Overview of the Reaction
When 50 mL of 0.1 M H₂SO₄ is mixed with 5 mL of 0.1 M NaOH, a neutralization reaction occurs. H₂SO₄, a strong acid, reacts with NaOH, a strong base, resulting in the formation of water and sodium sulfate (Na₂SO₄).
Calculating Moles of Reactants
- Moles of H₂SO₄:
- Volume = 50 mL = 0.050 L
- Molarity = 0.1 M
- Moles = Volume × Molarity = 0.050 L × 0.1 M = 0.005 moles
- Moles of NaOH:
- Volume = 5 mL = 0.005 L
- Molarity = 0.1 M
- Moles = Volume × Molarity = 0.005 L × 0.1 M = 0.0005 moles
Determining the Reaction Outcome
- Balanced Reaction:
- H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
- Stoichiometry:
- 1 mole of H₂SO₄ reacts with 2 moles of NaOH.
- Therefore, 0.005 moles of H₂SO₄ would require 0.010 moles of NaOH.
- Analyzing the Available Reactants:
- Available NaOH = 0.0005 moles, which is insufficient to completely neutralize the H₂SO₄.
Final Nature of the Solution
- Excess Reactant:
- Since NaOH is in lesser quantity, it will be completely consumed.
- H₂SO₄ remains unreacted.
- Nature of the Solution:
- The solution will be acidic due to the excess H₂SO₄.
Conclusion
The final solution formed will be acidic, as there is unreacted H₂SO₄ remaining after the neutralization of NaOH.