calculate [H30+],pH,[OH-] and pOH for .025 M nitric acid (HNO3)??

HNO3 is a strong acid and 100% ionized; therefore, the (H^+) = (HNO3).

From there, pH = -log(H^+).
pH + pOH = 14 to calculate pOH.
and OH from pOH = -log(OH^-).