A chemistry student weighs out 0.131g of acrylic acid HCH2CHCO2 into a 250.mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1800M NaOH solution.Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Round your answer to 3 significant digits.
Question
A chemistry student weighs out 0.131g of acrylic acid HCH2CHCO2 into a 250.mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1800M NaOH solution.Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Round your answer to 3 significant digits.
Solution
To solve this problem, we need to follow these steps:
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First, we need to calculate the number of moles of acrylic acid (HCH2CHCO2). The molar mass of acrylic acid is approximately 72.06 g/mol. So, the number of moles of acrylic acid is given by:
Moles of acrylic acid = mass / molar mass Moles of acrylic acid = 0.131 g / 72.06 g/mol = 0.00182 mol
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The balanced chemical equation for the reaction between acrylic acid and NaOH is:
HCH2CHCO2 + NaOH → NaCH2CHCO2 + H2O
From this equation, we can see that one mole of acrylic acid reacts with one mole of NaOH. Therefore, the number of moles of NaOH needed is also 0.00182 mol.
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Finally, we can calculate the volume of 0.1800 M NaOH solution needed using the formula:
Volume = moles / concentration Volume = 0.00182 mol / 0.1800 mol/L = 0.0101 L
Convert this volume to mL (since the question asks for the volume in mL) by multiplying by 1000:
Volume = 0.0101 L * 1000 = 10.1 mL
So, the student will need to add approximately 10.1 mL of NaOH solution to reach the equivalence point.
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