A 0.870 g sample of a monoprotic acid is dissolved in water and titrated with 0.300 M KOH.
What is the molar mass of the acid if 23.0 mL of the KOH solution is required to neutralize the sample?

The molar mass of monoprotic acid is 126.1 g/mol

Explanation:

To calculate the molarity of solution, we use the equation:

Molarity of KOH solution = 0.300 M

Volume of solution = 23.00 mL

Putting values in above equation, we get:

The chemical equation for the reaction of monoprotic acid and KOH follows:

By Stoichiometry of the reaction:

1 mole of KOH reacts with 1 mole of HA

So, 0.0069 moles of KOH will react with = moles of HA

To calculate the molar mass for given number of moles, we use the equation:

Given mass of HA = 0.870 g

Moles of monoprotic acid = 0.0069 mol

Putting values in above equation, we get:

Hence, the molar mass of monoprotic acid is 126.1 g/mol

dynamic equilibrium is the answer.

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