Question

The volume of 0.01M H2SO4 required to neutralize completely 100ml of 0.2M
NaOH solution is​

Answer 1

Answer:

The volume of $0.01M H_{2}SO_{4}$ required to neutralize completely $100$$ml$ of $0.2M$ NaOH solution is $1000ml$.

Explanation:

A neutralization reaction is a chemical reaction in which an acid and base quantitavely react together to form a salt and water.

Equation of a neutralization reaction, when Molarity and volume are mentioned is, $N_{1} V_{1} = N_{2} V_{2}$

Here, $N_{1} V_{1}$ is the product of normality and volume of the acid, sulphuric acid and $N_{2} V_{2}$ is the product of normality and volume of the base, sodium hydroxide.

Normality is the product of the molarity and basicity of acid or base.

It is given that, the molarity of sulphuric acid is $0.01$

the molarity of sodium Hydroxide is $0.2$ and the volume used by NaOH is $100 ml$

So, normality of  $0.01M H_{2}SO_{4}$ is $0.01 * 2$ (basicity of $H_{2}SO_{4}$)  $=$ $0.02$ N

        normality of  $0.2M NaOH$ is $0.2$ $* 1$ ( basicity of NaOH) $= 0.2$ N

Now, substituting these values in the equation,

         $N_{1} V_{1} = N_{2} V_{2}$

     $0.02 * V_{1}= 0.2* 100$

               $V_{1} = 1000 ml$

Therefore the volume required to neutralize is $1000ml$