Question

1 L solution of each 1 M H₂SO₄, HCl and H₃PO₄ are mixed together and then volume is made upto 4 L. In this solution 1 L of 2 M NaOH is added. Consider the 100% ionization of all the acids. Find out the [H⁺].

Answer 1

0.8 M

Answer 2

To determine the final concentration of H⁺ ions, we need to calculate the total moles of H⁺ from all acids, the moles of OH⁻ from the base, and then find the net moles of H⁺ after neutralization, finally dividing by the total volume.

1. Calculate moles of H⁺ from each acid:

• H₂SO₄:

  • Volume = 1 L, Concentration = 1 M
  • Moles of H₂SO₄ = $1 \text{ L} \times 1 \text{ mol/L} = 1 \text{ mol}$
  • Since H₂SO₄ is a diprotic acid and 100% ionized: $\text{H₂SO₄} \rightarrow 2\text{H⁺} + \text{SO₄²⁻}$
  • Moles of H⁺ from H₂SO₄ = $1 \text{ mol} \times 2 = 2 \text{ mol}$

• HCl:

  • Volume = 1 L, Concentration = 1 M
  • Moles of HCl = $1 \text{ L} \times 1 \text{ mol/L} = 1 \text{ mol}$
  • Since HCl is a monoprotic acid and 100% ionized: $\text{HCl} \rightarrow \text{H⁺} + \text{Cl⁻}$
  • Moles of H⁺ from HCl = $1 \text{ mol} \times 1 = 1 \text{ mol}$

• H₃PO₄:

  • Volume = 1 L, Concentration = 1 M
  • Moles of H₃PO₄ = $1 \text{ L} \times 1 \text{ mol/L} = 1 \text{ mol}$
  • Since H₃PO₄ is a triprotic acid and 100% ionization is considered for all protons: $\text{H₃PO₄} \rightarrow 3\text{H⁺} + \text{PO₄³⁻}$
  • Moles of H⁺ from H₃PO₄ = $1 \text{ mol} \times 3 = 3 \text{ mol}$

2. Calculate total moles of H⁺ from the acid mixture: Total moles of H⁺ = Moles H⁺(H₂SO₄) + Moles H⁺(HCl) + Moles H⁺(H₃PO₄) Total moles of H⁺ = $2 \text{ mol} + 1 \text{ mol} + 3 \text{ mol} = 6 \text{ mol}$

3. Calculate moles of OH⁻ from NaOH:

• Volume of NaOH solution = 1 L, Concentration = 2 M
• Moles of NaOH = $1 \text{ L} \times 2 \text{ mol/L} = 2 \text{ mol}$
• Since NaOH is a strong base and 100% ionized: $\text{NaOH} \rightarrow \text{Na⁺} + \text{OH⁻}$
• Moles of OH⁻ from NaOH = $2 \text{ mol} \times 1 = 2 \text{ mol}$

4. Determine the net moles of H⁺ after reaction: The reaction between H⁺ and OH⁻ is: $\text{H⁺} + \text{OH⁻} \rightarrow \text{H₂O}$

• Initial moles of H⁺ = 6 mol
• Moles of OH⁻ added = 2 mol Since moles of H⁺ are greater than moles of OH⁻, the OH⁻ will be completely neutralized, leaving an excess of H⁺. Net moles of H⁺ = Initial moles of H⁺ - Moles of OH⁻ Net moles of H⁺ = $6 \text{ mol} - 2 \text{ mol} = 4 \text{ mol}$

5. Calculate the final volume of the solution: The initial acid mixture (1L + 1L + 1L = 3L) was made up to 4 L. Then, 1 L of NaOH solution was added to this 4 L acid solution. Final Volume = Volume of diluted acid solution + Volume of NaOH solution Final Volume = $4 \text{ L} + 1 \text{ L} = 5 \text{ L}$

6. Calculate the final concentration of H⁺: $[\text{H⁺}] = \frac{\text{Net moles of H⁺}}{\text{Final Volume}}$ $[\text{H⁺}] = \frac{4 \text{ mol}}{5 \text{ L}} = 0.8 \text{ M}$

The final concentration of H⁺ is 0.8 M.

The final answer is $\boxed{\text{0.8 M}}$.

Explanation of the solution:

1. Calculate moles of H⁺ from each acid, accounting for stoichiometry (H₂SO₄: 2H⁺/mol, HCl: 1H⁺/mol, H₃PO₄: 3H⁺/mol, assuming 100% ionization for all).
   • H₂SO₄: $1 \text{ L} \times 1 \text{ M} \times 2 = 2 \text{ mol H⁺}$
   • HCl: $1 \text{ L} \times 1 \text{ M} \times 1 = 1 \text{ mol H⁺}$
   • H₃PO₄: $1 \text{ L} \times 1 \text{ M} \times 3 = 3 \text{ mol H⁺}$
2. Sum the moles of H⁺: $2 + 1 + 3 = 6 \text{ mol H⁺}$.
3. Calculate moles of OH⁻ from NaOH: $1 \text{ L} \times 2 \text{ M} = 2 \text{ mol OH⁻}$.
4. Determine net moles of H⁺ after neutralization: $6 \text{ mol H⁺} - 2 \text{ mol OH⁻} = 4 \text{ mol H⁺}$.
5. Calculate final total volume: $4 \text{ L}$ (diluted acid solution) + $1 \text{ L}$ (NaOH solution) = $5 \text{ L}$.
6. Calculate final $[\text{H⁺}]$: $\frac{4 \text{ mol}}{5 \text{ L}} = 0.8 \text{ M}$.

Answer: 0.8 M