Answer :
D) 80.0 mL volume of 5.00 × 10^-3 M HNO3 is needed to titrate 40.00 mL of 5.00 × 10^-3 M Ca(OH)2 to the equivalence point.
The equivalence point is the point at which the number of moles of acid and base are equal. Titration is the process of adding a known volume of a solution of known concentration to a known volume of a solution of unknown concentration until the reaction between the two is complete.
In this case, we have a solution of HNO3 of concentration 5.00 × 10^-3 M and a solution of Ca(OH)2 of concentration 5.00 × 10^-3 M. We want to know the volume of HNO3 needed to titrate 40.00 mL of Ca(OH)2 to the equivalence point.
The balanced chemical equation for the reaction between HNO3 and Ca(OH)2 is:
2HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O
From the equation, we can see that 2 moles of HNO3 react with 1 mole of Ca(OH)2. Therefore, the number of moles of HNO3 needed to react with 1 mole of Ca(OH)2 is 2.
To find the number of moles of Ca(OH)2 in 40.00 mL of a 5.00 × 10^-3 M solution, we use the formula:
moles of solute = concentration × volume (in liters)
moles of Ca(OH)2 = 5.00 × 10^-3 M × 40.00 mL / 1000 mL/L = 2.00 × 10^-4 moles
Since we need 2 moles of HNO3 to react with 1 mole of Ca(OH)2, we need:
2 × 2.00 × 10^-4 = 4.00 × 10^-4 moles of HNO3
To find the volume of 5.00 × 10^-3 M HNO3 needed to provide 4.00 × 10^-4 moles of HNO3, we use the formula:
moles of solute = concentration × volume (in liters)
4.00 × 10^-4 moles = 5.00 × 10^-3 M × volume / 1000 mL/L
volume = 8.00 × 10^-2 L = 80.0 mL
Therefore, the correct answer is D) 80.0 mL.
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