Answer:
pOH is 12.3
Explanation:
let's first get the pH of nitric acid:
[tex]{ \sf{pH = - log[H {}^{ + } ]}}[/tex]
but:
[H†] = 0.020 M
therefore:
[tex]{ \sf{pH = - log(0.020) }} \\ { \sf{pH = 1.7}}[/tex]
but, pOH = 14 - pH
[tex]{ \sf{pOH = 14 - 1.7}} \\ { \sf{pOH = 12.3}}[/tex]
