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I made an iodine solution referring to the methodology provided in this link: https://www.thoughtco.com/vitamin-c-determination-by-iodine-titration-606322

$$\ce{KIO3 + 5KI + 3H2SO4->3I2 + 3H2O + 3K2SO4}$$

I want to calculate the molarity of the iodine solution produced from reacting $\pu{0.268g}$ of potassium iodate and $\pu{5g}$ of potassium iodide. I'm not sure how to calculate that.

I tried finding the limiting reactant and calculating the number of moles based off of that, but I'm not sure if that's right. Also, in order to calculate molarity you need volume. Based on the procedure outlined in the link, first I dissolved $\ce{KI}$ and $\ce{KIO_3}$ in $\pu{200ml}$ of water, but then I'd to add $\pu{30ml}$ of sulfuric acid and make it up to $\pu{500ml}$. So, I wonder what volume should I take into account when calculating molarity, $\pu{200ml}$ or $\pu{500ml}$?

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You must calculate the molarity in the final $\pu{500 mL}$ solution. If you do it in the $\pu{200 mL}$ solution, it is no use, because there is no reaction, and so no iodine $\ce{I2}$ in the solution containing just $\ce{KI + KIO3}$, and no acid.

Now, if you want to solve this problem, the first thing to do is to calculate the amount of $\ce{KI}$ and $\ce{KIO3}$ in moles. Well ! $\ce{ 0.268 g KIO3}$ contains $0.268 g/214 = 1.252·10^{-3}$ mol $\ce{KIO3}$. Do the same calculation for $\ce{KI}$. You'll see that this product is present in rather large excess (about $25$ times too much). The only important product is the potassium iodate.

Mixed with $\ce{KI}$ and $\ce{H2SO4}$, $\ce{KIO3}$ reacts according to the equation given in your reference : $\ce{KIO3 + 5 KI + 3 H2SO4 -> 3 I2 + 3 K2SO4 + 3 H2O}$. As a final result, the amount of iodine $\ce{I2}$ is $3$ times the amount of iodate. So you will obtain $3·1.252·10^{-3}$ mol $\ce{I2}$. We will not do all the calculation. Do the calculation yourself ! You will obtain $a$ mol. This iodine reacts with the iodide ion in excess according to $\ce{I2 + I^- -> I3^-}$, which can also be written $\ce{I2 + KI -> KI3}$. As a consequence, you obtain a solution containing the same amount $a$ of $\ce{I3^-}$ (in moles) dissolved in the final volume $\pu{500 mL}$. Do the calculation of the concentration $\ce{[I3^-]}$ which is numerically equal to the concentration $\ce{[I2]}$. Some teachers wants the student calculate the amount of $\ce{I2}$ produced before reaction with $\ce{KI}$.

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  • $\begingroup$ Hello, I did the calculations and found the answer to be 0.0075M. Now, I wanna ask the possibility of calculating molarity of this iodine solution through titration. I've already done the experiment where I titrated it with standard ascorbic acid solution. I did some calculation and found the answer to be somewhere around 0.005M. Why is this so? I expected the answer to be the same (0.0075M) since it was the same iodine solution...do you think its just a calculation mistake or is it wrong to find molarity through titration for this problem? $\endgroup$
    – ninetysix
    Commented Jul 4, 2021 at 23:01
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    $\begingroup$ Of course the molarity of the iodine solution can be calculated by titration, provided you are sure of the concentration of the ascorbic acid. And you should get the same result. $\endgroup$
    – Maurice
    Commented Jul 5, 2021 at 12:33
  • $\begingroup$ I see, so there must be a mistake in my calculations. I'll check that. Btw, I'd like some clarification on your answer. You said that mixed with KI and H2SO4, KIO3 produces I2. And then this reacts again with KI to produce I3− (triiodide). First question, does KIO3 react with only H2SO4 to produce I2 or do all three KIO3, KI and H2SO4 react to produce I2? Secondly, why does I2 react again with KI to produce I3, is it because KI is in excess? If the final product is I3-, then why is I2 included in the equation and not I3-? $\endgroup$
    – ninetysix
    Commented Jul 6, 2021 at 12:59
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    $\begingroup$ In solution $\ce{KIO3}$ does not react with $\ce{H2SO4}$ or with $\ce{KI}$ alone. In solution $\ce{KI}$ does not react with $\ce{H2SO4}$ alone. But, if mixed together, the three substances $\ce{KIO3, KI, H2SO4}$ do react and produce some $\ce{I2}$. Now $\ce{KI}$ reacts with $\ce{I2}$ because it is in excess.. In fact, if you know that $\ce{KI}$ is in excess, you can add the two equations. But if $\ce{KI}$ is not in great excess, the reaction produces $\ce{I2}$ that makes a precipitate, as $\ce{I2}$ is poorly soluble in water. So you choose the equation if you know of the excess of $\ce{KI}$. $\endgroup$
    – Maurice
    Commented Jul 6, 2021 at 16:20
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    $\begingroup$ NO ! Add the equations, as if they were algebraic equations. Add $\ce{3 I2 + 3 I- ->3 I3^{-}}$ to your first equation. Add the two left-hand sides together. Add the two right-hand sides together. Suppress the $\ce{I2}$ in excess. It gives $$\ce{KIO3 + 8 KI + 3 H2SO4 -> 3 KI3 + 3 K2SO4 + 3 H2O}$$ $\endgroup$
    – Maurice
    Commented Jul 8, 2021 at 14:16

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