole = 115g/mole

second equiv point: 2n (NAOH) = n(Acid) beacause its the acid donates two protons we nee two mole of NaOH to react with one mole acid. ->0.5 n(acid) = n (NaOH)

Trial 1: n = 0.172mole/L * 0.0116L = 1.9*10-3 mole

M = 0.1056g / 0.5(1.9*10-3 mole) = 110 g/mole

Trial 2: n = 0.172mole/L * 0.01302L = 2.2*10-3 mole

M = 0.1272g/ 0.5(2.2*10-3 mole) = 114g/mole

average molar mass: (110+114)/2 = 112g/mole

M (maleic acid) = 116 -> must be the used acid because it is the closest to the calculated molecular mass. The first pka value is close but the second is not so close. But the pka Values of the other acids are also not so close.

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