Question

Q. 0.1M, 500ml NaOH is mixed with 0.2M, 200mL $H_2SO_4$ Soln find (i) nature of resullent solution. (ii) $[Na^+]$ (iii) $[SO_4^{2-}]$

Answer 1

Nature of resultant solution: Acidic, $[Na^+]$: $\frac{1}{14}$ M, $[SO_4^{2-}]$: $\frac{2}{35}$ M

Answer 2

To solve this problem, we need to determine the millimoles of reactants, identify the limiting and excess reactants, and then calculate the concentrations of the specified ions in the final volume.

1. Calculate initial millimoles of reactants:

• NaOH:
  Molarity = 0.1 M
  Volume = 500 mL
  Millimoles of NaOH = Molarity × Volume (mL) = 0.1 M × 500 mL = 50 millimoles

• H₂SO₄:
  Molarity = 0.2 M
  Volume = 200 mL
  Millimoles of H₂SO₄ = Molarity × Volume (mL) = 0.2 M × 200 mL = 40 millimoles

2. Write the balanced chemical equation:

The reaction between sulfuric acid (H₂SO₄) and sodium hydroxide (NaOH) is: $\text{H}_2\text{SO}_4 \text{(aq)} + 2\text{NaOH (aq)} \rightarrow \text{Na}_2\text{SO}_4 \text{(aq)} + 2\text{H}_2\text{O (l)}$

From the equation, 1 mole of H₂SO₄ reacts with 2 moles of NaOH.

3. Determine the limiting and excess reactants and the nature of the solution:

• We have 40 millimoles of H₂SO₄ and 50 millimoles of NaOH.
• To react completely with 40 millimoles of H₂SO₄, the required NaOH would be $40 \text{ mmol H}_2\text{SO}_4 \times \frac{2 \text{ mmol NaOH}}{1 \text{ mmol H}_2\text{SO}_4} = 80 \text{ mmol NaOH}$.
• Since we only have 50 millimoles of NaOH, NaOH is the limiting reactant.
• The 50 millimoles of NaOH will react with: $50 \text{ mmol NaOH} \times \frac{1 \text{ mmol H}_2\text{SO}_4}{2 \text{ mmol NaOH}} = 25 \text{ mmol H}_2\text{SO}_4$.
• Millimoles of H₂SO₄ remaining = Initial H₂SO₄ - Reacted H₂SO₄ = 40 mmol - 25 mmol = 15 mmol.
• Since H₂SO₄ (a strong acid) is in excess, the resultant solution will be acidic.

4. Calculate the total volume of the solution:

Total volume = Volume of NaOH + Volume of H₂SO₄ = 500 mL + 200 mL = 700 mL

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

All Na⁺ ions come from the initial NaOH. The total millimoles of Na⁺ ions in the solution will be the initial millimoles of NaOH, as Na⁺ is a spectator ion and does not get consumed in the reaction.

Millimoles of Na⁺ = 50 millimoles Total volume = 700 mL $[\text{Na}^+] = \frac{\text{Millimoles of Na}^+}{\text{Total volume (mL)}} = \frac{50 \text{ mmol}}{700 \text{ mL}} = \frac{5}{70} \text{ M} = \frac{1}{14} \text{ M}$ $[\text{Na}^+] \approx 0.0714 \text{ M}$

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

All SO₄²⁻ ions come from the initial H₂SO₄. The total millimoles of SO₄²⁻ ions in the solution will be the initial millimoles of H₂SO₄, as SO₄²⁻ is a spectator ion (it remains in solution, either as part of unreacted H₂SO₄ or as Na₂SO₄).

Millimoles of SO₄²⁻ = 40 millimoles Total volume = 700 mL $[\text{SO}_4^{2-}] = \frac{\text{Millimoles of SO}_4^{2-}}{\text{Total volume (mL)}} = \frac{40 \text{ mmol}}{700 \text{ mL}} = \frac{4}{70} \text{ M} = \frac{2}{35} \text{ M}$ $[\text{SO}_4^{2-}] \approx 0.0571 \text{ M}$

Solution:

(i) Nature of resultant solution: Acidic (ii) $[Na^+]$: $\frac{1}{14}$ M (or approximately 0.0714 M) (iii) $[SO_4^{2-}]$: $\frac{2}{35}$ M (or approximately 0.0571 M)

Explanation of the solution:

1. Calculated initial millimoles of NaOH (50 mmol) and H₂SO₄ (40 mmol).
2. Used the balanced equation $\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$ to determine the stoichiometric ratio (1:2).
3. Identified NaOH as the limiting reactant (50 mmol) and H₂SO₄ as the excess reactant (15 mmol remaining after reaction).
4. Concluded that the solution is acidic due to excess H₂SO₄.
5. Calculated total volume (700 mL).
6. Calculated $[\text{Na}^+]$ using initial millimoles of NaOH (50 mmol) and total volume: $[\text{Na}^+] = 50 \text{ mmol} / 700 \text{ mL} = 1/14 \text{ M}$.
7. Calculated $[\text{SO}_4^{2-}]$ using initial millimoles of H₂SO₄ (40 mmol) and total volume: $[\text{SO}_4^{2-}] = 40 \text{ mmol} / 700 \text{ mL} = 2/35 \text{ M}$.

Answer:

(i) Nature of resultant solution: Acidic (ii) $[Na^+]$: $\frac{1}{14}$ M (iii) $[SO_4^{2-}]$: $\frac{2}{35}$ M