A chemistry student weighs out 0.137g of citric acid H3C6H5O7, a triprotic acid, into a 250.mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1600M NaOH solution.Calculate the volume of NaOH solution the student will need to add to reach the final equivalence point. Round your answer to 3 significant digits.
Question
A chemistry student weighs out 0.137g of citric acid H3C6H5O7, a triprotic acid, into a 250.mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1600M NaOH solution.Calculate the volume of NaOH solution the student will need to add to reach the final equivalence point. Round your answer to 3 significant digits.
Solution
To solve this problem, we need to follow these steps:
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Calculate the moles of citric acid: Citric acid is a triprotic acid, meaning it can donate three protons (H+ ions) per molecule. The molar mass of citric acid is approximately 192.13 g/mol. So, the moles of citric acid can be calculated as follows:
Moles of citric acid = mass / molar mass = 0.137 g / 192.13 g/mol = 0.000712 mol
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Calculate the moles of NaOH needed: Since citric acid is triprotic and NaOH is monoprotic, it takes three moles of NaOH to neutralize one mole of citric acid. So, the moles of NaOH needed can be calculated as follows:
Moles of NaOH = 3 * moles of citric acid = 3 * 0.000712 mol = 0.002136 mol
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Calculate the volume of NaOH solution needed: The molarity (M) of a solution is defined as the moles of solute per liter of solution. So, the volume of NaOH solution needed can be calculated as follows:
Volume of NaOH solution = moles of NaOH / molarity of NaOH = 0.002136 mol / 0.1600 mol/L = 0.01335 L
Convert this volume to milliliters (since the problem asks for the volume in milliliters): = 0.01335 L * 1000 mL/L = 13.35 mL
So, the student will need to add approximately 13.4 mL of NaOH solution to reach the final equivalence point.
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