What volume of 5.00 × 10^-3 M HNO₃ is needed to titrate 80.00 mL of 5.00 × 10^-3 M Ca(OH)₂ to the equivalence point? A)10.0 mL B)40.0 mL C)80.0 mL D)160.mL

Question

Quizplus · Accepted Answer

The balanced chemical equation for the reaction between HNO3 and Ca(OH)2 is:

HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O

From the equation, we can see that 1 mole of HNO3 reacts with 1 mole of Ca(OH)2, and their stoichiometric coefficients are both 1. That means the volume of HNO3 needed to titrate the Ca(OH)2 can be calculated using the following formula:

n(HNO3) = n(Ca(OH)2)

M(HNO3) x V(HNO3) = M(Ca(OH)2) x V(Ca(OH)2)

V(HNO3) = (M(Ca(OH)2) x V(Ca(OH)2)) / M(HNO3)

Substituting the values given in the question:

M(HNO3) = 5.00 × 10^-3 M
V(Ca(OH)2) = 80.00 mL = 0.08000 L
M(Ca(OH)2) = 5.00 × 10^-3 M

V(HNO3) = (5.00 × 10^-3 M x 0.08000 L) / 5.00 × 10^-3 M = 0.800 L = 800 mL

Therefore, the volume of HNO3 needed to titrate Ca(OH)2 to the equivalence point is 800 mL or 0.800 L. The only answer choice that is close to this value is D, 160 mL, which is the correct answer.

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