Question

A sample of battery acid is to be analyzed for its sulfuric acid content. A 1.00-mL sample weighs 1.227 g . This 1.00-mL sample is diluted to 250.0 mL, and 10.00 mL of this diluted acid requires 35.05 mL of 4.462×10−3 M Ba(OH)2 for its titration. What is the mass percent of H2SO4 in the battery acid?

Answer 1

Answer: The mass percent of $H_2SO_4$ in the battery acid is 31.9 %

Explanation:

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

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

We are given:

Molarity of $Ba(OH)_2$ solution = $4.462\times 10^{-3}M$

Volume of solution = 35.05 mL

Putting values in above equation, we get:

$4.462\times 10^{-3}M=\frac{\text{Moles of }Ba(OH)_2\times 1000}{35.05mL}\\\\\text{Moles of }Ba(OH)_2=0.00016mol$

• The chemical equation for the reaction of barium hydroxide and sulfuric acid follows the reaction:

$Ba(OH)_2+H_2SO_4\rightarrow BaSO_4+2H_2O$

By Stoichiometry of the reaction:

1 mole of barium hydroxide reacts with 1 mole of sulfuric acid

So, 0.00016 moles of barium hydroxide will react with = $\frac{1}{1}\times 0.00016=0.00016mol$ of sulfuric acid.

The calculated moles of sulfuric acid is present in 10 mL of solution.

To calculate the number of moles in 250 mL of solution, we use unitary method:

In 10 mL, the number of moles of sulfuric acid are 0.00016 moles

So, in 250 mL, the number of moles of sulfuric acid will be = $\frac{0.00016}{10}\times 250=0.004moles$

The calculated moles of sulfuric acid are present in 1 mL of solution.

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

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

Molar mass of sulfuric acid = 98 g/mol

Moles of sulfuric acid = 0.004 moles

Putting values in above equation, we get:

$0.004mol=\frac{\text{Mass of }H_2SO_4}{98g/mol}\\\\\text{Mass of }H_2SO_4=0.398g$

• To calculate the percentage composition of sulfuric acid in battery acid, we use the equation:

$\%\text{ composition of }H_2SO_4=\frac{\text{Mass of }H_2SO_4}{\text{Mass of battery acid}}\times 100$

Mass of battery acid = 1.227 g

Mass of sulfuric acid = 0.398 g

Putting values in above equation, we get:

$\%\text{ composition of }H_2SO_4=\frac{0.398g}{1.227g}\times 100=31.9\%$

Hence, the mass percent sulfuric acid in battery acid is 31.9 %.