A student followed the procedure of this experiment to determine the K_sp of zin
ID: 506593 • Letter: A
Question
A student followed the procedure of this experiment to determine the K_sp of zinc(II) iodate, Zn(IO_5)_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)2: b. 0.0100 M Zn(NO_3)_2: Assuming that you performed the titrations described above using 100.00 mL samples of zinc nitrate, what volumes of 0.200 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. a. b. The actual volumes of 0.200 M potassium iodate required when the student performed the experiment were 7.54 mL and 17.82 mL, respectively. Explain.Explanation / Answer
Answer for question 2a:
it says 0.2 M potassium iodide is used.
But according to the question 1 the reaction is between zinc nitrate and potassium iodate. So the question is started on wrong foot.
It's assumed that potassium iodate is used instead of potassium iodide for finding volume of potassium iodate required for before precipitation starts happening. To find this we need concentrations of individual ions of Zinc Iodate to know its solubility product. According to solubility product give
Zn (IO3)2 -------------------------> Zn2+ + 2 IO3-
Ksp = [ Zn2+] [IO3- ]2
According to this we need to find initial and final concentration of Zn2+ ions and IO3- before and after dilutions.
According to the question Initial volume of Zn2+ ions = 0.1 M
and initial volume of Zn (NO3)2 used = 100 ml.
It is suggested final volume after adding potassium nitrate = 100 ml
Initial concentration of NO3- = 0.2 M
If no change in final volume of solution then initial concentrations used will be the final concentrations of ions. I think this assumption is wrong and takes us in wrong direction. Any way let's try to find initial concentration of Iodate ions with given data. According to dilution formula,
IC x IV = FC x FV
IV of potassium nitrate = FC x FV / IC
= 0.2 x 100 / 0.2
IV of potassium nitrate = 100 ml
However know that it's not right. So question is not structured properly.
However we may try to find Ksp of Zn ions and iodate ions with the given concentrations and see if precipitation happens at the given concentrations.
Accordingly,
Ksp of this Zinc Iodate solution = [0.1][0.2]2 = 0.004 = 4.0 x 10-3
Standard Ksp of Zinc iodate = 3.9 x 10-6
Experimental Ksp of Zinc iodate = 4.0 x 10-3
Since 4.0 x 10-3 > 3.9 x 10-6 much larger the precipitation happens immediately
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