How many milliliters of 0.0200 M Ca(OH)_2 are required to completely neutralize 75.0 mL of 0.0300 M HCl?

• A. 75.0 mL
• B. 112.5 mL
• C. 33.8 mL
• D. 56.3 mL

Answer: (D)

Verified step by step guidance

1. Write the balanced chemical equation for the neutralization reaction between calcium hydroxide and hydrochloric acid: \(\mathrm{Ca(OH)_2 + 2HCl \rightarrow CaCl_2 + 2H_2O}\).
2. Identify the mole ratio between \(\mathrm{Ca(OH)_2}\) and \(\mathrm{HCl}\) from the balanced equation. Here, 1 mole of \(\mathrm{Ca(OH)_2}\) reacts with 2 moles of \(\mathrm{HCl}\).
3. Calculate the moles of \(\mathrm{HCl}\) present using its concentration and volume: \(\text{moles HCl} = M_{HCl} \times V_{HCl} = 0.0300\,M \times 0.0750\,L\).
4. Use the mole ratio to find the moles of \(\mathrm{Ca(OH)_2}\) needed to neutralize the given moles of \(\mathrm{HCl}\): \(\text{moles Ca(OH)_2} = \frac{1}{2} \times \text{moles HCl}\).
5. Calculate the volume of \(\mathrm{Ca(OH)_2}\) solution required using its concentration: \(V_{Ca(OH)_2} = \frac{\text{moles Ca(OH)_2}}{M_{Ca(OH)_2}}\). Convert this volume to milliliters if necessary.