Question

Question 19 A chemist dissolves of pure hydroiodic acid in enough water to make up of solution. Calculate the pH of the solution. Round your answer to significant decimal places.

Answer 1

The question is incomplete , complete question is:

A chemist dissolves 640.0 mg of pure hydroiodic acid in enough water to make up 280.0 mL of solution . Calculate the pH of the solution. Round your answer to significant decimal places.

Answer:

1.8 is the pH of the solution.

Explanation:

$HI(aq)\rightarrow H^+(aq)+I^-(aq)$

Mass of hydroiodic acid = 640.0 mg = 0.640 g

1 mg = 0.001 g

Moles of hydroiodic acid =$\frac{0.640 g}{128 g/mol}=0.005 mol$

1 mole of hydroiodic acid gives 1 mole of hydrogen ions and 1 mole of iodide ions.

Then 0.005 moles of hydroiodic acid will give :

1 × 0.005 mol = 0.005 mol

Moles of hydrogen ions = 0.005 mol

Volume of the hydroiodic solution = 280.0 mL - 0.280 L ( 1 mL = 0.001 L)

$[Concentration]=\frac{Moles}{Volume (L}$

$[H^+]=\frac{0.005 mol}{0.280 L}=0.02$

The pH of the solution is given by :

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

$pH=-\log[0.01786 M]=1.8$

1.8 is the pH of the solution.