Question

How many milliliters of 0.1 N ${{H}_{2}}S{{O}_{4}}$ solution will be required for complete reaction with a solution containing 0.125 g of pure $N{{a}_{2}}C{{O}_{3}}$?
(A) 23.6 mL
(B) 25.6 mL
(C) 26.3 mL
(D) 32.6 mL

Answer 1

Hint: A reaction between an acid and base in which a salt and water is formed as a product is known as the neutralization reactions. The reaction between \[{{H}_{2}}S{{O}_{4}}\] and $N{{a}_{2}}C{{O}_{3}}$ is an example of a neutralization reaction.

\[{{H}_{2}}S{{O}_{4}}+N{{a}_{2}}C{{O}_{3}}\to N{{a}_{2}}S{{O}_{3}}+{{H}_{2}}C{{O}_{3}}\]

Complete step by step answer:
-For neutralization reactions, net ionic equations may include solid acids, solid bases, solid salts, and water.
-For the complete reaction to occur between ${{H}_{2}}S{{O}_{4}}$ and $N{{a}_{2}}C{{O}_{3}}$, the reaction equivalents must be equal.

-Calculating the number of moles for $N{{a}_{2}}C{{O}_{3}}$,
$\text{Number of moles of N}{{\text{a}}_{2}}C{{O}_{3}}=\dfrac{\text{Given mass}}{\text{Molar mass}}=\dfrac{0.125}{126}=0.001179moles$

-Now calculating the number of equivalents for $N{{a}_{2}}C{{O}_{3}}$,
$\text{Number of equivalents of N}{{\text{a}}_{2}}C{{O}_{3}}=\text{Number of moles}\times \text{Valence factor}=0.001179\times 2=0.002358$
 -Therefore, 0.002358 number of equivalents ${{H}_{2}}S{{O}_{4}}$ must be present for completing the reaction.

-Calculating the number of equivalents for ${{H}_{2}}S{{O}_{4}}$,
$\text{Number of equivalents for }{{\text{H}}_{2}}S{{O}_{4}}=Normality\times Volume(litres)$
-Therefore, the volume of ${{H}_{2}}S{{O}_{4}}$in liters $=\dfrac{0.002358}{0.1}=0.02358litres$

-Converting the volume of ${{H}_{2}}S{{O}_{4}}$ in milliliters from liters,
$=0.02358\times 1000mL=23.58mL\approx 23.6mL$
So, the correct answer is “Option A”.

Note: In neutralization reactions, there is a combination of hydrogen ions and hydroxide ions. There are many applications of neutralization reactions. The first application of neutralization reactions involves titration methods. Titration is a method employed for finding the concentration of unknown solutions by finding their neutralization point. With the help of simple stoichiometric calculations and knowledge of the volume and molarity of one solution, we can find the molarity of the unknown sample. A second application in the wastewater treatments. Neutralization reactions are used for neutralizing toxicities present in water by using different chemicals like sodium bicarbonate, magnesium hydroxide, calcium oxide, and calcium carbonate. The third application is in the synthesis of nanomaterials which can facilitate the chemical reduction of metal precursors. The fourth application is in human digestive systems for neutralizing the various nutrients by creating favorable conditions inside the body by the production of various acids and bases for the working of enzymes. The fifth application is in controlling the pH of the soil by neutralizing the acidity of the soil using calcium carbonate and calcium hydroxide.