What will be the normality of the solution obtained by mixing $100mL$ of $0.2M{H}_{2}S{O}_4$​ with $100mL$ of $0.2MNaOH$ ?

The correct option is D $0.1N$
$\text{Normality}=\text{Molarity}\times n_f$
$H_{2}S{O}_4\text{is dibasic acid}\therefore n-factor=2$
n-factor for $NaOH=1,\text{as it is a monoacidic base.}$
$N_{1}of{H}_{2}S{O}_4=0.2\times 2=0.4$
$N_{2}ofNaOH=0.2\times 1=0.2$
The required formula is -
$N_1{V}_1-N_2{V}_2=N_3{V}_3$​

where, $V_1$​ and $V_2$​ are the volumes of $H_{2}S{O}_{4}andNaOH$ respectively.
Given , $V_1=V_2=100mL$
$V_3=100+100=200mL$
Where, $N_3$​ and $V_3$​ are the normality & volume of the final solution.

$(0.4\times 100)-(0.2\times 100)=N_3\times \left(200\right)N_3=\frac{20}{200}{N}_3=0.1N$
The normality of solution obtained is $0.1N$.

Titrations :

Types of titrations :

1. Acid - base titrations 2. Redox titrations

Equivalence point is the point where:

No. of equivalent of the analyte = No. of equivalent of the titrant

1. Analyte: Solution with unknown concentration

2. Titrant: Solution with known concentration

Redox titration :

In this titration, one oxidizing agent (O.A.) reacts with one reducing agent (R.A.).

$O.A+R.A⟶products$

At the equivalence point :

Equivalence of O.A. = Equivalence of R.A