Achemistry 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.

27.06 mL.

Explanation:

Molarity is the no. of moles of solute dissolved in a 1.0 L of the solution.

M = (no. of moles of acrylic acid)/(V of the solution (L))

M = (mass/molar mass)acrylic acid / (V of the solution (L))

mass of acrylic acid = 0.0975 g, molar mass of acrylic acid = 72.06 g/mol, V of the solution = 250 mL = 0.25 L.

∴ M = (0.0975 g/72.06 g/mol)/(0.25 L) = 0.0054 M.

The no. of millimoles of acid is equal to that of the base at the neutralization.

∴ (XMV) NaOH = (XMV) acrylic acid.

X is the no. of reproducible H⁺ (for acid) or OH⁻ (for base),

M is the molarity.

V is the volume.

X = 1, M = 0.05 M, V = ??? mL.

X = 1, M = 0.0054 M, V = 250.0 mL.

∴ V of NaOH = (XMV) acrylic acid/(XM) NaOH = (1)(0.0054 M)(250.0 mL)/(1)(0.05 M) = 27.06 mL.

pie chart because it is in percents ( % )