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Question: A volume of \(10\,ml\) of a \({H_2}S{O_4}\) solution is diluted to \(100\,ml\). Twenty-five millilit...

A volume of 10 ml10\,ml of a H2SO4{H_2}S{O_4} solution is diluted to 100 ml100\,ml. Twenty-five milliliters of this diluted solution is mixed with 50 ml50\,ml of 0.5 N NaOH0.5\,N\,NaOH solution. The resulting solution requires 0.205 g Na2CO30.205\,g\,N{a_2}C{O_3} for complete neutralization. The normality of the original H2SO4{H_2}S{O_4} solution is.
A)12 N12\,N.
B)11 N11\,N.
C)3N3N.
D)0.275 N0.275\,N.

Explanation

Solution

We know, normality of a solution is gram equivalent of solute in one liter of solution. The unit of normality is eq−1.e{q^{ - 1}}.
The mathematical expression of normality is,
N=weight of Solute (g)Equivalent weight × Volume (L)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.
N1V1=N1V2{N_1}{V_1} = {N_1}{V_2}
Where,
The normality of the acidic solution is N1{N_1}.
The volume of the acidic solution is V1{V_1}.
The normality of the basic solution is N2{N_2}.
The volume of the basic is V2{V_2}.
A volume of 10 ml10\,ml of sulfuric acid is dissolved in 100 ml. Thus, the normality of sulfuric acid can be found by using the relation.
NV1=N1V2N{V_1} = {N_1}{V_2}
N(10 ml)=N(100 ml) N1=0.1N  N\left( {10\,ml} \right) = N\left( {100\,ml} \right) \\\ {N_{_1}} = 0.1N \\\
Now, a volume of 25 ml25\,ml 0.1 N H2SO40.1\,N\,{H_2}S{O_4} react with 50 ml50\,ml of 0.5 N NaOH0.5\,N\,NaOH
After that, sodium carbonate required 0.265 g0.265\,g for complete neutralization.
Milli equivalent of Na2CO3=0.265106/2×1000=5N{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×25−50×0.5=5 2.5N−25=5 2.5 N=30 N=302.5=12 N  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×Basicity=Molarity×AcidityN = Molarity \times Basicity = Molarity \times Acidity