Question

The amount of ozone in a mixture of gases can be determined by passing the mixture through an acidic aqueous solution of potassium iodide, where the ozone reacts according to

${\text{O}}_3\left(g\right)+3{\text{I}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(\ell \right)\to {\text{O}}_2\left(g\right)+{\text{I}}_3^{-}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)$

to form the triiodide ion${\text{I}}_3^{-}$. The amount of triiodide produced is then determined by titrating with thiosulfate solution:

${\text{I}}_3^{-}\left(aq\right)+2{\text{S}}_2{\text{O}}_3^{2-}\left(aq\right)\to 3{\text{I}}^{-}\left(aq\right)+{\text{S}}_4{\text{O}}_6^{2-}\left(aq\right)$

The solution requires 26.2 mL of a 0.1359-M solution of thiosulfate ion to titrate to the endpoint. Calculate the mole fraction of ozone in the original gas sample. A small amount of starch solution is added as an indicator because it forms a deep-blue complex with the triiodide solution. Disappearance of the blue color thus signals the completion of the titration. Suppose 53.2 L of a gas mixture at a temperature of${18}^{°}\mathit{C}$and a total pressure of 0.993 atm is passed through a solution of potassium iodide until the ozone in the mixture has reacted completely. The solution requires 26.2 mL of a 0.1359-M solution of thiosulfate ion to titrate to the endpoint. Calculate the mole fraction of ozone in the original gas sample.

Short answer

The mole fraction of the ozone in the original gas sample is$0.9968$.

Step-by-step solution

Given information

The ozone reaction after passing through the acidic solution is,

${\text{O}}_3\left(g\right)+3{\text{I}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(\ell \right)\to {\text{O}}_2\left(g\right)+{\text{I}}_3^{-}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)$

According to the ideal gas law:.

$\mathbf{Pressure}\left(\mathbf{P}\right)\mathbf{\times }\mathbf{Volume}\left(\mathbf{V}\right)\mathbf{=}\mathbf{numberofmoles}\left(\mathbf{n}\right)\mathbf{\times }\mathbf{gasconstant}\left(\mathbf{R}\right)\mathbf{\times }\mathbf{Temperature}\left(\mathbf{T}\right)$

Where the values of$T$, $R$,$P$and $V$are:

Rearrange the equation to find the number of moles:

$$\begin{gathered} \text{n}=\frac{PV}{RT} \\ n=\frac{0.993\text{atm}\times 53.2\text{K}}{0.0821{\text{Katmmol}}^{-1}{\text{K}}^{-1}\times 291\text{K}} \\ =2.2112\text{mol} \end{gathered}$$

Concept of acid-base titrations

The basic idea behind titration is that if you know the stoichiometry of a reaction and the quantity of one species, you can calculate the quantity of the other.

There are various sorts of titrations (acid-base, redox), but they all require a reaction in which we incrementally add one reactant until they are in stoichiometric proportions, and then we may compute the moles of the analyte using the titrant's moles.

Calculate the value of moles

The formula for the number of moles (n) of thiosulphate ion:

$n=Molarity\times Volume$

role="math" localid="1663403809162" $$\begin{gathered} \text{Molarity(thiosulfate ion)}=0.1359\text{M} \\ =0.1359\text{mol}/\text{L} \\ Volume=\frac{26.2\text{mL}}{1000\text{mL}/\text{L}}=0.0262\text{L} \end{gathered}$$

Substitute the values in the above formula.

$$\begin{gathered} n=0.1359\text{mol}/\text{L}\times 0.0262\text{L} \\ =0.00356\text{mol} \end{gathered}$$

Reaction of ozone with iodine

In chemical reaction, 1 mole of ozone reacts with 3 moles of iodine forms one mole of $I_3^{-}$and $l_3$one mole of reacts with 2 moles of sodium thiosulfate.

$$\begin{gathered} n\left(1\right)=2\times n\left(thiosulphate\right) \\ =2\times 0.00356\text{mol} \\ =0.00712\text{mol} \end{gathered}$$

Calculation of molar fraction

The formula to calculate mole ratio is,

$\text{Mole ratio =}\frac{\text{moles of Ozone}}{\text{moles of Ozone + moles of lodide lon}}$

The known values are substituted.