0.4535 g of an unknown (organic/inorganic) acid was dissolved in 50 ml of approp
ID: 963689 • 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. 0.4535 g of an unknown organic acid was dissolved in 50 ml of appropriate solvent, after addition of 1 drop of phenolphthalein indicator, it was titrated to the end point with 0.1020 M NaOH solution by using buret. The intiial reading of buret was 15.45 ml and the final reading of buret was 44.35 ml. The total volume of NaOH is 28.90 ml.
2. If all of the unknown solid was not transferred during the experiment, the resultant equivalent weight would be lower than the true value. Equivalent weight is actual amount reacted, for monoprotic acid it is directly dependent upon the mass of unknown solid taken. Thus as mass decreases, the equivalent weight also reduces.
3. For the graph
3a. End point of titration = 20 ml
3b. pKa value = 4.38
3c. equivalent weight of unknown acid = 0.2720 g/(0.1010 M x 0.020 L) = 134.6535
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