In a titration experiment, the concentration 0.1202 mol/L of Na 2 S 2 O 3 is tit

Question

In a titration experiment, the concentration 0.1202 mol/L of Na2S2O3 is titrated in an analyte with a mixture of 5ml of 0.2043 mol/L of Kio3, 10ml of 0.5mol/l of KI and 10ml of 1.0 mol/l of HCl.

In this experiment titration will be used to determine the coefficients of a chemical equation for the reaction between iodate (IO), iodide (I) and acid (H'). When these three species are mixed, iodine is produced and the solution turns dark brown. The unbalanced net ionic equation is: 10,-(aq) + I(aq) + H+(aq) 12(aq) + H2O(l) An accurately known number of moles of IO, (as potassium iodate; the limiting reagent) is mixed with an excess of I (as potassium iodide) and H' (as hydrochloric acid). Because it is the "limiting reagent", the reaction completely consumes the IO,. The number of moles of IO consumed by the reaction is the same as the number of moles of IO added. The number of moles of iodine produced by reaction (2.1) will be determined by titration. lodine reacts with thiosulphate according to: 12(aq) + 2S2032-(aq) S4062-(aq) + 21(aq) S,O, -(aq), S,O(a) and I are all colourless while L2(aq) is dark brown. 1. Calculate the number ofmoles of Na2S20, present in the average volume of Na,S20,. From this calculate the number of moles of molecular iodine produced by reaction (2.1). 2. All of the iodine produced by reaction (2.1) is in the form of I2 (molecular iodine). Calculate the number of moles of atomic iodine (I) produced by reaction (2.1). 3.Calculate the number of moles of KIO, consumed by reaction (2.1). Remember, it is the limiting reagent. 4. Calculate the number of moles of I that must have been consumed by reaction (2.1). 5. Determine the ratio for IO,:I: I2 for reaction (2.1). Express this as a ratio of integers. Using the results for question 5 complete the balancing of: IO,(aq) + 1-(aq) +--H'(aq) 12(aq) + ___1120(1) In this experiment you were given a precise concentration for the KIO, solution but not for the KI solution. Explain why it is not necessary to know precisely the number of moles of KI added.

Explanation / Answer

KIO3 + 5 KI + 6 HCl ---> 3 I2 + 6 KCl + 3 H2O (1)
2 Na2S2O3 + I2 ---> 2 NaI + Na2S4O6 (2)

You usually take an excess of KI and you might not necessarily know the amount of it, as you're interested in the amount of I2 which you determine by the second reaction when titrating it with Na2S2O3. Then when you find the number of moles of I2 from the second reaction, you can go back to the first one and calculate the number of moles of KIO3 without considering the amount of KI.

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