We’re being asked to **calculate the molar mass of a weak monoprotic acid**.

Recall that at the ** equivalence point** of a titration:

$\overline{){\mathbf{moles}}{\mathbf{}}{\mathbf{acid}}{\mathbf{=}}{\mathbf{moles}}{\mathbf{}}{\mathbf{base}}}$

Also, recall that **moles = molarity × volume**.

This means:

$\overline{){\left(\mathbf{MV}\right)}_{{\mathbf{acid}}}{\mathbf{=}}{\left(\mathbf{MV}\right)}_{{\mathbf{base}}}}$

We're going to calculate for the molar mass of the weak monoprotic acid using the following steps:

A 0.70 g sample of a weak monoprotic acid was dissolved in 150 mL of water and titrated with a 0.055 M solution of KOH at 25°C. At the half-equivalence point, after addition of 25.5 mL of the KOH solution, the pH was 3.8.

Determine the molecular weight (in g/mol) of the acid.

A. 84.8 g/mol

B. 249.6 g/mol

C. 499.1 g/mol

D. 269.4 g/mol

E. 49.9 g/mol

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Equivalence Point concept. If you need more Equivalence Point practice, you can also practice Equivalence Point practice problems.