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

  • chamistry -

    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^-)

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