Final answer:
To neutralize 265 mL of 1.18 M Ca(OH)2, 355.34 mL of 1.76 M CH3COOH is required. This was calculated using the stoichiometry of the balanced chemical equation between acetic acid and calcium hydroxide.
Step-by-step explanation:
To calculate the volume of 1.76 M CH3COOH (aqueous acetic acid) that is needed to neutrailize 265 mL of 1.18 M Ca(OH)2 (aqueous calcium hydroxide), we use the chemical reaction and stoichiometry. The balanced chemical equation for the reaction between acetic acid and calcium hydroxide is:
2 CH3COOH(aq) + Ca(OH)2(aq) → (CH3COO)2Ca(aq) + 2 H2O(l)
From the equation, we can see that it takes 2 moles of acetic acid to neutralize 1 mole of calcium hydroxide. First, we calculate the number of moles of calcium hydroxide:
Moles of Ca(OH)2 = 1.18 M × 0.265 L = 0.3127 moles
Next, we use the stoichiometry of the reaction to find moles of CH3COOH required:
Moles of CH3COOH needed = 0.3127 moles of Ca(OH)2 × 2 moles CH3COOH/mole Ca(OH)2 = 0.6254 moles
Finally, we calculate the volume of 1.76 M CH3COOH required:
Volume of CH3COOH = 0.6254 moles / 1.76 M = 0.35534 L = 355.34 mL
Therefore, the volume of 1.76 M CH3COOH required to completely neutralize 265 mL of 1.18 M Ca(OH)2 is 355.34 mL.