A sample of an unknown monoprotic organic acid is dissolved in water and titrated with a sodium hydr — Ionic Equilibrium Chemistry Question
Question
A $0.28 \text{ g}$ sample of an unknown monoprotic organic acid is dissolved in water and titrated with a $0.1 \text{ M}$ sodium hydroxide solution. After the addition of $17.5 \text{ ml}$ of base, a pH of 5.0 is recorded. The equivalence point is reached when a total of $35.0 \text{ ml}$ of $\text{NaOH}$ is added. The molar mass of the organic acid is
💡 Solution & Explanation
At the equivalence point, the moles of base added equals the initial moles of the monoprotic organic acid. Moles of NaOH added = $35.0 \text{ ml} \times 0.1 \text{ M} = 3.5 \text{ mmol} = 0.0035 \text{ moles}$. Therefore, there were $0.0035 \text{ moles}$ of the unknown acid. The molar mass of the acid is defined as mass divided by moles: $\text{Molar mass} = 0.28 \text{ g} / 0.0035 \text{ mol} = 80 \text{ g/mol}$. Therefore, correct answer is B.