Question

In: Chemistry

Find [NH3], [NH4+], [OH–], [H3O+] and pH of a 0.080 M NH3 solution NH3(aq) + H2O(l)...

Find [NH3], [NH4+], [OH–], [H3O+] and pH of a 0.080 M NH3 solution

NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)

Kb = 1.76·10–5

Solutions

Expert Solution

NH3(aq) + H2O(l) ------------->NH4+(aq) + OH–(aq)

I         0.8                                          0                 0

C         -x                                           +x               +x

E         0.8-x                                       +x              +x

             Kb   = [NH4^+][OH^-]/[NH3]

            1.76*10^-5   = x*x/0.8-x

          1.76*10^-5*(0.8-x) = x^2

             x   = 0.00374

          [NH4^+] = x= 0.00374M

          [OH^-] = x   = 0.00374M

         [NH3] = 0.8-x = 0.8-0.00374    = 0.79626M

        [H3O+] = Kw/[OH^-]

                     = 1*10^-14/0.79626   = 1.25*10^-14 M

        PH   = -log[H3O+]

                = -log1.25*10^-14

               = 13.9030 >>>>>answer


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