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Bad White [126]
3 years ago
14

An analytical chemist weighs out 0.188 g of an unknown triprotic acid into a 250 mL volumetric flask and dilutes to the mark wit

h 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.
Chemistry
1 answer:
Anon25 [30]3 years ago
7 0

<u>Answer:</u> The molar mass of unknown triprotic acid is 97.66 g/mol

<u>Explanation:</u>

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

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of triprotic acid

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=3\\M_1=?M\\V_1=250mL\\n_2=1\\M_2=0.0600M\\V_2=95.9mL

Putting values in above equation, we get:

3\times M_1\times 250=1\times 0.0600\times 95.9\\\\M_1=0.0077M

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

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

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:

0.0077M=\frac{0.188\times 1000}{\text{Molar mass of triprotic acid}\times 250}\\\\\text{Molar mass of triprotic acid}=97.66g/mol

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

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