Question

A chemist dissolves 614. mg of pure hydrochloric acid in enough water to make up 400 mL of solution. Calculate the pH of the solution. Round your answer to 3 significant decimal places. X 5 ? 1 Don't Know Suomi

Answer 1

Answer:

The pH of the solution is 1.38.

Explanation:

Mass of HCl = 614 mg = 0.614 g

Moles of HCl = $\frac{0.614 g}{36.5 g/ mol}=0.01682 mol$

Concentration of HCl :

$Concentration =\frac{Moles}{\text{Volume of solution in L}}$

On adding 0.01682 moles to 400 mL of water that 0.4 L of water.

$[HCl]=\frac{0.01682 mol}{0.4 L}=0.04205 M$

$HCl(s)+H_2O(l)\rightarrow H_3O^+(aq)+Cl^-(aq)$

1 mole of HCl gives 1 mole of hydronium ion and 1 mole of chloride ions in an aqueous solution.

Then 0.04205 mol/L of HCl will give:

$1\times 0.04205 M =0.04205 mol/L$ of hydronium ions.

$[H_3O^+]=0.04205 M$

$pH=-\log [H_3O^+]$

$pH=-\log [0.04205 M]=1.38$

The pH of the solution is 1.38.