Pre-Lab Questions A studen followedhoeperiment to determine the K, of inctt iode

Question

Pre-Lab Questions A studen followedhoeperiment to determine the K, of inctt iode Zn(10h. Solutions of Zn(NO), of known concentrations were titrated with 0.200 M Kio, solutions to the first appearane of a white precipitate. For each of the zincan nitrate solution concentrations below, calculate the expected concentration of iodate that would be required to initiate precipitation of zinc(I) iodate. Show all calculations. (Assume that K-3.9 x 106 at 25°C a 0.100 M Zn(NO Lo 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 caculations, neglect dilution effects from the added potassium iodate and assume a total final solution volume of 100.00 mL 2. Vito con v: 7.22 V1.nya.. 0.01 ? yo. 2 t 110 mL Kl - X 0.2 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. 3. Determining Ksp of Lead II) lodide 5

Explanation / Answer

Given the same concentration of potassium iodate solution, which is stated to be 0.200M, you would need a smaller volume (7.54 ml) for the first solution to precipitate, and a bigger one (17.82 ml) for the second solution to do so.

This is mainly because of the concentrations for each initial solution. Given that the first is 0.100M, it can be seen that it is more concentrated than the other second one. Given so, it would take less amount of the other ionic solution for it to start precipitation. So, with a smaller amount of ions, titration would be complete, observing the beginning of precipitation, because the initial concentration of the solution is higher.

In the second solution, the concentration is way lower, (0.010 M). In fact, it's 10% of the first. In this case, it would take a higher volume of ionic solution for it to start precipitation. As it is not so concentrated initially, more ions need to be added to increase concentration and therefore induce precipitation in the solution.

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