Question

By titration, 15.0 mLmL of 0.1008 MM sodium hydroxide is needed to neutralize a 0.2053-gg sample of an organic acid. What is the molar mass of the acid?

Answer 1

The question is incomplete, here is the complete question:

By titration, 15.0 mL of 0.1008 M sodium hydroxide is needed to neutralize a 0.2053-g sample of an organic acid. What is the molar mass of the acid if it is monoprotic.

Answer: The molar mass of monoprotic acid is 135.96 g/mol

Explanation:

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

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

• For sodium hydroxide:

Molarity of NaOH solution = 0.1008 M

Volume of solution = 15.0 mL

Putting values in above equation, we get:

$0.1008M=\frac{\text{Moles of NaOH}\times 1000}{15.0}\\\\\text{Moles of NaOH}=\frac{(0.1008\times 15.0)}{1000}=0.00151mol$

As, the acid is monoprotic, it contains 1 hydrogen ion

1 mole of $OH^-$ ion of NaOH neutralizes 1 mole of $H^+$ ion of monoprotic acid

So, 0.00151 moles of $OH^-$ ion of NaOH will neutralize $\frac{1}{1}\times 0.00151=0.00151mol$ of $H^+$ ion of monoprotic acid

In monoprotic acid:

1 mole of $H^+$ ion is released by 1 mole of monoprotic acid

So, 0.00151 moles of $H^+$ ion will be released by $\frac{1}{1}\times 0.00151=0.00151mol$ of monoprotic acid

To calculate the number of moles, we use the equation:

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

Moles of monoprotic acid = 0.00151 mole

Given mass of monoprotic acid = 0.2053 g

Putting values in above equation, we get:

$0.00151mol=\frac{0.2053g}{\text{Molar mass of monoprotic acid}}\\\\\text{Molar mass of monoprotic acid}=\frac{0.2053g}{0.00151mol}=135.96g/mol$

Hence, the molar mass of monoprotic acid is 135.96 g/mol