Problem: 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/molB. 249.6 g/molC. 499.1 g/molD. 269.4 g/molE. 49.9 g/mol

FREE Expert Solution

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



Recall that at the equivalence point of a titration:


moles acid=moles base


Also, recall that moles = molarity × volume

This means:


MVacid=MVbase


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

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Problem Details

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.