Subject: Chemistry SL
Title: Preparation of a salt (lead iodide) by using precipitation method. Aim: To produce 1.50 g of lead iodide, PbI2, assuming 100 % conversion, using 1.00 g solid potassium iodide, KI, and 0.1 M lead nitrate, Pb(NO3)2 solution of volume 30 cm3 and hence calculate percentage yield. Calculations:
2KI(s) + Pb(NO3)2( aq) —> PbI2 (s) + 2KNO3 (aq)
Mass of PbI2 to be produced = 1.5 g
Moles of PbI2 = = 0.003
Moles of KI = 0.003*2 = 0.006
Mass of KI = 0.006*166 = 1.00 g
Moles Pb(NO3)2 = 0.003
Volume Pb(NO3)2 = = 0.03 dm3 = 30 cm3
Reaction: 2KI(s) + Pb(NO3)2( aq) —> PbI2 (s) + 2KNO3 (aq)
Lead (II) iodide (PbI2) or plumbous iodide is a bright yellow solid at room temperature. Lead iodide is toxic due to its lead content. Lead iodide can be obtained as a yellow precipitate by mixing solutions of lead (II) nitrate and potassium iodide. It is sparingly soluble in cold water but quite soluble in hot water, yielding a colorless solution; on cooling it crystallizes as yellow hexagonal platelets. Source: http://www.britannica.com/EBchecked/topic/333615/lead-iodide Apparatus:
1 measuring cylinder (50cm3)
1 weighing scale
1 conical flask
1 filter paper
1 petri dish
Solid potassium iodide (KI)
0.1 M lead nitrate [Pb(NO3)2]
Wear safety glasses to prevent lead iodide entering your eye. 2.
Wear rubber gloves while handling the lead iodide to avoid skin contact. 3.
Make sure the lab is well-ventilated so that you do not inhale the lead iodide powder. Procedure:
Measure 30 cm3 of 0.1 M lead nitrate, using a measuring cylinder, and pour it into a conical flask. 2.
Weigh 1.00 grams of solid potassium iodide powder and using a weighing paper put the powder into the flask. 3.
Wait for the reaction to get completed. Shake the flask regularly. 4.
Weigh the filter paper.
Repeatedly filter the yellow precipitate of lead iodide formed by...
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