Question

When a 17.9 mL sample of a 0.458 M aqueous nitrous acid solution is titrated with a 0.368 M aqueous potassium hydroxide solution, what is the pH after 33.4 mL of potassium hydroxide have been added

Answer 1

pH = 12.90

Step-by-step explanation:

THe reaction of HNO₃ with KOH is:

HNO₂ + KOH → KNO₂ + H₂O

That means 1 mole of nitrous acid reacts per mole of potassium hydroxide.

To solve this question, we need to find the moles of each reactant:

Moles HNO₂:

0.0179L * (0.458mol / L) = 0.00820 moles

Moles KOH:

0.0334L * (0.368mol / L) = 0.01229 moles

That means KOH is in excess. The moles in excess are:

0.01229 moles - 0.00820 moles = 0.00409 moles KOH = Moles OH⁻

The [OH⁻] is -Total volume = 17.9mL+33.4mL = 51.3mL = 0.0513L-:

0.00409 moles / 0.0513L =

0.0797M =[OH⁻]

pOH = -log[OH⁻] = 1.098

pH = 14 - pOH

pH = 12.90