Question

Equal volumes of aqueous solution of 0.1 M HCl and 0.2 M H₂SO₄ are mixed. The concentration of H+ ions in the resulting solution is:

• A. 0.15 M
• B. 0.30 M
• C. 0.10 M
• D. 0.25 M

Answer 1

The Correct Option is D

• HCl is a monoprotic acid, so 0.1 M HCl provides 0.1 M H⁺ ions.
• H₂SO₄ is a diprotic acid, so 0.2 M H₂SO₄ provides 2 × 0.2 = 0.4 M H⁺ ions.

When equal volumes of the two solutions are mixed, the total volume doubles, and the concentration of each acid is halved:

New concentration of HCl = \(\frac{0.1}{2} = 0.05 \text{M H}^+\)

New concentration of H₂SO₄ = \(\frac{0.4}{2} = 0.2 \text{M H}^+\)

The total H⁺ concentration is:

\( 0.05 \text{M} + 0.2 \text{M} = 0.25 \text{M} \)

Answer 2

Correct Answer:

Option 4: 0.25 M

Explanation:

Let's assume the volume of each solution is V liters.

1. Moles of H+ from HCl:

Moles of HCl = 0.1 M * V L = 0.1V moles

HCl → H+ + Cl- (1 mole HCl gives 1 mole H+)

Moles of H+ from HCl = 0.1V moles

2. Moles of H+ from H2SO4:

Moles of H2SO4 = 0.2 M * V L = 0.2V moles

H2SO4 → 2H+ + SO4^2- (1 mole H2SO4 gives 2 moles H+)

Moles of H+ from H2SO4 = 2 * 0.2V moles = 0.4V moles

3. Total Moles of H+:

Total moles of H+ = 0.1V moles + 0.4V moles = 0.5V moles

4. Total Volume of Solution:

Total volume = V L + V L = 2V L

5. Concentration of H+:

Concentration of H+ = (0.5V moles) / (2V L) = 0.25 M