Question

Assume you mixed 20.00mL of 0.040M KI with 20.00mL of 0.060M (NH4)2S2O8, 10.00mL of 0.00070M Na2S2O3, and a few drops of strach. The point of mixing sets time=0. b) The basis of the "Method of Initial Rates" used in this experiement, is that the concentrations of reagents are essentially unchanged during the measurement time period. Calculate the % of initial (NH4)2S2O8 that has reacted when the blue colour appears... Need help specifically with part b. :-)

Answer 1

It should be desirable to look into underlying theoretical principle to solve this problem ,as summarized below-

The reaction involved in this case is between persulphate ion S2O8^2- and iodide ion I^- in aqueous solution . The net ionic equation for this reaction is

S2O8^2- (aq) + 3I^- (aq) <----------> 2SO4^2- + I3^- (aq)

The source of reactant ions in this case will be ammonium per sulphate (NH4)2 S2O8 and KI . Both are soluble salts ,so ammonium ions NH4⁺ & K⁺ are spectator ions and do not participtate in the reaction.

Again , since all the species involved in this reaction are colorless, starch solution is added to the reaction mixture. As soon as the tri-iodide ion I3 ^- reaches an appreciable concentration it will react with any starch present in solution to produce a dep blue complex. Thiosulphate ion , S2O3 ^-2 is added to react with the tri-iodide ion produced and thus prevent the tri-iodide ion from reacting with starch to form blue colored complex until all thiosulphate ions are consumed.

2S2O3 ^-2 (aq) + I3^- (aq) ,<---------------> S4O6 ^-2 (aq) + 3I ^-

Thus , the blue color cannot form until the thiosulphate is consumed and the triiodide ion concentration can then build up very quickly ,reacting with starch to give to form a dark blue color. In this experiment the amount of thiosulphate ion will be constant.

In order to get the change in persulphate ion concentration [S2O8 ^-2  ] we will need to start with the knwledge of of the amount of thiosulphate ion, S2O3^-2 consumed and use stoichiometry to convert that to the amount of per sulphate consumed. Since the volume and concentration of thiosulphate added is constant (10.00 ml of 0.00070 M Na2S2O3 ) , the moles of thiosulphate can be calculated as -

moles S2O3^-2   = ( 10 x 10^-3 L ) ( 7 x 10^-4 M ) = 7.0 x 10^-6 moles of S2O3 ^-2

Again the number of moles of persulphate consumed is also constant which is calculated using stoichiometric relations in equation given above.

Thus ,

moles S2O8 ^-2  = 7.0 x 10 ^-6 moles S2O3 ^-2   x { 1 mole I3 ^-  / 2moles S2O3 ^2-  } x { 1mole S2O8 ^-2  / 1mole I3^ - }

--------------------- = 3.5 x 10 ^-5  moles of S2O8 ^-2

Further ,the desired percentage of initial (NH4)2S2O8 when the blue color appears would be = 50%