Question #31f57

I will tell the same with acid base neutralisation reaction

`HNO_3 + NaOH rarr NaNO_3 + H_2O`

where M is the molarity and V volume of solution

`V_1M_1 = V_2 M_2`

The above equation is applicable to calculate volume (`V_2`) of known sodium hydroxide concentration `(M_2`) required to neutralise a known concentration of nitric acid (`M_1` ) and volume of solution (`V_1`)

in the above balanced reaction one mole of or equivalent of sodium hydroxide neutralises one mole or equivalent of nitric acid

`H_2SO_4 + 2NaOH rarr Na2SO_4 +2 H_2O`

In this example mole of sulphuric acid requires 2 moles of sodium hydroxide

`V_1M_1 = 2V_2 M_2`

The above equation to be used to calculate volume (`V_2`) sodium hydroxide of `M_2` concentration to neutralise a volume of `V_1` and concentration of `M_1` sulphuric acid n a given volume of solution
If write the same in terms of equivalents i.e., normality then the equation is same for both case

`V_1N_1 = V_2 N_2`
where N is the normality of solution

For acid base reactions
Equivalent weight of a acid or base = Molecular weight of acid or base / Number of replaceable H or OH
for example in 1 M sulphuric acid is 2 N acid solution, hence consumption of sodium hydroxide for neutralising acid is two times as compared to same volume and concentration of Nitric acid of 1 M

Hence normality based solutions are employed during such titrations