Question

3. In an experiment, 10.00 mL of Cu(IO3)2 were mixed with a scoop of KI, a scoop of sodium citrate, and small amount of sulfuric acid. 15.22 mL of 0.0473 M Na S203 were required to titrate to the endpoint. What is the concentration of [103] in the original solution? Continued on back 265

Answer 1

Below is balanced chemical reaction;

IO₃^- + 6S₂O₃^2- + 6H⁺= 3S₄O₆^2- + I^- + 3H₂O

Mili Moles of S2O3 2- = 0.0473 * 15.22 = 0.7199

From reaction;

6 milimoles of S2O3 2- requires 1 milimole IO3-

So, 0.7199 milimoles S2O3 2- will require = 0.7199/6 = 0.11998 milimoles of IO3 -

So, milimoles of IO3 - = 0.11998

Volume of IO3 - = 10 ml

So, concentration of IO3- in original solution = 0.11998/10 = 0.011998 = 0.012 M .....Answer

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