calculate the pH pOH and (OH(aq)) of 0.006mol/l HI(aq)

Answer 1

HI is a strong acid; i.e., it ionizes 100%; therefore, (H^+) = 0.006M
pH = -log(H^+)
Get pOH from this.
pH + pOH = pKw = 14
Then OH^- from pOH = -log(OH^-)