[H₃O⁺]=4.4 x 10⁻⁴
pH=3.357
[OH⁻] = 2.28 x 10⁻¹¹
Given
HNO3 concentration = 0.00044 M
Required
[H3O+],[OH-] and the pH
Solution
HNO₃ = strong acid
HNO₃ ⇒ H₃O⁺ + NO₃⁻
[H₃O⁺]=[HNO₃]=4.4 x 10⁻⁴
For strong acid pH=-log[H₃O⁺]
pH=-log 4.4 x 10⁻⁴ = 4-log 4.4 =3.357
pOH+pH=pKw
pOH+3.357=14
pOH=14-3.357=10.643
pOH=-log[OH⁻]
[tex]\tt [OH^-]=10^{-10.643}=2.28\times 10^{-11}[/tex]