Question

A mixture of $NaOH$ and $Mg{(OH)_2}$ weights $2.325g$. It requires $3g$ of ${H_2}S{O_4}$ for its neutralization. What is the percentage composition of mixture?

Answer 1

Hint: The chemical reactions in which there is a reaction between an acid and a base are termed as neutralization reactions. If the reaction is carried out in water then neutralization results in there being no excess of hydrogen or hydroxide ions, present in the solution.

Complete step by step answer:
A neutralization reaction can be stated as a reaction between an acid and an alkali (base), resulting in the formation of salt and water. The general form of the reaction can be represented as, $acid + base \to salt + water$. The $ \to $ shows that the reaction is complete, i.e., quantitative in nature. The $pH$ of the neutralized solutions depend upon the strength of acid used.
Let us consider the case of sulphuric acid ${H_2}S{O_4}$ . For this acid, there would be two possible reactions, resulting in its partial neutralization.
${H_2}S{O_4} + O{H^ - } \to HS{O_4}^ - + {H_2}O$
$HS{O_4}^ - + O{H^ - } \to S{O_4}^{2 - } + {H_2}O$

For an acid-base reaction, equivalent weight of an acid or a base can be defined as the mass which reacts with one mole of ${H^ + }$ ions. For an acid, it is calculated as molar mass divided by the number of protons. For a base, it is calculated as molar mass divided by the number of hydroxyl ions.
For the above question, let us try to calculate the equivalent weight of all the three reacting species present.
Equivalent weight of $NaOH$$ = \dfrac{{40}}{1} = 40g$
Equivalent weight of ${H_2}S{O_4}$ $ = \dfrac{{98}}{2} = 49g$
Equivalent weight of $Mg{(OH)_2}$ $ = \dfrac{{58.3}}{2} = 29.15g$
Let us consider the weight of sodium hydroxide taken be $a$ $grams$, i.e. ${W_{NaOH}} = a$ $g$
Weight of magnesium hydroxide taken be b, \[{W_{Mg(OH)}}_2 = b\] $g$
Therefore, $a + b = 2.325g$ …… (1)
For neutralization, equivalent mass of sodium hydroxide + equivalent mass of magnesium hydroxide= equivalent mass of sulphuric acid.
  ${M_{eq.(NaOH)}} + {M_{eq.Mg(OH)2}} = {M_{eq.(H2SO4)}}$
$ \Rightarrow \dfrac{{a \times 1000}}{{40}} + \dfrac{{b \times 1000}}{{29.15}} = \dfrac{{3 \times 1000}}{{49}}$
$ \Rightarrow 0.025a + 0.034b = 0.061$ ……….. (2)
On solving equations (1) and (2), we will get the values of a and b as, $a = 2g$ $b = 0.32g$
Therefore, percentage composition of $NaOH$ =$\dfrac{2}{{2.325}} \times 100 = 86.02\% $
Percentage composition of $Mg{(OH)_2}$ = $\dfrac{{0.32}}{{2.325}} \times 100 = 13.76\% $


Note:
Applications of neutralization reactions:
The wastewater coming from industries needs to be neutralized in order to remove their toxicity, before they are thrown out in the environment.
When the food moves from our stomach to the intestines, it needs to be neutralized. Antacid bicarbonate is thus produced to create this favorable condition.