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Determine the [OH-], pH, and pOH of a solution with a [H+] of 0.080 M at 25 °C. [OH-] = 1 pOH = pOH = Determine the [H+), pH,

Determine the (H+). [OH-], and pOH of a solution with a pH of 7.41 at 25 °C. [OH-] =( pOH = Determine the fH+], [OH-], and pH

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Answer #1

[H^+]   = 0.08M

[OH^-]    = Kw/[H^+]

              = 1*10^-14/0.08

              = 1.25*10^-13M

PH   = -log[H^+]

        = -log0.08

       = 1.097

POH   = 14-PH

         = 14-1.097

         = 12.903

b. [OH^-]   = 1.9*10^-9M

     [H^+]   = Kw/[OH^-]

                = 1*10^-14/(1.9*10^-9)

                 = 5.26*10^-6M

   PH      = -log[H^+]

              = -log(5.26*10^-6)

               = 5.28

POH    = 14-PH

          = 14-5.28

            = 8.72

c. PH = 7.41

-log[H^+] = 7.41

       [H^+]   = 10^-7.41   = 3.9*10^-8M

      [OH^-]   = Kw/[H^+]

                   = 1*10^-14/(3.9*10^-8)

                   = 2.56*10^-7M

       POH   = 14-PH

                 = 14-7.41

                  = 6.59

d. POH = 5.45

-log[OH^-] = 5.45

       [OH^-] = 3.55*10^-6M

   [H^+]   = Kw/[OH^-]

              = 1*10^-14/(3.55*10^-6)

               = 2.8*10^-9M

    PH   = 14-POH

           = 14-5.45

            = 8.55

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