Question

A volume of $10\,ml$ of a ${H_2}S{O_4}$ solution is diluted to $100\,ml$. Twenty-five milliliters of this diluted solution is mixed with $50\,ml$ of $0.5\,N\,NaOH$ solution. The resulting solution requires $0.205\,g\,N{a_2}C{O_3}$ for complete neutralization. The normality of the original ${H_2}S{O_4}$ solution is.
A)$12\,N$.
B)$11\,N$.
C)$3N$.
D)$0.275\,N$.

Answer 1

Hint: We know, normality of a solution is gram equivalent of solute in one liter of solution. The unit of normality is $e{q^{ - 1}}.$
The mathematical expression of normality is,
$N = \dfrac{{weight\,of\,Solute\,\left( g \right)}}{{Equivalent\,weight\, \times \,Volume\,\left( L \right)}}$

Complete step by step answer:Given,
We know titration is the process of the addition of a solution of known concentration and volume with other solutions of unknown concentration until the reaction attains neutralization. To find the normality of the acid and base titration we can use the relation.
${N_1}{V_1} = {N_1}{V_2}$
Where,
The normality of the acidic solution is ${N_1}$.
The volume of the acidic solution is ${V_1}$.
The normality of the basic solution is ${N_2}$.
The volume of the basic is ${V_2}$.
A volume of $10\,ml$ of sulfuric acid is dissolved in 100 ml. Thus, the normality of sulfuric acid can be found by using the relation.
$N{V_1} = {N_1}{V_2}$
$
  N\left( {10\,ml} \right) = N\left( {100\,ml} \right) \\
  {N_{_1}} = 0.1N \\
 $
Now, a volume of $25\,ml$ $0.1\,N\,{H_2}S{O_4}$ react with $50\,ml$ of $0.5\,N\,NaOH$
After that, sodium carbonate required $0.265\,g$ for complete neutralization.
Milli equivalent of $N{a_2}C{O_3} = \dfrac{{0.265}}{{106/2}} \times 1000 = 5$
The normality of original sulfuric acid is calculated by,
$
  0.1\,N \times 25 - 50 \times 0.5 = 5 \\
  2.5N - 25 = 5 \\
  2.5\,N = 30 \\
  N = \dfrac{{30}}{{2.5}} = 12\,N \\
 $
$\therefore $The option A is correct.


Note:
We can use normality in precipitation reactions to find the number of ions which are precipitated in a reaction.
It is also used in redox reactions to determine the number of electrons that a reducing or oxidizing agent can accept or give.
We can calculate normality from molarity using the formula,
$N = Molarity \times Basicity = Molarity \times Acidity$