Final answer:
The volume of 0.842 mol/L calcium hydroxide needed to neutralize 44.8 mL of 0.483 mol/L hydrochloric acid is calculated by dividing the required moles of calcium hydroxide by its concentration, which results in 12.8565 mL, although this answer is not listed among the provided choices.
Step-by-step explanation:
To determine the volume of 0.842 mol/L calcium hydroxide needed to neutralize 0.483 mol/L hydrochloric acid, we first need to understand the balanced chemical equation:
Ca(OH)2 + 2HCl → CaCl2 + 2H2O
From the equation, we can see that one mole of calcium hydroxide reacts with two moles of hydrochloric acid. Next, we calculate the number of moles of HCl present in 44.8 mL of the solution:
moles HCl = volume (L) × concentration (mol/L)
= 0.0448 L × 0.483 mol/L
= 0.0216384 mol
Since it takes two moles of HCl to react with one mole of Ca(OH)2, the moles of Ca(OH)2 needed would be half the moles of HCl, which is 0.0108192 mol. Finally, we can calculate the volume of Ca(OH)2 needed:
volume Ca(OH)2 = moles Ca(OH)2 / concentration
= 0.0108192 mol / 0.842 mol/L
= 0.0128565 L
= 12.8565 mL
However, we see that none of the answer choices match this calculated value. Therefore, we must reassess our calculations or the given answer choices to ensure we did not make a mistake.