Please find the 75% equivalence point, 25% equivalence point, Difference in equi
ID: 488905 • Letter: P
Question
Please find the 75% equivalence point, 25% equivalence point, Difference in equivalence points, and half equivalence point(pKa1), and molecular weight range? Then based on the results, determine the unknown acid (possibilities: acetic acid, oxalic acid, benzoic acid, maleic acid, malic acid, and KHP).
standardized NaOH(M) is 0.0840±0.0005 and Used unknown acid was 0.2005g. I did titration curve graph, and found the end point as 44.40.
Below is a data of pH vs. NaOH titration:
V(NaOH), mL pH 0 1.69 0.5 1.67 1 1.68 1.5 1.68 2 1.69 2.5 1.7 3 1.71 3.5 1.73 4 1.74 4.5 1.75 5 1.76 5.5 1.77 6 1.78 6.5 1.79 7 1.8 7.5 1.81 8 1.82 8.5 1.84 9 1.85 9.5 1.86 9.5 1.87 10 1.88 10.1 1.89 10.2 1.89 10.6 1.92 11 1.93 11.2 1.94 11.5 1.95 12 1.96 12.1 1.98 12.3 1.99 12.5 1.99 12.7 2 12.9 2 13.1 2.01 13.3 2.01 13.5 2.02 13.7 2.03 13.9 2.04 14.1 2.05 14.3 2.06 14.5 2.07 14.7 2.08 14.9 2.09 15 2.13 15.5 2.14 16 2.16 16.5 2.2 17 2.23 17.5 2.26 18 2.3 18.5 2.34 19 2.39 19.5 2.43 20 2.48 20.5 2.54 20.7 2.6 20.9 2.63 21.1 2.65 21.2 2.66 21.4 2.69 21.6 2.72 21.8 2.74 22 2.77 22.2 2.8 22.5 2.83 23 2.89 23.5 2.94 24 3.01 24.5 3.08 25 3.13 25.5 3.19 26 3.24 26.5 3.28 27 3.34 27.5 3.39 28 3.44 28.5 3.48 29 3.53 29.5 3.56 30 3.61 30.2 3.64 30.4 3.65 30.6 3.66 30.8 3.67 31 3.69 31.2 3.7 31.4 3.72 31.6 3.74 31.8 3.76 32 3.77 32.2 3.79 32.4 3.8 32.6 3.82 32.8 3.83 33 3.83 33.2 3.86 33.4 3.87 33.6 3.89 33.8 3.91 34 3.92 34.2 3.93 34.4 3.94 34.6 3.96 34.8 3.99 35 4 35.2 4.01 35.4 4.03 35.6 4.05 35.8 4.06 36 4.09 36.2 4.1 36.4 4.12 36.6 4.13 36.8 4.15 37 4.17 37.2 4.19 37.4 4.21 37.6 4.23 37.8 4.25 38 4.27 38.2 4.29 38.4 4.31 38.6 4.34 38.8 4.35 39 4.38 39.2 4.4 39.4 4.43 39.6 4.45 39.8 4.48 40 4.5 40.2 4.54 40.4 4.57 40.6 4.58 40.8 4.62 41 4.65 41.2 4.69 41.4 4.71 41.6 4.78 41.8 4.81 42 4.84 42.2 4.89 42.4 4.98 42.6 5.05 42.8 5.13 43 5.2 43.2 5.28 43.4 5.46 43.6 5.57 43.8 5.81 44 6.11 44.2 6.74 44.4 8.42 44.6 9.17 44.8 10.14 45 10.33 45.2 10.5 45.4 10.65 45.6 10.79 45.8 10.89 46 10.92 46.2 10.98 46.4 11.05 46.6 11.12 46.8 11.16 47 11.21 47.2 11.23 47.4 11.26 47.6 11.3 47.8 11.33 48 11.35 48.2 11.37 48.4 11.4 48.6 11.42 48.8 11.44 49 11.46 49.2 11.48 49.4 11.5 49.6 11.51 49.8 11.53 50 11.54Explanation / Answer
Please find the 75% equivalence point, 25% equivalence point, Difference in equivalence points, and half equivalence point(pKa1), and molecular weight range? Then based on the results, determine the unknown acid (possibilities: acetic acid, oxalic acid, benzoic acid, maleic acid, malic acid, and KHP).
standardized NaOH(M) is 0.0840±0.0005 and Used unknown acid was 0.2005g. I did titration curve graph, and found the end point as 44.40.
This has only 1 equivalence point, so no "difference between points" exist
Volume in equivlaence point = 44 mL approx
Half equivalence point = 1/2*44 = 22 mL
pH in half equivalenc epoint = 2.8 approx
25% poiunt --> 44/4 = 11 mL
75% = 3/4*44 = 33 mL
so
pK = 2.8
molar mass = mass of acid / mol
mol of base = MV = 0.0840(44.4/1000) = 0.003729 mol of base
mol of acid = 0.003729
so...
MW = 0.2005/0.003729 = 53.767 g/mol
MWof:
acetic acid = 60
oxalic acid = N/A since diprotic
benzoic acid= 122.123
maleic acid = 116.072
malic acid = 134.0874
and KHP) =204.22
nearest answer is that of acetic acid