A chemistry student weighs out 0.0975 g of acrylic acid (HCH2CHCO2) into a 250.

Question

A chemistry student weighs out 0.0975 g of acrylic acid (HCH2CHCO2) into a 250. mL volumetric flask and diluted to the mark with distilled water. He plans to titrate the acid with 0.0500 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equilvalence point. Be sure your answer has the correct number of significant digits. A chemistry student weighs out 0.0975 g of acrylic acid (HCH2CHCO2) into a 250. mL volumetric flask and diluted to the mark with distilled water. He plans to titrate the acid with 0.0500 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equilvalence point. Be sure your answer has the correct number of significant digits.

Explanation / Answer

chemical equation,

HCH2CHCO2 + NaOH --> NaCH2CHCO2 + H2O

1 mole of acrylic acid requires 1 mole of NaOH to neutralize

moles of acrylic acid = grams/molar mass

                                 = 0.0975 g/72.063 g/mol

                                 = 0.00135 mol

moles of NaOH required = 0.00135 mol

Volume of NaOH needed = moles/molarity

                                         = 0.00135 mol/0.0500 M

                                         = 0.027 L

                                         = 27.00 mL

Thus, 0.027 L (or 27 ml) of 0.05 M NaOH is needed to neutralize arylic acid in solution to reach equivalence point of titration.

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