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What volume of 0.0105-M HBr solution is required to titrate 125 mL of a 0.0100-M Ca(OH)₂ solution? Ca(OH)₂(aq) + 2HBr(aq) ⟶ CaBr₂(aq) + 2H₂O(l)

a) 118.5 mL
b) 125.0 mL
c) 131.0 mL
d) 140.2 mL

1 Answer

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Final answer:

To titrate 125 mL of a 0.0100-M Ca(OH)₂ solution, 238 mL of 0.0105-M HBr solution is required.

Step-by-step explanation:

To determine the volume of 0.0105-M HBr solution required to titrate 125 mL of a 0.0100-M Ca(OH)₂ solution, we can use the balanced chemical equation and stoichiometry.

The equation shows that 1 mole of Ca(OH)₂ reacts with 2 moles of HBr. First, calculate the moles of Ca(OH)₂ in the given volume. Moles of Ca(OH)₂ = volume (L) × concentration (M) = 0.125 L × 0.0100 M = 0.00125 mol. Since the molar ratio of Ca(OH)₂ to HBr is 1:2, the moles of HBr needed is 2 times the moles of Ca(OH)₂.

Therefore, moles of HBr needed = 2 × 0.00125 mol = 0.0025 mol. Finally, calculate the volume of the 0.0105-M HBr solution needed to provide 0.0025 mol. Volume = moles (HBr) / concentration (HBr) = 0.0025 mol / 0.0105 M = 0.238 L = 238 mL.

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