During the electrolysis of 400 ml of 0.5 M copper(II) nitrate solution, 1.12 L of gas (STP) was released at the anode. What is the concentration of the acid remaining in the solution?
- 0.25 M
- 0.5 M
- 1.0 M
- 0.125 M
📝 Detailed explanation
Anode reaction: 2H₂O → O₂ + 4H⁺ + 4e⁻.Moles of O₂ = 1.12 / 22.4 = 0.05 mol.
Moles of H⁺ produced = 0.05 × 4 = 0.2 mol.
Molarity of H⁺ = 0.2 / 0.4 = 0.5 M.