Question

An analytical chemist is titrating 54.8 mL of a 0.8900 M solution of hydrazoic acid (HN3) with a 0.3500 M solution of KOH. The pKa of hydrazoic acid is 4.72. Calculate the pH of the acid solution after the chemist has added 116. mL of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added. Round your answer to decimal places.

Answer 1

Answer:

The pH of this solution is 5.4

Explanation:

Step 1: Data given

Volume of a 0.8900 M hydrazoic acid (HN3) solution = 54.8 mL = 0.0548 L

The pKa of hydrazoic acid is 4.72.

Molarity of KOH = 0.3500 M

Volume of KOH solution added = 116 mL = 0.116 L

Step 2: The balanced equation

HN3 + KOH → KN3 + H2O

Step 3: Calculate moles HN3

Moles HN3 = molarity * volume

Moles HN3 = 0.8900 M * 0.0548 L

Moles HN3 = 0.048772 moles

Step 4: Calculate moles KOH

Moles KOH = 0.3500 M * 0.116 L

Moles KOH = 0.0406 moles

Step 5: Calculate the limiting reactant

For 1 mol HN3 we need 1 mol KOH to produce 1 mol KN3 and 1 mol H2O

KOH is the limiting reactant. There will react 0.0406 moles

HN3 is in excess. There will react 0.0406 moles. There will remain 0.048772 - 0.0406 = 0.008172 moles HN3

Step 6: Calculate moles KN3 produce

For 1 mol HN3 we need 1 mol KOH to produce 1 mol KN3 and 1 mol H2O

There will be 0.0406 moles KN3 produced

Step 7: Calculate molarity

Molarity = moles / volume

[HN3] = 0.008172 moles / 0.1708 L

[HN3] = 0.0478 M

[KN3] = 0.0406 moles / 0.1708 L

[KN3] = 0.2377 M

Step 8: Calculate the pH

pH = pKa + log ([salt] / [acid])

pH = 4.72 + log (0.2377/0.0478)

pH = 5.4

The pH of this solution is 5.4