You dissolve 1.22 g of an unknown diprotic acid in 155.0 mL of H₂O. This solution is just neutralized by 6.22 mL of a 1.23 M NaOH solution. What is the molar mass of the unknown acid?

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Quizplus · Accepted Answer

First, calculate the moles of NaOH used: $6.22 \, \text{mL} \times \frac{1.23 \, \text{mol}}{1000 \, \text{mL}} = 0.00765 \, \text{mol} \, \text{NaOH}$. Since the acid is diprotic, the moles of acid neutralized by NaOH are half the moles of NaOH, so $0.00765 \, \text{mol} \, \text{NaOH} / 2 = 0.003825 \, \text{mol} \, \text{acid}$. The molar mass of the acid is then $1.22 \, \text{g} / 0.003825 \, \text{mol} = 319 \, \text{g/mol}$, which corresponds to $3.19 \times 10^{2} \, \text{g/mol}$.

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