How many milliliters of 0.534 M H₂SO₄ are required to neutralize 245 mL of 0.633 M KOH?

• A. 205 mL
• B. 145 mL
• C. 182 mL
• D. 230 mL

Answer: (C)

Verified step by step guidance

1. Start by writing the balanced chemical equation for the neutralization reaction between sulfuric acid (H₂SO₄) and potassium hydroxide (KOH): ${\text{H}}_2{\text{SO}}_4+2{\text{KOH}}_{}\to {\text{K}}_{}{\text{2}}_{}{\text{SO}}_4+2{\text{H}}_{}{\text{O}}_{}$.
2. Identify the mole ratio from the balanced equation. For every 1 mole of H₂SO₄, 2 moles of KOH are required.
3. Calculate the moles of KOH present using its concentration and volume: $\text{moles of KOH =}\text{0.633 M}\times \text{0.245 L}$. Remember to convert milliliters to liters by dividing by 1000.
4. Use the mole ratio to find the moles of H₂SO₄ needed: $\text{moles of H₂SO₄ =}\frac{\text{moles of KOH}}{2}$.
5. Calculate the volume of H₂SO₄ required using its concentration: $\text{Volume of H₂SO₄ =}\frac{\text{moles of H₂SO₄}}{\text{0.534 M}}$. Convert the result from liters to milliliters by multiplying by 1000.