# Calculating molarity of iodine solution after reaction

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 $$\pu{500ml}$$?

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 $$\ce{KI}$$.
• 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 $\ce{KI}$. Jul 6 at 16:20
• 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 $$\ce{KIO3 + 8 KI + 3 H2SO4 -> 3 KI3 + 3 K2SO4 + 3 H2O}$$ Jul 8 at 14:16