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

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.

answer: mole of the gas

answer: mole of the gasexplanation: