0.4535 g of an unknown (organic/inorganic) acid was dissolved in 50 ml of approp
ID: 963688 • Letter: 0
Question
0.4535 g of an unknown (organic/inorganic) acid was dissolved in 50 ml of appropriate solvent, after addition of one drop of phenolphthalein (indicator/marker), it was titrated to the end point with a 0.1020 M MaOH solution by using (buret/pipette). The initial reading is 15.45 ml; the final reading is at end point is 44.35ml. The total volume of NaOH is ml. Please calculate the equivalent weight of the acid to four significant figures. If the student didn't transfer all the unknown solid into the titration container (flask or beaker), the calculated equivalent weight will be bigger or smaller than the true value? Please explain why ? Following graph is a PH titration cure.Explanation / Answer
1.
Volume of NaOH required = 44.35 - 15.45 = 28.9 mL
Moles of NaOH added = M * V = 0.102 * 28.9 mmoles
= 2.95 mmoles
Moles of NaOH added = Moles of unknown acid
2.95 x 10-3 = 0.4535 / MW
Equivalent weight of acid, MW = 153.8 g/equivalent of H+
2.
In that case the moles of NaOH required will be less. Hence the equivalent weight of acid calculated will be bigger.
3.
At the end point, the pH titration curve rises steeply.
3a. End point = 20 mL
3b.
HAc --> H+ + Ac-
pH = pKa when [HAc] = [Ac-] i.e. halfway equivalence point (10 mL)
pKa = 4.38
3c.
Initially x = [H+] = 10-3.19 = 6.46 x 10-4 M
Ka = 10-4.38 = 4.17 x 10-5
Ka = x2 / (C- x)
4.17 x 10-5 = (6.46 x 10-4)2 / (C - 6.46 x 10-4)
C = 0.01 M
Moles of acid = C * V = 0.01 * 100 / 1000 = 0.001 moles = 0.272 / MW
Equivalent weight of acid, MW = 272 g/equivalent of H+
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