Question

Problem Page Question An analytical chemist weighs out of an unknown monoprotic acid into a volumetric flask and dilutes to the mark with distilled water. He then titrated this solution with solution. When the titration reaches the equivalence point, the chemist finds he has added of solution. Calculate the molar mass of the unknown acid. Round your answer to significant digits.

Answer 1

The question is incomplete. The complete question is:

An analytical chemist weighs out 0.093g of an unknown monoprotic acid into a 250mL volumetric flask and dilutes to the mark with distilled water. He then titrates this solution with 0.1600M NaOH solution. When the titration reaches the equivalence point, the chemist finds he has added 6.5mL of NaOH solution. Calculate the molar mass of the unknown acid. Round your answer to 2 significant digits.

Answer: The molar mass of the unknown acid is 89 g/mol

Explanation:

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

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Molarity of NaOH solution = 0.1600 M

Volume of  NaOH solution = 6.5 mL

Putting values in above equation, we get:

$0.1600M=\frac{\text{Moles of NaOH}\times 1000}{6.5}\\\\\text{Moles of NaOH}=\frac{0.1600\times 6.5}{1000}=1.04\times 10^{-3}moles$

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

$HA+NaOH\rightarrow NaA+H_2O$

By Stoichiometry of the reaction:

1 mole of NaOH reacts with 1 mole of HA

So,  moles of NaOH will react with =$\frac{1}{1}\times 1.04\times 10^{-3}=1.04\times 10^{-3}$ moles of HA

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of HA = 0.093  g

Moles of monoprotic acid =

Putting values in above equation, we get:

$1.04\times 10^{-3}=\frac{0.093g}{\text{Molar mass of monoprotic acid}}\\\\\text{Molar mass of monoprotic acid}=\frac{0.093g}{1.04\times 10^{-3}mol}=89g/mol$