A chemist dissolves of pure hydroiodic acid in enough water to make up of solution. Calculate the pH of the solution. Be sure your answer has the correct number of significant digits.

1.76

Explanation:

There is some info missing. I think this is the original question.

A chemist dissolves 660.mg of pure hydroiodic acid in enough water to make up 300.mL of solution. Calculate the pH of the solution. Be sure your answer has the correct number of significant digits.

Step 1: Calculate the molarity of HI(aq)

M = mass of solute / molar mass of solute × liters of solution

M = 0.660 g / 127.91 g/mol × 0.300 L

M = 0.0172 M

Step 2: Write the acid dissociation reaction

HI(aq) ⇄ H⁺(aq) + I⁻(aq)

HI is a strong acid, so [H⁺] = 0.0172 M

Step 3: Calculate the pH

pH = -log [H⁺]

pH = -log 0.0172

pH = 1.76

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