Calculate the volume of 0.5 mol dm^-3 H₂SO₄ that is neutralized by 25 cm³ of 0.1 mol dm^-3 NaOH 

Step 1: Write the neutralization equation

\[ H_2SO_4 + 2NaOH \;\;\rightarrow\;\; Na_2SO_4 + 2H_2O \]

• 1 mole of H₂SO₄ reacts with 2 moles of NaOH.

Step 2: Calculate moles of NaOH given

\[ \text{Moles} = \text{Concentration} \times \text{Volume (dm³)} \]

\[ = 0.1 \times \frac{25}{1000} = 0.1 \times 0.025 = 0.0025 \;\text{mol} \]

Step 3: Find moles of H₂SO₄ required

• From the equation, 1 mol H₂SO₄ reacts with 2 mol NaOH.
• So, moles of H₂SO₄ = \(\frac{0.0025}{2} = 0.00125\) mol

Step 4: Calculate volume of H₂SO₄

\[ \text{Volume (dm³)} = \frac{\text{Moles}}{\text{Concentration}} = \frac{0.00125}{0.5} = 0.0025 \;\text{dm³} \]

Convert to cm³: \[ 0.0025 \times 1000 = 2.5 \;\text{cm³} \]

Correct Answer: 2.5 cm³