# 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:

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:

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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