Procedure: Data: Questions: Part C- Titration of Equilibrium Solutions Pipette 1

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Procedure:

Data:

Questions:

Part C- Titration of Equilibrium Solutions Pipette 10.00 mL of the first filtrate prepared in Part A into a clean Erlenmeyer flask. Add 20 mL DI Water, 2 mL HCl and 1.5 grams KI I. 2. Titrate the KIO, solution with the Na2S.O, until the solution turns pale yellow (the color of hay or straw). Add 1 dropper full starch solution. Carefully continue the titration until the solution turns clear. Record the total volume of Na2S2Os used Repeat this process for a 10.00 mL aliquot of each filtrate prepared in Part A. If time permits, titrate a second aliquot of each filtrate to0i o Pat A. If time 3.

Explanation / Answer

step 1: Calculate moles of KIO3 = molarity x volume

1) 0.1 x 0.0153 = 1.53 x 10^-3 mols

2) 0.1 x 0.0149 = 1.49 x 10^-3 mols

3) 0.1 x 0.0149 = 1.49 x 10^-3 mols

step 2: Calculate moles of Na2S2O3 = moles of KIO3 x 6

1) 1.53 x 10^-3 x 6 = 9.18 x 10^-3 mols

2) 1.49 x 10^-3 x 6 = 8.94 x 10^-3 mols

3) 1.49 x 10^-3 x 6 = 8.94 x 10^-3 mols

step 3 : Calculate Molarity of Na2S2O3 = moles/L of solution

1) 9.18 x 10^-3/0.0153 = 0.6 M

2) 8.94 x 10^-3/0.0149 = 0.6 M

3) 8.94 x 10^-3/0.0149 = 0.6 M

step 4 : Calculate moles of Na2S2O3 in IInd titration

1) 0.6 x 0.0079 = 4.74 x 10^-3 mols

2) 0.6 x 0.0079 = 4.74 x 10^-3 mols

3) 0.6 x 0.0099 = 5.94 x 10^-3 mols

4) 0.6 x 0.0141 = 8.46 x 10^-3 mols

7) step 5 : Calculate [Ca2+]

moles of Ca2+ = moles of S2O3^2-

[Ca2+] = moles/volume

1) 4.74 x 10^-3/0.0079 = 0.6 + 0.1 = 0.7 M

2) 4.74 x 10^-3/0.0079 = 0.6 + 0.1 = 0.7 M

3) 5.94 x 10^-3/0.0099 = 0.6 + 0.1 = 0.7 M

4) 8.46 x 10^-3/0.0141 = 0.6 + 0.1 = 0.7 M

8) Molar solubility of Ca(IO3)2

Ksp = [Ca2+][IO3^-]^2

7.1 x 10^-7 = (0.7)[IO3^-]^2

[IO3^-] = 0.10 M

So the molar solubility of Ca(IO3)2 = 0.1 M

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