Question

Excess of KI and dil. were mixed in 50mL ${{H}_{2}}{{O}_{2}}$. The liberated ${{I}_{2}}$ required 20 mL of 0.1N $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$. Find the strength of ${{H}_{2}}{{O}_{2}}$ in g per litre.

Answer 1

Hint: Volume strength is the volume of oxygen supplied by one volume hydrogen peroxide sample on heating at STP. Here, Volume strength of hydrogen peroxide is a term used to express concentration of ${{H}_{2}}{{O}_{2}}$ in terms of volumes of oxygen gas based on its decomposition to form water and oxygen.

Complete step by step answer:
Normality is defined as the number of gram or mole equivalents of solute present in one liter of a solution. It is denoted by the letter N. Equivalent weight is known as the mass of one equivalent, i.e. the mass of a substance taken under consideration which will combine with or displace a definite quantity of another substance in a chemical reaction. For an element the equivalent weight is nothing but the mass which combines with or displaces 1.008 gram of hydrogen or 8.0 gram of oxygen or 35.5 gram of chlorine. It is also known as gram equivalents.
According to given information in the question,
 ${{M}_{eq}}$ of ${{H}_{2}}{{O}_{2}}$ = N $\times $ 50
Where N is the Normality of ${{H}_{2}}{{O}_{2}}$.
${{M}_{eq}}$ of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$ = 0.1 $\times $ 20
Therefore, ${{M}_{eq}}$ of ${{H}_{2}}{{O}_{2}}$ = ${{M}_{eq}}$ of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$
N $\times $ 50 = 0.1$\times $ 20
Hence, N = 0.04N
Therefore, Volume strength of ${{H}_{2}}{{O}_{2}}$ = N 5.6
= 0.04 $\times $ 5.6
= 0.224 g/L

Hence, the strength of ${{H}_{2}}{{O}_{2}}$ in g per litre is 0.224 g/L.

Note: To avoid calculation mistakes, convert all the quantities into SI units. Normality is not a proper unit of concentration in some situations. It is an ambiguous measure and molarity and molality are better options for units. A defined equivalence factor is essentially required by Normality.