Q.Sulphuric acid reacts with sodium hydroxide as follows: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$ When 1 L of 0.1M sulphuric acid solution is allowed to react with 1 L of 0.1M sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution obtained is (Note: this is a multiple-correct question; two or more options may be correct.)

Concept: Limiting reagent stoichiometry and molarity after mixing

The stoichiometry shows 1 mole of $H_2SO_4$ requires 2 moles of $NaOH$.

Step 1: Calculate moles available.

• Moles of $H_2SO_4 = 0.1 \times 1 = 0.1$ mol
• Moles of $NaOH = 0.1 \times 1 = 0.1$ mol

Step 2: Identify the limiting reagent.

For 0.1 mol $H_2SO_4$, we need $0.1 \times 2 = 0.2$ mol $NaOH$. Since only 0.1 mol $NaOH$ is available, $NaOH$ is the limiting reagent.

Step 3: Calculate $Na_2SO_4$ formed.

From stoichiometry, 2 mol $NaOH$ produces 1 mol $Na_2SO_4$.

So 0.1 mol $NaOH$ produces $\frac{0.1}{2} = 0.05$ mol $Na_2SO_4$. …

Moles of each reactant

Limiting reagent

The equation needs $2$ mol $NaOH$ per mol $H_2SO_4$, so $0.1$ mol $H_2SO_4$ would require $0.2$ mol $NaOH$. Only $0.1$ mol $NaOH$ is present, so $NaOH$ is the limiting reagent.

Moles and mass of $Na_2SO_4$

From the stoichiometry, $2$ mol $NaOH$ give $1$ mol $Na_2SO_4$:

With molar mass $M_{Na_2SO_4}=2(23)+32+4(16)=142\ \text{g mol}^{-1}$:

Molarity in the final solution …

Concept: Limiting Reagent & Molarity in Mixture

Method: Limiting Reagent Method — Identify the reactant that gets fully consumed first; product amount is based on that.

Steps

Step 1: Calculate moles of each reactant

• Moles of $H_2SO_4$:

  $M = 0.1\ \text{mol/L}$, $V = 1\ \text{L}$

  $\text{moles} = 0.1 \times 1 = 0.1\ \text{mol}$

• Moles of $NaOH$:

  $M = 0.1\ \text{mol/L}$, $V = 1\ \text{L}$

  $\text{moles} = 0.1 \times 1 = 0.1\ \text{mol}$

Step 2: Use stoichiometry to find limiting reagent

From the balanced equation:

$H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$

• 1 mol $H_2SO_4$ requires 2 mol $NaOH$
• We have only 0.1 mol $NaOH$, which would require $0.1/2 = 0.05\ \text{mol}$ $H_2SO_4$
• We have 0.1 mol $H_2SO_4$ — more than needed

Conclusion: $NaOH$ is the limiting reagent.

Step 3: Calculate moles of $Na_2SO_4$ formed

From equation:

2 mol $NaOH$ → 1 mol $Na_2SO_4$

So, 0.1 mol $NaOH$ → $0.1/2 = 0.05\ \text{mol}$ $Na_2SO_4$

Step 4: Calculate mass of $Na_2SO_4$

Molar mass of $Na_2SO_4$:

$Na = 23 \times 2 = 46$

$S = 32$ …

Let’s break this down step-by-step — first the concept, then the common mistakes, and finally how to avoid each.

The Core Concept: Limiting Reagent & Molarity After Reaction

The balanced equation:

• 1 mole of $H_2SO_4$ reacts with 2 moles of $NaOH$.
• We have equal volumes (1 L each) and equal molarities (0.1 M each).
• Moles of $H_2SO_4$ = $0.1 \times 1 = 0.1$ mol
• Moles of $NaOH$ = $0.1 \times 1 = 0.1$ mol

But the stoichiometry requires twice as much NaOH as $H_2SO_4$.

So NaOH is the limiting reagent.

Step-by-Step Correct Calculation

• From 0.1 mol NaOH, moles of $Na_2SO_4$ formed = $\frac{0.1}{2} = 0.05$ mol
• Molar mass of $Na_2SO_4$ = $2(23) + 32 + 4(16) = 142$ g/mol
• Mass of $Na_2SO_4$ = $0.05 \times 142 = 7.10$ g
• Total volume after mixing = $1 + 1 = 2$ L
• Molarity of $Na_2SO_4$ = $\frac{0.05}{2} = 0.025$ mol/L

Correct options: (ii) 7.10 g and (iii) $0.025\ \text{mol L}^{-1}$

Common Mistakes & How to Avoid Each

✗ Mistake 1: Assuming both reactants are completely used (no limiting reagent check)

• What students do: They see equal volumes and equal molarities and think both react fully. They then calculate $Na_2SO_4$ from $H_2SO_4$ directly: $0.1$ mol $H_2SO_4$ → $0.1$ mol $Na_2SO_4$ → mass = $14.2$ g, molarity = $0.05$ M.
• Why it’s wrong: The equation shows 1:2 ratio, not 1:1. NaOH runs out first.
• How to avoid: Always compare mole ratios — not just volumes or molarities. Write the balanced equation and check which reactant gives the smaller product amount.

✗ Mistake 2: Forgetting to add volumes for final molarity

• What students do: They calculate molarity as $\frac{0.05}{1} = 0.05$ M (using only one solution’s volume).
• Why it’s wrong: After mixing, the total volume is 2 L, not 1 L.
• How to avoid: Whenever two aqueous solutions are mixed, total volume = sum of individual volumes (unless stated otherwise). Always divide moles of product by final total volume.

✗ Mistake 3: Using wrong molar mass for $Na_2SO_4$

• What students do: They forget to multiply atomic masses correctly — e.g., using 23 for Na but forgetting there are 2 atoms.
• Why it’s wrong: Leads to mass like 3.55 g (half of correct) or other wrong numbers.
• How to avoid: Write the formula clearly: $Na_2SO_4$ → 2 Na, 1 S, 4 O. Calculate stepwise: …