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Question: Q. 0.1M, 500ml NaOH is mixed with 0.2M, 200mL $H_2SO_4$ soln find (i) nature of resullent solution. ...

Q. 0.1M, 500ml NaOH is mixed with 0.2M, 200mL H2SO4H_2SO_4 soln find (i) nature of resullent solution. (ii) [Na+][Na^+] (iii) [SO42−][SO_4^{2-}]

Answer

(i) Nature of resultant solution: Acidic

(ii) [Na+][Na^+]: 114 M\frac{1}{14} \text{ M}

(iii) [SO42−][SO_4^{2-}]: 235 M\frac{2}{35} \text{ M}

Explanation

Solution

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

  1. Calculate the initial millimoles of each reactant:

    • For NaOH: Molarity = 0.1 M Volume = 500 mL Millimoles of NaOH = Molarity × Volume (in mL) = 0.1 M × 500 mL = 50 millimoles
    • For H₂SO₄: Molarity = 0.2 M Volume = 200 mL Millimoles of H₂SO₄ = Molarity × Volume (in mL) = 0.2 M × 200 mL = 40 millimoles

  2. 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)}

  3. 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.

    • To react completely with 40 millimoles of H₂SO₄, the required NaOH would be: 40 mmol H2SO4×(2 mmol NaOH1 mmol H2SO4)=80 mmol NaOH40 \text{ mmol H}_2\text{SO}_4 \times \left(\frac{2 \text{ mmol NaOH}}{1 \text{ mmol H}_2\text{SO}_4}\right) = 80 \text{ mmol NaOH}
    • We only have 50 millimoles of NaOH. Since the available NaOH (50 mmol) is less than the required NaOH (80 mmol), NaOH is the limiting reactant.
    • The amount of H₂SO₄ that reacts with 50 millimoles of NaOH is: 50 mmol NaOH×(1 mmol H2SO42 mmol NaOH)=25 mmol H2SO450 \text{ mmol NaOH} \times \left(\frac{1 \text{ mmol H}_2\text{SO}_4}{2 \text{ mmol NaOH}}\right) = 25 \text{ mmol H}_2\text{SO}_4
    • Millimoles of H₂SO₄ remaining = Initial H₂SO₄ - Reacted H₂SO₄ Millimoles of H₂SO₄ remaining = 40 mmol - 25 mmol = 15 mmol
    • Since H₂SO₄ (an acid) is remaining in excess, the resultant solution is acidic.

  4. Calculate the total volume of the solution: Total volume = Volume of NaOH + Volume of H₂SO₄ Total volume = 500 mL + 200 mL = 700 mL

  5. Calculate the concentration of Na⁺ ions ([Na⁺]): All the Na⁺ ions come from the initial NaOH. Initial millimoles of Na⁺ = Millimoles of NaOH = 50 millimoles. These 50 millimoles of Na⁺ are now present in the total volume of 700 mL. [Na+]=Millimoles of Na+Total volume (in mL)=50 mmol700 mL=570 M=114 M[\text{Na}^+] = \frac{\text{Millimoles of Na}^+}{\text{Total volume (in mL)}} = \frac{50 \text{ mmol}}{700 \text{ mL}} = \frac{5}{70} \text{ M} = \frac{1}{14} \text{ M}

  6. Calculate the concentration of SO₄²⁻ ions ([SO₄²⁻]): All the SO₄²⁻ ions come from the initial H₂SO₄. Initial millimoles of SO₄²⁻ = Millimoles of H₂SO₄ = 40 millimoles. These 40 millimoles of SO₄²⁻ are now present in the total volume of 700 mL. [SO42−]=Millimoles of SO42−Total volume (in mL)=40 mmol700 mL=470 M=235 M[\text{SO}_4^{2-}] = \frac{\text{Millimoles of SO}_4^{2-}}{\text{Total volume (in mL)}} = \frac{40 \text{ mmol}}{700 \text{ mL}} = \frac{4}{70} \text{ M} = \frac{2}{35} \text{ M}

Explanation of the solution:

  1. Calculated initial millimoles of NaOH (50 mmol) and H₂SO₄ (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 H₂SO₄ is in excess (15 mmol remaining) after reacting with all of the NaOH. Thus, the solution is acidic.
  4. Calculated total volume (700 ml).
  5. Calculated [Na+][\text{Na}^+] using initial millimoles of NaOH (50 mmol) and total volume: [Na+]=50 mmol700 ml=114 M[\text{Na}^+] = \frac{50 \text{ mmol}}{700 \text{ ml}} = \frac{1}{14} \text{ M}.
  6. Calculated [SO42−][\text{SO}_4^{2-}] using initial millimoles of H₂SO₄ (40 mmol) and total volume: [SO42−]=40 mmol700 ml=235 M[\text{SO}_4^{2-}] = \frac{40 \text{ mmol}}{700 \text{ ml}} = \frac{2}{35} \text{ M}.