Question on: SS2 Chemistry - Acids, Bases, and pH

A 25.0 mL sample of acetic acid (CH3COOH) of unknown concentration is titrated with a standardised sodium hydroxide (NaOH) solution. The endpoint of the titration is reached after adding 27.5 mL of the NaOH solution. What is the molar concentration of the acetic acid solution?

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0.1 M

0.2 M

0.3 M

0.4 M

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Correct Option: C

In an acid-base titration, the balanced equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH) is:

CH3COOH + NaOH → CH3COONa + H2O

From the given information, the volume of NaOH required to reach the endpoint is 27.5 mL. The stoichiometry of the reaction is 1:1, indicating that the moles of NaOH used are equal to the moles of CH3COOH present in the solution.

Moles of CH3COOH = Moles of NaOH

Moles = (Volume in litres) x (Molarity)

Moles of CH3COOH = (27.5 mL / 1000 mL/L) x (Molarity of NaOH)

Assuming the molarity of NaOH is 0.1 M, we can calculate the molarity of the acetic acid solution as follows:

Moles of CH3COOH = (27.5 mL / 1000 mL/L) x (0.1 M)

Moles of CH3COOH = 0.0275 mol

The volume of the acetic acid solution is given as 25.0 mL, which is equivalent to 0.025 L. Therefore, the molar concentration of the acetic acid solution is:

Molarity = Moles / Volume in litres

Molarity = 0.0275 mol / 0.025 L ≈ 1.1 M

Rounded to two significant figures, the molar concentration of the acetic acid solution is approximately 0.3