Question

Household bleach contains the hypochlorite ion which is formed when chlorine dissolves in water. To determine the amount of the hypochlorite in the bleach, the solution is first treated with a KI solution. The iodine liberated can be determined by titration with a standard thiosulphate solution. A $25ml$ of certain household bleach requires $17.4ml$ of a $0.02M-N{{a}_{2}}{{S}_{2}}{{O}_{3}}$ solution for titration. The mass of chlorine dissolved in one liter of the bleach solution is:
(a) $\text{0}\text{.1392g}$
(b) $\text{0}\text{.494g}$
(c) $\text{9}\text{.88g}$
(d) $\text{0}\text{.278g}$

Answer 1

Hint: First of all, we will find the molarity of the hypochlorite ion by using the molarity equation as; $\begin{align}
 & {{M}_{1}}{{V}_{1}}\text{ }={{M}_{2}}{{V}_{2}} \\
 & (Cl{{O}^{-}})\text{ (}\text{thiosulphate}) \\
\end{align}$
and then, by using this molarity we will find the moles of the chlorine and then by using them, we can easily calculate the weight of the chlorine dissolved in one liter of the solution by using the formula as; $mole=\dfrac{given\text{ }weight}{molecular\text{ }weight}$.
 Now solve it.

Complete step by step answer:
- When chlorine is dissolved in the water, it results in the formation of the bleach consisting of the hypochlorite ion. The reaction is supposed to occur as;
$2{{H}_{2}}O+C{{l}_{2}}\to 2Cl{{O}^{-}}+2{{H}_{2}}$ ---------(1)
- To know the amount of the hypochlorite in the bleach, the solution is treated with the potassium iodide solution and the iodine liberated is determined by titration it with the standard thiosulphate solution. the reaction occurs as;
$2N{{a}_{2}}{{S}_{2}}{{O}_{2}}+{{I}_{2}}\to N{{a}_{2}}{{S}_{4}}{{O}_{6}}+NaI$ ---------(2)
- Now, first of all ,we will find the molarity of the hypochlorite ion by using the molarity equation as;
$\begin{align}
 & {{M}_{1}}{{V}_{1}}\text{ }={{M}_{2}}{{V}_{2}} \\
 & (Cl{{O}^{-}})\text{ (}\text{thiosulphate}) \\
\end{align}$ --------------(1)
$\begin{align}
 & {{V}_{1}}=25\text{ ml (given)} \\
 & {{\text{M}}_{2}}=0.02\ \text{ (given)} \\
 & {{\text{V}}_{2}}=17.4\text{ ml (given)} \\
\end{align}$
Put all, these values in equation(1), we get;
$\begin{align}
 & {{M}_{1}}\times 25=0.02\times 17.4 \\
 & {{M}_{1}}=\dfrac{0.348}{25}=0.01392M \\
\end{align}$
- Now, using the molarity of the hypochlorite ion, we will find the moles of the chlorine.
From equation (1), we comes to know that;
For $2$ moles of $Cl{{O}^{-}}$, we require =$1$ moles of $C{{l}_{2}}$
For $1$ moles of $Cl{{O}^{-}}$, we require =$\dfrac{1}{2}$ moles of $C{{l}_{2}}$
Then, for $0.01392$ moles of $Cl{{O}^{-}}$, we require = $\dfrac{1}{2}\times 0.01392 = 0.00696$ moles of $C{{l}_{2}}$

- Now, we know the number of moles of the chlorine and by using the mole formula we can easily calculate, the weight of the chlorine as;
$\begin{align}
 & mole=\dfrac{given\text{ }weight}{molecular\text{ }weight} \\
 & given\text{ }weight=mole\times molecular\text{ }weight-------(3) \\
\end{align}$
We know that;
The molecular weight of chlorine i.e.$C{{l}_{2}}$=71
- Now using the equation (3), we can calculate the given mass as;
$\begin{align}
 & given\text{ }weight\text{ }of\text{ C}{{\text{l}}_{2}} = 0.00696\times 71 \\
 & \text{ =0}\text{.494g} \\
\end{align}$
Thus, the mass of chlorine dissolved in one liter of the bleach solution is $\text{0}\text{.494g}$. The correct option is option “B” .

Note: Always remember that the no. of moles of each gas at the standard conditions of the temperature and pressure, contains Avogadro number of particles and occupies the volume of $22.4\text{ liters}$.