An analytical chemist weighs out 0.188 g of an unknown triprotic acid into a 250 ml volumetric flask and dilutes to the mark with distilled water. he then titrates this solution with 0.0600 m naoh solutions. when the titration reaches the equivalence point, the chemist finds he has added 95.9 ml of naoh solution.

calculate the molar mass of the unknown acid.

The molar mass of unknown triprotic acid is 97.66 g/mol

Explanation:

To calculate the molarity of acid, we use the equation given by neutralization reaction:

where,

are the n-factor, molarity and volume of triprotic acid

are the n-factor, molarity and volume of base which is NaOH.

We are given:

Putting values in above equation, we get:

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

We are given:

Molarity of solution = 0.0077 M

Given mass of triprotic acid = 0.188 g

Volume of solution = 250 mL

Putting values in above equation, we get:

Hence, the molar mass of unknown triprotic acid is 97.66 g/mol

a police officer and a ambulance

the dependant variable is spilled water

the independent is the heat