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Question: 0.1M, 200ml of H2SO4 is mixed with 0.2M, 200 ml NaOH solution, find nature of solution and [Na+] and...

0.1M, 200ml of H2SO4 is mixed with 0.2M, 200 ml NaOH solution, find nature of solution and [Na+] and [SO4 2-]

Answer

Nature of solution: Neutral, [Na+] = 0.1 M, [SO4 2-] = 0.05 M

Explanation

Solution

To determine the nature of the solution and the concentrations of the specified ions, we follow these steps:

  1. Calculate the millimoles of H₂SO₄:
    Given, Molarity of H₂SO₄ = 0.1 M
    Volume of H₂SO₄ = 200 ml
    Millimoles of H₂SO₄ = Molarity × Volume (in ml) = 0.1 M × 200 ml = 20 millimoles

  2. Calculate the millimoles of NaOH:
    Given, Molarity of NaOH = 0.2 M
    Volume of NaOH = 200 ml
    Millimoles of NaOH = Molarity × Volume (in ml) = 0.2 M × 200 ml = 40 millimoles

  3. Write the balanced chemical equation for the reaction:
    The reaction between sulfuric acid (a strong acid) and sodium hydroxide (a strong base) is:
    H2SO4(aq)+2NaOH (aq)→Na2SO4(aq)+2H2O (l)\text{H}_2\text{SO}_4 \text{(aq)} + 2\text{NaOH (aq)} \rightarrow \text{Na}_2\text{SO}_4 \text{(aq)} + 2\text{H}_2\text{O (l)}

  4. Determine the limiting and excess reactants (and nature of solution):
    From the balanced equation, 1 mole of H₂SO₄ reacts with 2 moles of NaOH.
    This means 1 millimole of H₂SO₄ reacts with 2 millimoles of NaOH.

    We have 20 millimoles of H₂SO₄.
    The amount of NaOH required to react completely with 20 millimoles of H₂SO₄ is:
    20 millimoles H2SO4×(2 millimoles NaOH1 millimole H2SO4)=40 millimoles of NaOH20 \text{ millimoles H}_2\text{SO}_4 \times \left(\frac{2 \text{ millimoles NaOH}}{1 \text{ millimole H}_2\text{SO}_4}\right) = 40 \text{ millimoles of NaOH}

    We have 40 millimoles of NaOH initially. Since exactly 40 millimoles of NaOH are required to react with 20 millimoles of H₂SO₄, both reactants are completely consumed. There is no excess acid or base.
    Therefore, the resulting solution is Neutral.

  5. Calculate the total volume of the solution:
    Total volume = Volume of H₂SO₄ + Volume of NaOH
    Total volume = 200 ml + 200 ml = 400 ml

  6. Calculate the concentration of Na⁺ ions ([Na⁺]):
    All the Na⁺ ions come from the initial NaOH.
    Initial millimoles of Na⁺ = Millimoles of NaOH = 40 millimoles.
    These 40 millimoles of Na⁺ are now present in the total volume of 400 ml.
    [Na+]=Millimoles of Na+Total volume (in ml)=40 millimoles400 ml=0.1 M[\text{Na}^+] = \frac{\text{Millimoles of Na}^+}{\text{Total volume (in ml)}} = \frac{40 \text{ millimoles}}{400 \text{ ml}} = 0.1 \text{ M}

  7. Calculate the concentration of SO₄²⁻ ions ([SO₄²⁻]):
    All the SO₄²⁻ ions come from the initial H₂SO₄.
    Initial millimoles of SO₄²⁻ = Millimoles of H₂SO₄ = 20 millimoles.
    These 20 millimoles of SO₄²⁻ are now present in the total volume of 400 ml.
    [SO42−]=Millimoles of SO42−Total volume (in ml)=20 millimoles400 ml=0.05 M[\text{SO}_4^{2-}] = \frac{\text{Millimoles of SO}_4^{2-}}{\text{Total volume (in ml)}} = \frac{20 \text{ millimoles}}{400 \text{ ml}} = 0.05 \text{ M}

Explanation of the solution:

  1. Calculated initial millimoles of H₂SO₄ (20 mmol) and NaOH (40 mmol).
  2. Used the balanced equation H2SO4+2NaOH→Na2SO4+2H2O\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} to find the stoichiometric ratio (1:2).
  3. Determined that 20 mmol of H₂SO₄ requires 40 mmol of NaOH for complete neutralization. Since exactly 40 mmol of NaOH were present, both reactants are completely consumed, making the solution neutral.
  4. Calculated total volume (400 ml).
  5. Calculated [Na+][\text{Na}^+] using initial millimoles of NaOH (40 mmol) and total volume: [Na+]=40 mmol/400 ml=0.1 M[\text{Na}^+] = 40 \text{ mmol} / 400 \text{ ml} = 0.1 \text{ M}.
  6. Calculated [SO42−][\text{SO}_4^{2-}] using initial millimoles of H₂SO₄ (20 mmol) and total volume: [SO42−]=20 mmol/400 ml=0.05 M[\text{SO}_4^{2-}] = 20 \text{ mmol} / 400 \text{ ml} = 0.05 \text{ M}.