Question

Calculate the volume of 0.1 N ${H_2}S{O_4}$ required to neutralize 200 ml of 0.2 N NaOH solution.

Answer 1

Hint: We simply know that molar mass is the weight of the substance multiplied by the number of atoms and every mole or gram equivalent contains an equal number of atoms that is $6.022 \times {10^{23}}$.

Complete step by step solution:
We are given with a neutralization reaction in the question and normality of the acid and base. Normality can be defined as the number of gram equivalents of solute in the volume of solution in litres.
In a simple titration like above solution of sulfuric acid of unknown concentration is made to react with a solution of sodium hydroxide whose concentration is known, so that the concentration of sodium hydroxide can be calculated. The two are reacted in such a way that the volume of sodium hydroxide required to completely react with sulfuric acid can be found out by using some indicators.
Since the normality of sodium hydroxide is 0.2 N let it be ${N_1}$ then its volume will be ${V_1}$, then the gram equivalents of sodium hydroxide reacted =${N_1} \times {V_1}$
According to the law of equivalents, the gram equivalents of sulfuric acid will be equal to that of sodium hydroxide. Therefore gram equivalent of sulfuric acid = N*V .
If the normality of sulfuric acid is N then its volume will be the equation given below after substituting the values from the question
 $
\Rightarrow V\,= \dfrac{{{N_1}{V_1}}}{N} = \dfrac{{0.2 \times 0.2(litres)}}{{0.1}} \\
\Rightarrow V \,= 0.4l = 400ml \\
 $ (since we know that 1000ml=1l)
Hence the volume of sulfuric acid required to neutralize 200 ml of sodium hydroxide is 400 ml.

Note:
From Avogadro’s law it is known that equal volumes of all gases contain equal numbers of molecules under similar conditions. Now since one mole of molecules of all gases contain the same number (6.022$ \times {10^{23}}$ ) of molecules, therefore they occupy the same volume under similar pressure and temperature. The mass of one mole of atoms is exactly equal to the atomic mass in grams of that element.