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

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} (or approximately 0.0714 M0.0714 \text{ M}) (iii) [SO42−][SO_4^{2-}]: 235 M\frac{2}{35} \text{ M} (or approximately 0.0571 M0.0571 \text{ M})

Explanation

Solution

To solve this problem, we need to calculate the initial millimoles of each reactant, determine the limiting reactant, find the nature of the resulting solution, and then calculate the concentrations of the specified ions in the final volume.

Step-by-step Derivations:

  1. Calculate initial millimoles of NaOH and H₂SO₄:

    • For NaOH: Molarity = 0.1 M Volume = 500 mL Millimoles of NaOH = Molarity × Volume (in mL) = 0.1 M×500 mL=50 mmol0.1 \text{ M} \times 500 \text{ mL} = 50 \text{ mmol}
    • For H₂SO₄: Molarity = 0.2 M Volume = 200 mL Millimoles of H₂SO₄ = Molarity × Volume (in mL) = 0.2 M×200 mL=40 mmol0.2 \text{ M} \times 200 \text{ mL} = 40 \text{ mmol}

  2. Write the balanced chemical equation for the reaction: Sulfuric acid (H₂SO₄) reacts with sodium hydroxide (NaOH) as follows: 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)} From the equation, 1 mole of H₂SO₄ reacts with 2 moles of NaOH.

  3. Determine the limiting reactant and the nature of the resultant solution:

    • We have 40 mmol of H₂SO₄.
    • To completely neutralize 40 mmol of H₂SO₄, we would need 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 mmol of NaOH. Since the available NaOH (50 mmol) is less than the required NaOH (80 mmol), NaOH is the limiting reactant.
    • All 50 mmol of NaOH will react.
    • The amount of H₂SO₄ that reacts with 50 mmol 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 = 40 mmol−25 mmol=15 mmol40 \text{ mmol} - 25 \text{ mmol} = 15 \text{ mmol}.
    • Since there is an excess of H₂SO₄ (a strong acid) remaining, the resultant solution will be acidic.

  4. Calculate the total volume of the solution:

    • Total volume = Volume of NaOH solution + Volume of H₂SO₄ solution
    • Total volume = 500 mL+200 mL=700 mL500 \text{ mL} + 200 \text{ mL} = 700 \text{ mL}

  5. Calculate the concentration of Na⁺ ions ([Na⁺]):

    • All Na⁺ ions originate from the initial NaOH. The number of millimoles of Na⁺ remains constant, only its concentration changes due to dilution.
    • Initial millimoles of Na⁺ = Millimoles of NaOH = 50 mmol.
    • These 50 mmol 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≈0.0714 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} \approx 0.0714 \text{ M}

  6. Calculate the concentration of SO₄²⁻ ions ([SO₄²⁻]):

    • All SO₄²⁻ ions originate from the initial H₂SO₄. The number of millimoles of SO₄²⁻ remains constant, as it is a spectator ion in the reaction (it does not participate in the acid-base reaction itself, only its counter-ion H⁺ does).
    • Initial millimoles of SO₄²⁻ = Millimoles of H₂SO₄ = 40 mmol.
    • These 40 mmol 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≈0.0571 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} \approx 0.0571 \text{ M}

Explanation of the solution:

  1. Calculated initial millimoles of NaOH (50 mmol) and H₂SO₄ (40 mmol).
  2. Used the balanced reaction 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 determine the stoichiometric ratio (1:2).
  3. Identified NaOH as the limiting reactant, as 40 mmol H₂SO₄ requires 80 mmol NaOH, but only 50 mmol NaOH is available.
  4. Calculated the excess H₂SO₄ (15 mmol), indicating the solution is acidic.
  5. Determined the total volume by summing the initial volumes (700 mL).
  6. Calculated [Na+][\text{Na}^+] using initial millimoles of NaOH (50 mmol) and total volume: [Na+]=50 mmol/700 mL=1/14 M[\text{Na}^+] = 50 \text{ mmol} / 700 \text{ mL} = 1/14 \text{ M}.
  7. Calculated [SO42−][\text{SO}_4^{2-}] using initial millimoles of H₂SO₄ (40 mmol) and total volume: [SO42−]=40 mmol/700 mL=2/35 M[\text{SO}_4^{2-}] = 40 \text{ mmol} / 700 \text{ mL} = 2/35 \text{ M}.