A chemistry student weighs out of acetic acid into a volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with solution. Calculate the volume of solution the student will need to add to reach the equivalence point. Be sure your answer has the correct number of significant digits.

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Question: A chemistry student weighs out 0.112g 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 0.1600 M NaOH solution. Calculate the volume of solution the student will need to add to reach the equivalence point. Be sure your answer has the correct number of significant digits.

Volume of NaOH is 11.6 mL

Explanation:

The reaction of acetic acid with NaOH is as follows:

CH3COOH + NaOH > CH3COONa + H2O

M1V1 = M2V2

Here M1 V1 are molarity and volume of acetic acid.

M2, V2 are molarity and volume of NaOH.

Number of moles of acetic acid:

0.112 g CH3COOH × (1 mol / 60.05 g) = 0.001865 mol

Molarity = moles of solute / Liters of solution

Molarity = 0.001865 mol / 0.250 L = 0.00746 M

Hence,

M1 = 0.00746 M

V1 = 250 mL

M2 = 0.160 M

V2 = ?

V2 = M1V1 / M2

V2 = 0.00746 M × 250 mL / 0.160 M

V2 = 11.6 mL

Hence the volume of NaOH is 11.6 mL

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