To measure the mass of the precipitate (lead iodide) when potassium iodide solution is reacted with lead nitrate solution.

Hypothesis: - The given equation -

Pb(NO3)2(aq)        +        2KI(aq)         →        2KNO3(aq)        +        PbI2(ppt)

According to the given equation when 1 part of Lead Nitrate reacts with 2 parts of Potassium Iodide we get 1 part of Lead Iodide. Therefore if we use 20ml of Potassium Iodide then we will need to use more than 10ml of Lead Nitrate as it is supposed to be in excess. According to the equation then we would get around 0.46 g of Lead Iodide.

Procedure: - Pipette Potassium Iodide solution as well as Lead Nitrate solution into a beaker in the ratio 2:1. Stir the mixture continuously. Weigh a piece of filter paper and fold it into 4 parts and then make it in a conical shape. Fit the folded filter paper into the funnel and place a beaker below it. Now distill the mixture slowly to get the precipitate on the filter paper. Add distilled water to the solution to remove all the precipitate from the mixture in the beaker. Once this is done remove the wet filter paper and put it in the china dish to dry. It can even be put in the oven at around 60o C to dry for some time. After it is completely dry re-weigh it and find out the difference in the weights of the filter paper and the one with the precipitate on it to get the weight of the Lead Iodide precipitate.

Evaluation: -

Average = 0.445 g = 0.45 g

There is as error in the reading as the reading is not as expected-

% error = ((0.46 - 0.45)/0.46) * 100 = 2.17 % error

Conclusion: - The precipitate had an average weight of 0.45 g which was just 1 g lesser than the expected value of the precipitate and hence there was an error of 2.17%. This error can be corrected if more care is taken while performing the experiment so as to avoid the systematic errors. There are always random errors some of which cant be avoided.