TOLC-med 2021 Q 45

100 mL of an aqueous solution containing 0,006 moles of K2S2O3 and 0,09 moles of KI are added to an acidic aqueous solution of KBrO3 (100 mL, 0,1 mol/L). The following quantitative reactions take place:

(1) 9I⁻ + BrO3⁻ + 6H⁺ — 3I3⁻ + Br⁻ + 3H2O

(2) I3⁻ + 2S2O3²⁻ — 3I⁻ + S4O6²⁻

Reaction (2) is instantaneous, while reaction (1) is relatively slow. By the time all the S2O3²⁻ anion present in the solution has reacted, how many moles of BrO3⁻ will have been consumed?

Hi Ari,
This question was very difficult for me, how can it be solved?
Thank you!

Hey! What is the answer to this question?

Sorry i forgot to add it!
it’s 0,001 moles

I found this worked solution online:
BrO3⁻ ---- 3I3⁻ ----- 3 * 2SO3²⁻
Ratio 1 : 3 : 6
so n = 0,001 : 0,003 : 0,006

It still feels strange to use so little information from the question to solve

Hey! Sorry, I didn’t get to review it during the weekends, let me share and explanation for the worked solution you saw online

From reaction (1), we can see that 9 moles of I⁻ react with 1 mole of BrO3⁻ to produce 3 moles of I3⁻ and other products.

From reaction (2), we can see that 1 mole of I3⁻ reacts with 2 moles of S2O3²⁻ to produce 3 moles of I⁻ and 1 mole of S4O6²⁻.

Now, the question tells us that by the time all the S2O3²⁻ has reacted (we have 0.006 moles of S2O3²⁻), we want to know how many moles of BrO3⁻ will have been consumed.

So before everything, let’s find out how many moles of I3⁻ would be produced from the 0.006 moles of S2O3²⁻. From reaction (2), the ratio of I3⁻ to S2O3²⁻ is 1:2. So, the amount of I3⁻ that would react with 0.006 moles of S2O3²⁻ is half that amount, or 0.003 moles.

Now, we need to find out how many moles of BrO3⁻ would be consumed to produce these 0.003 moles of I3⁻. From reaction (1), the ratio of BrO3⁻ to I3⁻ is 1:3. So, the amount of BrO3⁻ that would be consumed to produce 0.003 moles of I3⁻ is a third of that amount, or 0.001 moles.

So, by the time all the S2O3²⁻ has reacted, 0.001 moles of BrO3⁻ will have been consumed.

This is the stoichiometric relationship as given in the answer:
BrO3⁻ : I3⁻ : S2O3²⁻
Ratio 1 : 3 : 6
so n = 0.001 : 0.003 : 0.006

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