Solubility Product of Calcium Iodate

    What's another name for "The Great Terrain Robbery?" "Erosion!" As you are aware, water will eventually dissolve most things. Need proof? The Grand Canyon is the product of both dissolution and particulate erosion. In the preceding weeks you have seen how some substances are more soluble than others - lead chloride is much more soluble than is silver chloride. If we take a look at this process:

PbCl2 ↔ Pb++ + 2 Cl-

    This sure looks like a set-up ripe with mathematics!

    But there is a problem! What's the concentration of the precipitate? It doesn't matter whether there's a speck or a tonne.* So, by convention, we ignor the denominator and set it to "1", and get:

for saturated solutions.

    Because we are familiar with lead chloride, we'd like to use it in today's exercise. BUT there are some problems. Chloride is hard to quantitate, and lead is expensive. Furthermore, when we're done for the day, dumping lead down the drain isn't very environmentally friendly. So that's why we're switching to calcium iodate, which is environmentally friendly, cheap, AND - very importantly, we can easily quantitate iodate due to its oxidizing potentials. Hence:

  1. Solubility in Water

    1. Obtain 1 mL of 1M calcium nitrate, Ca(NO3)2, and 10 mL of 0.2M potassium iodate, KIO3, in separate, clean beakers. Add the KIO3 solution to the Ca(NO3)2 solution. Stir the mixture vigorously with a glass rod, You should obtain a white precipitate (ppt) of calcium iodate, Ca(IO3)2.
    2. Let the mixture stand for about ten minutes. Then filter it through a filter paper in a filter funnel. Rinse all the precipitate from the beaker into the paper using a distilled-water wash bottle.
    3. Rinse the ppt on the filter with two small portions of distilled water. This rinses away excess soluble salts from the ppt. (A so-called "distilled-water wash.")
    4. Remove the paper from the funnel, unfold it, and using a stirring rod, carefully scrape the ppt into a clean, DRY, labeled 125-mL Erlenheyer (conical) flask that is labeled #1.
    5. Add about 50 mL of distilled water to the flask, swirl.

  2. Solubility in Salt Solutions

    1. Repeat steps 1, 2 and 3, above, in order to prepare a total of three MORE samples of Ca(IO3)2 on filters. But this time you are going to rinse each of the filters with other solutions instead of distilled water. The purpose of these "washes" is to make sure than no excess water is present in the interstitial spaces among the grains of ppt BECAUSE you want to see the effects of superimposed ions on the dissolution. Namely, do these added ions in solution "push" the reaction in the opposite, leftward, direction if effect causing the ppt to dissolve less? Do the following with these three solutions:
      1. Rinse one with 0.004M KIO3
      2. Rinse one with 0.020M Ca(NO3)2
      3. Rinse one with 0.040M Ca(NO3)2
    2. Transfer the wet ppts to individual 125-mL flasks labeled #2, #3, and #4.
    3. Add about 50 mL of their appropriate salt solutions to the ppts and swirl.

  3. Equilibration and Analysis

    1. Let the flasks stand for 30 to 60 minutes, swirling them frequently. While they are coming to dissociative equilibrium, carry out the qualitative steps.
    2. After the 30 to 60 minute period has passed, assume that your mixtures have indeed reached equilibration. Now proceed as follows:
    3. The solutions you are going to analyze must NOT be diluted with even the tiniest amount of water. It is important to use dry apparatus ONLY. Using a dry funnel and dry filter paper, filter some of the mixtures into clean, and appropriately numbered beakers.
    4. Using a pipet (which is far more precise than a graduated cylinder), transfer 10 mL into a numbered 125 mL flask.
    5. Add about 10 ml water, and 10 mL of 0.2M KI (also with a pipet) and a few drops of starch solution.
    6. From a buret, titrate with continuous swirling with standardized 0.025M sodium thiosulfate, Na2S2O3 until the blue-black color just disappears. Your instructor will give you the actual accurate concentration of the thiosulfate, the odor of which you might recognize.
    7. If you overshoot your titrations, do them again!
    8. You have now titrated all four of your preparations.
    9. Determine how many millimoles, mmoles, of thiosulfate were used for each preparations. (Just multiply the mLs used times the concentration of the thiosulfate → mmoles.)
    10. So-, what's the concentration of the iodate? Let's look at the chemistry involved in this titration (the second equation has been tripled so the 3 I3- cancel when the two equations are added together!):

      IO3- + 8 I- + 6 H+ → 3 I3- + 3 H2O

      3 I3- + 6 S203-2 → 9 I- + 3 S4O6-2

      and summing them:

      IO3- + 6 S2O3-2 + 6 H+ → I- + 3 S4O6-2 + H2O

          We thus see that we needed 6 thiosulfates for every iodate. Hence, the amount of iodate was one-sixth the amount of thiosulfate used in the titration.

    11. And what is the amount of calcium ion? Obviously, one-half that of the iodate.
    12. And what were the volumes of solution analyzed? 10.0 mL
    13. Now what are the concenrations of the Ca++ and the iodate?
      mmoles/10 mL ⇒ how many moles per liter? (Hint: if you figure the mmoles/mL, the numbers are the same for moles/liter!)
    14. Finally, plug your numbers into this equation:

      and fill in all the numbers on the data table

    * "tonne" is the "long ton" = 2,200 lbs = 1,000 kilograms = 1 megagram. The short ton = 2,000 lbs.