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

5 0

3 years ago

A student determines that she used 0.0665 mol of sodium hydroxide (NaOH) to completely titrate 25.00 mL of sulfuric acid solutio

AysviL [449]

To determine the concentration of one solution which is specifically basic or acidic solution through taking advantage on its points of equivalence, titration analysis is done.

Let us determine the reaction for the titration below:

2NaOH +2H2SO4 = Na2SO4 +2H2O

So,

0.0665 mol NaOH (2 mol H2SO4/ 2mol NaOH) / .025 L solution

= 2.62 M H2SO4

The answer is the fourth option:

2.62 M

7 0

3 years ago

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