Calculate the pH of the following solutions:

Question

A) 0.03 M HNO3

B) 0.02M H3PO4

C) 0.03M NaOH

Answer 1

pH = -log [H⁺]

pOH = -log [OH^-]

pH = pOH = 14

HNO₃ is a strong acid and dissociates as follows:

HNO₃ ==> H⁺ + NO₃^-

0.03 M HNO₃ = 0.03 M H⁺

pH = -log 0.03

pH = 1.52

H₃PO₄ is actually not a strong acid, but we'll treat the first dissociation as such, since no Ka is provided.

H₃PO₄ ==> H⁺ + H₂PO₄^2-

0.02 M H³PO₄ = 002 M H⁺

pH = -log 0.02

pH = 1.70

NaOH is a strong base and ionizes as follows:

NaOH ==> Na+ + OH-

0.03 M NaOH = 0.03 M OH-

pOH = -log 0.03

pOH = 1.52

pH = 14 - 1.52

pH = 12.48