A student followed the procedure of this experiment to determine the K_sp of zin

Question

A student followed the procedure of this experiment to determine the K_sp of zinc (II) iodate, Zn(IO_3)_2 Solutions of Zn(NO_3)_2 of known concentrations were titrated with 0.200 M KIO_3 solutions to the first appearance of a white precipitate. For each of the zinc(II) nitrate solution concentrations below, calculate the expected concentration of iodate that would be required to initiate precipitation of zinc(II) iodate. Show all calculations. (Assume that K_sp = 3.9 times 10^-6 at 25 degree C for zinc (II) iodate). a. 0.100 M Zn(NO_3)_r. b. 0.0100 MZ(NO_3) Assuming that you performed the titrations described above using 100.00 mL samples of zinc nitrate, what volumes of0.200 M potassium iodide would be required to initiate precipitation in each case? To simplify the calculations, neglect dilution effects from the added potassium iodate and assume a total final solution volume of 100.00 mL. The actual volumes of 0.200 M potassium iodate required when the student performed the experiment were 734 mL and 17.82 mL, respectively. Explain.

Explanation / Answer

2.(a) concentration of iodate has to be 0.00624 M to start precipitation. Total volume of the solution = 100mL

     So, volume of 0.2M iodate required = 0.00624 M * 100mL/0.2M = 3.12 mL

(b)volume of KIO3 required = 1.97*10^-2 M * 100mL /0.2M = 9.85 mL

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Actual volume will be greater as in the above calculation we have neglected the dilution factor.

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