Questions: A 0.2321 g sample of a monoprotic acid neutralizes 57.08 mL of a 0.06453 M KOH solution. Calculate the molar mass of the acid. Be sure your answer has the correct number of significant digits.

Solution Steps

To find the moles of KOH used in the reaction, use the formula:

\[ \text{moles of KOH} = \text{volume (L)} \times \text{molarity (M)} \]

Convert the volume from mL to L:

\[ 57.08 \, \text{mL} = 0.05708 \, \text{L} \]

Now calculate the moles:

\[ \text{moles of KOH} = 0.05708 \, \text{L} \times 0.06453 \, \text{M} = 0.003682 \, \text{mol} \]

Since the acid is monoprotic, it reacts with KOH in a 1:1 molar ratio. Therefore, the moles of acid are equal to the moles of KOH:

\[ \text{moles of acid} = 0.003682 \, \text{mol} \]

The molar mass of the acid is calculated using the formula:

\[ \text{molar mass} = \frac{\text{mass of acid (g)}}{\text{moles of acid (mol)}} \]

Substitute the known values:

\[ \text{molar mass} = \frac{0.2321 \, \text{g}}{0.003682 \, \text{mol}} = 63.03 \, \text{g/mol} \]

Final Answer