Question

1. State and Explain how each of these errors would affect (high, low, no change) your calculated values of solubility and Ksp.

a) The saturated solution of Cu(IO₃)₂ is not filtered to remove undissolved Cu(IO₃)_2.

b) The saturated solution of Cu(IO₃)₂ is filtered into a wet flask.

c) You use only 5.00 mL of the saturated solution instead of 10.00 mL.

2. At the end point of a titration in the standardization of the Na₂S₂O₃ solution (using KIO₃), you have a colorless (and clear) solution. At the end point in the determination of Cu(IO₃)₂ solubility, you have a colorless (but cloudy) heterogeneous mixture. Explain this difference. (Hint: look at the reaction equations on the first page of the handout.)

http://www3.chem21labs.com/labfiles/UofC_GL22_Lab.pdf?rf=5018

Answer 1

1. Influence of experimental errors on the calculated values of solubility and Ksp of Cu(IO3)2

a) When the saturated solution of Cu(IO3)2 is not filtered to remove undissolved Cu(IO3)2, solubility calculated would be higher and thus the Ksp would also increase and will be higher than actual values.

b) When the saturated solution of Cu(IO3)2 is filtered into a wet flask, the water is the flask would dilute the solution and thus the solubility value would go down which would give a net low valow for the calculated Ksp.

c) If we take onlt 5.00 ml of saturated solution instead of required 10 ml, their will not be any change to the calculated solubility and the Ksp values as the concentation of Cu(IO3)2 remains the same in both volumes.

2. Cu(IO3)2 presence makes the solution cloudy due to lower solubility of it in solution as compared to the relative greater solubility of NaIO3 formed from the standardization reaction of Na2S2O3.