Question

1 Suppose in an experiment to determine the amount of sodium hypochlorite in bleach, 0.0000443 mol KIO30.0000443 mol KIO3 were titrated with an unknown solution of Na2S2O3 and the endpoint was reached after 17.80 mL17.80 mL.

How many moles of Na2S2O3 did this require?

2. How many moles are present in a 25.00 mL sample of 0.076 M acetate?

3. A 25.00 mL solution containing 0.035 M sodium acetate is titrated with a 0.098 M solution of HCl. How many milliliters of HCl are required to reach the endpoint?

Answer 1

1. The reaction between IO3 – and KI under acidic conditions is

The reaction between 10z -and Kl under acidic conditionsis 103 - +5- + 6 H+ ---------> 31, + 3 H2O In this reaction, lodine is liberated. The liberated lodine is titrated sodium thiosulphate 12 + 2 5,03 -2 ---------> 21- +5,06-2 312 + 6 5,03 -2 ---------> 61- +35406-2 The overall reaction is 103 - +51-+6H+ + 3 12 + 6 S203 -2 ------> 3 12 + 3 H20 +61 - +35406-2 103 - +6H+ +6 S205-2 ------> 35,06-2+ 3 H20+1- Number of moles of 102 - = M*V = 0.0000443 mol According to balanced equation, One mole of 103 - reacts with 6 moles of thiosulphate So 0.0000443 mol of 103 - would react with 0.0000443 mol * 6/1 = 0.0002658 mol of thiosulphate moles of Na2S2O3 = 0.0002658 mol Volume of Na,s,o, solution = 17.8 mL = 0.0178 L Molarity of Na2S2O3 = moles/volume = 0.0002658 mol /0.0178 L = 0.014933 M Number of moles of Na2S2O3 = 0.0002658 mol