What will be the pH of the resulting solution after 25.0 cm³ of 0.100 mol dm⁻³ sulfuric acid solution, H₂SO₄(aq) has been added to 25.0 cm³ of 0.200 mol dm⁻³ potassium hydroxide solution, KOH(aq)?

To solve this problem, we need to understand the reaction between sulfuric acid (H₂SO₄) and potassium hydroxide (KOH). This is a neutralization reaction, where an acid reacts with a base to produce a salt and water. The balanced chemical equation for this reaction is:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

From the balanced equation, we can see that one mole of H₂SO₄ reacts with two moles of KOH.

Step 1: Calculate the number of moles of H₂SO₄ and KOH

Number of moles = concentration x volume

For H₂SO₄: Number of moles = 0.100 mol/dm³ x 25.0 cm³ x (1 dm³/1000 cm³) = 0.0025 mol

For KOH: Number of moles = 0.200 mol/dm³ x 25.0 cm³ x (1 dm³/1000 cm³) = 0.0050 mol

Step 2: Determine the limiting reactant

From the balanced equation, we know that 1 mole of H₂SO₄ reacts with 2 moles of KOH. Therefore, the 0.0025 mol of H₂SO₄ will react with 0.0050 mol of KOH, which is exactly the amount we have. Therefore, all the H₂SO₄ and KOH will react completely, and there will be no excess acid or base.

Step 3: Calculate the pH of the resulting solution

Since all the acid and base have reacted completely, the resulting solution will be neutral, with a pH of 7.