Question

A chemist dissolves 581.1mg of pure perchloric acid in enough water to make up 150 ml of solution. Calculate the pH of the solution. Round your answer to significant decimal places.

Answer 1

Answer: The pH of the solution is 1.41

Explanation:

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

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

Given mass of perchloric acid = 581.1 mg = 0.5811 g   (Conversion factor:  1 g = 1000 mg)

Molar mass of perchloric acid = 100.5 g/mol

Volume of solution = 150 mL

Putting values in above equation, we get:

$\text{Molarity of perchloric acid}=\frac{0.5811\times 1000}{100.5\times 150}\\\\\text{Molarity of perchloric acid}=0.0385M$

1 mole of perchloric acid produces 1 mole of $H^+$ ions and 1 mole of $ClO_4^-$ ions

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

$pH=-\log[H^+]$

We are given:

$[H^+]=0.0385M$

Putting values in above equation, we get:

$pH=-\log (0.0385)=1.41$

Hence, the pH of the solution is 1.41