A chemistry student weighs out 0.172 grams of acrylic acid (HCH2CHCO2) into a volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1800 moles/liter NaOH solution.

Calculate the volume of NaOH solution the student will need to add to reach the equivalence point.

moles acrylic acid = grams/molar mass

moles acrylic acid = moles NaOH.
M NaOH = moles NaOH/L NaOH.
You know M NaOH and moles NaOH, solve for L NaOH and convert to mL.

To calculate the volume of NaOH solution needed to reach the equivalence point, we need to determine the number of moles of acrylic acid in the 0.172 grams of HCH2CHCO2.

1. Calculate the molar mass of HCH2CHCO2:
H = 1.01 g/mol x 3 = 3.03 g/mol
C = 12.01 g/mol x 4 = 48.04 g/mol
O = 16.00 g/mol x 2 = 32.00 g/mol

Molar mass of HCH2CHCO2 = 3.03 g/mol + 48.04 g/mol + 32.00 g/mol = 83.07 g/mol

2. Convert the mass of acrylic acid to moles:
moles = mass / molar mass

moles = 0.172 g / 83.07 g/mol ≈ 0.00207 mol

3. The balanced chemical equation for the reaction between acrylic acid (HCH2CHCO2) and NaOH is:
HCH2CHCO2 + NaOH → CH3CHOHCO2Na + H2O

From the equation, we can see that 1 mole of acrylic acid reacts with 1 mole of NaOH.

4. Calculate the volume of NaOH solution needed:
moles = volume (L) × concentration (mol/L)

Given concentration = 0.1800 mol/L

volume (L) = moles / concentration

volume (L) = 0.00207 mol / 0.1800 mol/L ≈ 0.0115 L

Therefore, the student will need to add approximately 0.0115 liters (or 11.5 milliliters) of NaOH solution to reach the equivalence point.