Question

VNaOH pH 2.25 0.100M H3PO4 Titration Curve 0.7 2.28 1 2.27 14 1.5 2.29 12 2 2.3 10 2.5 2.33 8 2.35 6 3.5 2.38 2,42 4 4.5 2.45 2 2.49 0 5.5 2.53 5 10 15 20 25 30 35 6 2.58 Volume 0.10OM NaOH (mL) 6.5 2.63 7 2.69 2.75 7.5 8 2.83 8.5 2.93 3.06 9.5 3.22 9.7 3.32 9.9 3.43 10.2 3.68 10.4 3.95 10.6 4.59 10.7 4.98 10.8 5.29 10.9 5.49 11.1 5.69 11.4 5.86 11.7 6.01 12 6.14 12.5 6.29 13 6.41 13.5 6.51 14 6.62 14.8 6.76 15.5 6.86 16.3 6.99 17 7.1 17.8 7.25 7.34 18.3 18.8 7.44 19.5 7.63 20.3 7.93 20.9 8.45 21.1 8.86

1. Determine the buffer zones on the graph.

2. Determine the volumes at the first two stoichiometric points of the acid titration.

3. Determine the average volumes for the first two stoichiometric points.

4. Use the average volume to obtain pKa 1 and pka 2 of the acid.

5. From the pKa 1 and pKa 2, obtain the values of ka 1 and ka 2.

6. Determine the % of error according to your results for the constant of dissociation for the first and second equilibrium.

Answer 1

1) The buffer zones are the two that are before the equivalence points (the pH jumps) in which the pH remains almost constant.

2) The volume in the first two equivalence points are:

V1 = 11 mL

V2 = 22 mL

3) The average between the two volumes is 16.5 mL.

4) The pKa1 and pKa2 are read at the average equivalence points, at 5.5 and 16.5 mL:

pKa 1 = 2.53

pKa2 = 7

5) The Ka are calculated:

Ka1 = 10 ^ -pKa1 = 10 ^ -2.53 = 0.003

Ka2 = 10 ^ -7

6) The error percentages are calculated with the real Ka:

Real Ka1 = 0.007

Real Ka2 = 6.3x10 ^ -8

% error Ka1 = (0.007 - 0.003) * 100 / 0.007 = 57.14%

% error Ka2 = (6.3x10 ^ -8 - 10 ^ -7) * 100 / 6.3x10 ^ -8 = 58.73%

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