The purpose of this lab was to understand and observe the solubility constant. In order to do this, we set up a dilution experiment with Ca(NO3)2. In the first well, A1, there was 5 drops of Ca(NO3)2. In well A2-A7, 4 drops of distilled water was added. A fine tip pipet was then used to put 2 mini drops from well A1 into A2 and then the excess solution was put back in the orin=gnial well. Well A2 was mixed and the process was repeated. By doing this, the solutions from A1 to A7 was progressively diluted by ⅕. Next, 1 drop of Na2C2O4 was put into each well and mixed with a toothpick. We then observed for the formation of precipitants in each of the wells. We discovered that as the concentration of Ca(NO3)2 decreased, less precipitate formed. …show more content…
It is important to determine which was the first well without a precipitate because it can give us the Ksp value. We used the concentration of A6, 3.2*10^-5, to get a ksp of 1*10^-9, very close to the accepted value of 2.3*10^-9. This gives me a 17.4% error. In our own experiment, we test how ph would affect solubility. Our team predicted that increasing the PH would decrease the amount of precipitate. We put 4 drops of Ca(NO3)2 and 1 drop of Na2C2O4 in wells C1-C5. Next, 1 drop of KOH was added to well C2 and an extra drop was put into the proceeding wells until we reached well C5. We observed that when we increased the PH of the solution, it did not have an effect on the formation of precipitates, which disproved out hypothesis. It seems like if the original compound had an H+ or OH-, then PH would have an effect on the solubility, probably due to the common ion effect. This procedure is quite accurate when determining the solubility product constant. In order to get the exact Ksp value, we would need a smaller dilution factor for the future. Also, to prove that PH did not have an effect on these compound due to nonsimilar ions, an experiment should be set up with compounds that have H+ or OH- in them and compare them to compounds that