Problem 103 - Chemistry: An Atoms-Focused Approach

Question

The solubility of slaked lime, \(\mathrm{Ca}(\mathrm{OH})_{2},\) in water at \(20^{\circ} \mathrm{C}\) is \(0.185 \mathrm{g} / 100.0 \mathrm{mL} .\) What volume of \(0.00100 M \mathrm{HCl}\) is needed to neutralize \(10.0 \mathrm{mL}\) of a saturated \(\mathrm{Ca}(\mathrm{OH})_{2}\) solution?

Step-by-Step Solution

Answer

Answer: 500 mL

Step 1: Calculate the moles of Ca(OH)₂ in the 10.0 mL saturated solution.

First, we need to convert the solubility of Ca(OH)₂ from grams per 100 mL to moles per 10 mL. Solubility of Ca(OH)₂ = 0.185 g / 100.0 mL Mass of Ca(OH)₂ in 10.0 mL = (0.185 g / 100.0 mL) * 10.0 mL = 0.0185 g Next, we need to find the molar mass of Ca(OH)₂: Ca(OH)₂ = Ca + 2(O) + 2(H) Molar mass of Ca(OH)₂ = 40.08 (Ca) + 2 * 16.00 (O) + 2 * 1.01 (H) = 74.10 g/mol Now, we can find the moles of Ca(OH)₂ in 10.0 mL saturated solution: Moles of Ca(OH)₂ = Mass / Molar Mass Moles of Ca(OH)₂ = 0.0185 g / 74.10 g/mol = 2.50 * 10⁻⁴ mol

Step 2: Determine the stoichiometry of the reaction between Ca(OH)₂ and HCl.

The balanced chemical equation between Ca(OH)₂ and HCl is: Ca(OH)₂ + 2 HCl → CaCl₂ + 2 H₂O From the balanced equation, we can see that 1 mole of Ca(OH)₂ reacts with 2 moles of HCl.

Step 3: Calculate the moles of HCl required to neutralize the amount of Ca(OH)₂.

Using the stoichiometry from Step 2, we can find the moles of HCl needed to neutralize the given moles of Ca(OH)₂: Moles of HCl = 2 * Moles of Ca(OH)₂ Moles of HCl = 2 * 2.50 * 10⁻⁴ mol = 5.00 * 10⁻⁴ mol

Step 4: Find the volume of 0.00100 M HCl needed for the neutralization reaction.

Now, we can find the volume of 0.00100 M HCl solution required to provide the needed moles of HCl using the relationship: Volume = Moles of HCl / Molarity of HCl Volume = (5.00 * 10⁻⁴ mol) / 0.00100 M = 500 mL Thus, 500 mL of 0.00100 M HCl is needed to neutralize 10.0 mL of a saturated Ca(OH)₂ solution.