Precipitation Titration Aim The purpose of this experiment was to determine the number of moles of water of hydration in hydrated barium chloride by calculating the value of x in the formula: BaCl2, xH2O Materials
Procedure In this experiment we prepared a solution of hydrated barium chloride by weighing out approximately 1.5g of salt. We then dissolved this in 250mL of distilled water. After this we set up the burette and poured 50mL of silver nitrate into it. We then transferred 10mL of the barium chloride solution and transferred it into a conical flask with a pipette. After weighing out approximately 1g of sodium sulphate crystals we added it to the flask and swirled it until it completely dissolved, as well as adding 3 drops of the potassium chromate (VI) indicator. Finally we titrated the solution until we saw a faint red precipitate of silver chromate (VI). We then repeated the titration 3 more times.
Jennie Sansom Ms. Hay
September 30th, 2011 AP Chemistry
Results Mass of BaCl2, x H2O = 1.56 g
Final Initial Volume used Mean Titre
Trial 1 9.95 mL 0.00 mL 9.95 mL
Trial 2 19.93mL 9.95mL 9.98 mL
Trial 3 29.99 mL 19.93 mL 10.06 mL
Trial 4 41.15 mL 29.99 mL 11.16 mL
10.29 mL
Calculation (9.95+9.98+10.06+11.16) 4 Mean titre = 10.29 mL Molarity of AgNO3 = 0.05M Mols Ag+ = concentration x volume 0.05 x 0.01029 mols Ag+ = 5.145 x 10-4 Mol Ag+ = Mol Cl5.145 x 10-4 = 5.145 x 10-4 mol Cl- = 5.145 x 10-4
BaCl2, xH2O 1.58 g Mol Cl- = 5.145 x 10-4 Mol BaCl2 =5.145 x 10-4 2 -4 = 2.573 x 10 mol BaCl2 Mass= Moles x Molar Mass Mass BaCl2 anhydrous = 2.573 x 10-4 x 208.236 = 0.0536 g in 10 mL Mass of BaCl2 anhydrous in 250 mL = 0.0536 x 25 = 1.339 g Mass of x H2O = mass of hydrate – mass of anhydrous =(1.58 - 1.339) = 0.241 g H2O
Jennie Sansom Ms. Hay
September 30th, 2011 AP Chemistry
Mols of crystalline water = 0.241/18.02 = 0.013 mol H2O
Mol anhydrous BaCl2 1.339/208.236 = 0.0064
Mol H2O 0.013
0.0064 = 1 0.013 X X = 2.08 Mol H2O = 2
Conclusion After calculations we determined the number of moles of water of hydration in hydrated barium chloride by found out that the moles of H2O was 2. The actual value calculated was 2.08 mols. To calculate the percent error we used the formula : (2.08-2.00) x 100 = 4%. This could have been because of dirty conical 2.00 flasks, and inaccurate readings of measurements. However, because the percent error was very small, few errors took place. Overall the experiment was very successful and an invaluable lesson about precipitates.
