Please find 75% equivalence point, 25% equivalence point, difference in equivale
ID: 488914 • Letter: P
Question
Please find 75% equivalence point, 25% equivalence point, difference in equivalence points, and molecular weight? and based on the results, determine the unknown acid (possibilities: acetic acid, oxalic acid, benzoic acid, maleic acid, malic acid, and KHP).
I did pH vs. NaOH titration curve graph and found the end point as 38.20. The standrdized NaOH(M) was 0.084±0.002 and used unknown acid was 0.6505g.
Below is a data of pH vs. NaOH titration:
V(NaOH), mL pH 0 3.8 2 3.92 2.5 3.92 3 4.02 3.5 4.06 4 4.08 4.5 4.13 5 4.15 5.5 4.18 6 4.22 6.5 4.26 7 4.29 7.5 4.31 8 4.34 8.5 4.37 9 4.41 9.5 4.44 10 4.45 10.5 4.45 11 4.53 11.5 4.55 12 4.57 12.5 4.61 13 4.62 13.5 4.65 14 4.67 14.5 4.68 15 4.71 15.5 4.75 16 4.77 16.5 4.8 17 4.83 17.5 4.83 18 4.85 18.5 4.9 19 4.9 19.5 4.92 20 4.95 20.5 4.98 21 5.01 21.5 5.04 22 5.04 22.5 5.05 23 5.07 23.5 5.09 24 5.11 24.5 5.14 25 5.17 25.5 5.19 26 5.22 26.5 5.26 27 5.3 27.5 5.31 28 5.33 28.5 5.36 29 5.39 29.2 5.42 29.4 5.43 29.6 5.44 29.8 5.45 30 5.45 30.2 5.48 30.4 5.5 30.6 5.51 30.8 5.52 31 5.54 31.2 5.56 31.4 5.59 31.6 5.61 31.8 5.62 32 5.64 32.2 5.66 32.4 5.68 32.6 5.71 32.8 5.73 33 5.85 33.2 5.77 33.4 5.79 33.6 5.8 33.8 5.82 34 5.83 34.2 5.85 34.4 5.88 34.6 5.89 34.8 5.93 35 5.96 35.2 5.98 35.4 6.02 35.6 6.05 35.8 6.08 36 6.13 36.2 6.19 36.4 6.24 36.6 6.26 36.8 6.32 37 6.37 37.2 6.44 37.3 6.53 37.4 6.64 37.5 6.68 37.6 6.76 37.7 6.68 37.8 7.07 37.9 7.23 38 7.57 38.05 8.07 38.1 8.14 38.15 8.42 38.2 8.7 38.25 8.92 38.3 9.16 38.35 9.38 38.4 9.51 38.45 9.64 38.5 9.74 38.55 9.85 38.6 9.93 38.65 9.97 38.7 10 38.8 10.27 39 10.32 39.2 10.36 39.4 10.45 39.6 10.56 39.8 10.68 40 10.78 40.5 10.92 41 11.06 41.5 11.18 42 11.24 42.5 11.32 43 11.37 43.5 11.4 44 11.47 44.5 11.52 45 11.57 45.5 11.61Explanation / Answer
The equivalence point --> 38 mL approx...
25% of equiv. point -- Z 1/4*38 = 9.5 mL
50% of equiv. point -- Z 1/2*38 = 19 mL
75% of equiv. point -- Z 3/4*38 = 28.5 mL
the pH in 1/2 point --> 4.8 approx
so
pH = pKa in 50% point... pKa = 4.8
MW = mass/mol
find moles
mol of base = MV = 0.084*(38.2/1000) = 0.0032088
mol of acid = 0.0032088
MW = 0.6505 / 0.0032088 = 202.7237 g/mol
acetic acid --> can't be since MW = 60
oxalic acid
benzoic acid = can't be since M;W = 122.12
maleic acid can0t be, MW = 116.072
malic acid = 134.0874 cant be
so.. nearest option_
KHP = 204.22 , pKa = 5.4