Question

100 g of a mixture of $NaOH$ and $Na_{2}SO_{4}$ is neutralized by 100 ml of 0.5M $H_{2}SO_{4}$. What is the amount of $Na_{2}SO_{4}$ present in the mixture?}

• A. 82 g
• B. 96 g
• C. 88 g
• D. 92 g

Hint:

Only the base ($NaOH$) is neutralized by the acid; the salt ($Na_{2}SO_{4}$) remains inert in this reaction.

Answer 1

The Correct Option is B

Step 1: Concept
$Na_{2}SO_{4}$ is a salt and does not react with $H_{2}SO_{4}$. Only $NaOH$ reacts with $H_{2}SO_{4}$ in a neutralization reaction: $2NaOH + H_{2}SO_{4} \rightarrow Na_{2}SO_{4} + 2H_{2}O$.

Step 2: Meaning
Moles of $H_{2}SO_{4} = Molarity \times Volume(L) = 0.5 \times 0.1 = 0.05$ moles.

Step 3: Analysis
From the balanced equation, 1 mole of $H_{2}SO_{4}$ reacts with 2 moles of $NaOH$. So, moles of $NaOH = 2 \times 0.05 = 0.1$ moles. Mass of $NaOH = 0.1 \times 40$ g/mol $= 4$ g.

Step 4: Conclusion
Mass of $Na_{2}SO_{4} = Total\ mass - Mass\ of\ NaOH = 100\ g - 4\ g = 96\ g$. Final Answer: (B)