Questions: An analytical chemist weighs out 0.165 g of an unknown diprotic acid into a 250 mL volumetric flask and dilutes to the mark with distilled water. He then titrates this solution with 0.0600 M NaOH solution. When the titration reaches the equivalence point, the chemist finds he has added 41.0 mL of NaOH solution. Calculate the molar mass of the unknown acid. Be sure your answer has the correct number of significant digits.

Solution Steps

First, we need to calculate the number of moles of NaOH used in the titration. We use the formula:

\[ \text{moles of NaOH} = \text{molarity of NaOH} \times \text{volume of NaOH (in liters)} \]

Given:

• Molarity of NaOH, \( M = 0.0600 \, \text{M} \)
• Volume of NaOH, \( V = 41.0 \, \text{mL} = 0.0410 \, \text{L} \)

\[ \text{moles of NaOH} = 0.0600 \, \text{M} \times 0.0410 \, \text{L} = 0.002460 \, \text{mol} \]

Since the acid is diprotic, each mole of acid reacts with 2 moles of NaOH. Therefore, the moles of the diprotic acid can be calculated as:

\[ \text{moles of acid} = \frac{\text{moles of NaOH}}{2} \]

\[ \text{moles of acid} = \frac{0.002460 \, \text{mol}}{2} = 0.001230 \, \text{mol} \]

The molar mass \( M \) of the diprotic acid is given by:

\[ M = \frac{\text{mass of acid}}{\text{moles of acid}} \]

Given:

• Mass of the acid, \( m = 0.165 \, \text{g} \)
• Moles of the acid, \( n = 0.001230 \, \text{mol} \)

\[ M = \frac{0.165 \, \text{g}}{0.001230 \, \text{mol}} = 134.1463 \, \text{g/mol} \]

Rounding to the correct number of significant digits (3 significant digits, based on the given data):

\[ M = 134 \, \text{g/mol} \]

Final Answer