determining a volume of a base

Question

A chemistry student weighs out 0.131g of acetic acid (HCH3CO2) into a 250.mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1100 M 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.

Answer 1

For this problem, I would start by writing the balanced equation:

NaOH_(aq) + CH₃COOH_(aq) → NaCH₃COO_(aq) + H₂O_(l)

Now, I will solve for to moles of CH₃COOH:

0.131 g CH₃COOH (1 mol CH₃COOH / 60.052 g) = 0.00218 mol CH₃COOH

Since NaOH and CH₃COOH react in 1:1 mol ratios, and at equivalence, mol H⁺ = mol OH^-:

0.00218 mol CH₃COOH (1 mol NaOH / 1 mol CH₃COOH) = 0.00218 mol NaOH

Now, using the molarity of NaOH (0.1100 mol NaOH / 1 L), we can determine the liters or mL of NaOH:

0.00218 mol NaOH (1 L / 0.110 mol NaOH) = 0.0198 L NaOH (1000 mL / 1 L) = 19.8 mL NaOH

Note that we were given unnecessary information, the total volume of the analyte (250 mL of CH₃COOH)