# 10 cm^3 of H_2O_2 solution when reacted with KI solution produced 0.5 g of iodine. Calculate the percentage purity of H_2O_2?

Jul 18, 2015

Weight by volume percent concentration: 0.67%

#### Explanation:

Let me start by saying that reacting hydrogen peroxide with potassium iodide will not produce iodine, it will only accelerate the decomposition of hydrogen peroxide to water and oxygen gas.

$2 {H}_{2} {O}_{2 \left(a q\right)} \stackrel{\textcolor{red}{K I}}{\to} 2 {H}_{2} {O}_{\left(l\right)} + {O}_{2 \left(g\right)}$

In this reaction, potassium iodide acts as a catalyst and increases the decomposition rate of the hydrogen peroxide.

To produce iodine, you need the reaction to take place in acidic medium, more specifically in a dilute sulfuric acid solution.

The balanced chemical reaction looks like this

${H}_{2} {O}_{2 \left(a q\right)} + K {I}_{\left(a q\right)} + {H}_{2} S {O}_{4 \left(a q\right)} \to {K}_{2} S {O}_{4 \left(a q\right)} + {I}_{2 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)}$

Notice the $1 : 1$ mole ratio that exists between hydrogen peroxide and iodine. This means that, regardless of how many moles of hydrogen peroxide react, the reaction will produce the same number of moles of iodine.

Assuming that the potassium iodide is not acting as a limiting reagent, you can use the mass of iodine produced to determine how many moles of hydrogen peroxide reacted.

So, use iodine's molar mass to see how many moles were produced

0.5cancel("g") * ("1 mole "I_2)/(253.809cancel("g")) = "0.00197 moles" ${I}_{2}$

This of course means that the hydrogen peroxide solution contained

0.00394cancel("moles"I_2) * ("1 mole "H_2O_2)/(1cancel("mole"I_2)) = "0.00197 moles" ${H}_{2} {O}_{2}$

Use hydrogen peroxide's molar mass to determine how many grams would contain this many moles

0.00197cancel("moles") * "34.015 g"/(1cancel("mol")) = "0.067 g" ${H}_{2} {O}_{2}$

Now, you don't have enough information to actually determine the purity of the hydrogen peroxide sample.

Here's why.

When you prepared your ${\text{10-dm}}^{3}$ solution, you presumably used a stock solution of hydrogen peroxide of known concentration.

If the solution you prepared contains 0.067 g of hydrogen peroxide, then you'd need to know the mass of hydrogen peroxide you've used to prepare this solution.

That's the only way you can determine the purity of the hydrogen peroxide.

Now, if the problem actually asked for the weight by volume percent concentration of the solution, then you can solve for it by

$\text{%w/v" = m_(H_2O_2)/V_"solution} \cdot 100$

"%w/v" = "0.067 g"/("10 cm"""^3) * 100 = color(green)("0.67% w/v")

Your solution contains 0.67 g of hydrogen peroxide for every 100 mL of solution.