Final answer:
The pH of the solution obtained from mixing 50 cm³ of 0.4M H₂SO₄ with 50 cm³ of 0.1M NaOH is approximately 0.7, as the remaining unreacted H₂SO₄ leads to an acidic solution.
Step-by-step explanation:
To calculate the pH of the resulting solution from mixing 50 cm³ of 0.4M H₂SO₄ with 50 cm³ of 0.1M NaOH, we must first find out how the two solutions react. Since sulfuric acid (H₂SO₄) is a strong acid and sodium hydroxide (NaOH) is a strong base, they will neutralize each other according to the reaction:
H₂SO₄ (aq) + 2 NaOH (aq) → Na₂SO₄ (aq) + 2 H₂O (l)
However, since the mixing volume is the same, we need to consider the number of moles of each reactant:
Each mole of H₂SO₄ will react with two moles of NaOH. Only 0.005 mol of NaOH are available to react with 0.01 mol of H₂SO₄ (half the moles of H₂SO₄), therefore, there will be remaining H₂SO₄ after the reaction.
The remaining moles of H₂SO₄ will be 0.02 mol - (2 × 0.005 mol) = 0.01 mol. This is because the ratio of H₂SO₄ to NaOH is 1:2, and there is not enough NaOH to completely neutralize the H₂SO₄. The concentration of the unreacted H₂SO₄ in the resulting mixture of 100 cm³ (0.05 L + 0.05 L) is 0.01 mol / 0.1 L = 0.1 M.
The pH of a 0.1 M sulfuric acid solution can be found using the formula:
pH = -log[H+]
Since H₂SO₄ is a strong acid and dissociates completely in water, the concentration of H+ ions would be twice the concentration of H₂SO₄, which is 0.2 M. Therefore, the pH of the mixture is:
pH = -log(0.2) ≈ 0.7
Hence, none of the options a) 1.6, b) 7.0, c) 9.4, or d) 13.0 is correct. The pH of the resulting solution is approximately 0.7, which is acidic.