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What is the pH of a 0.445 M ammonia solution? Express your answer numerically to two...

What is the pH of a 0.445 M ammonia solution? Express your answer numerically to two decimal places.

Solutions

Expert Solution

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

I              0.445                                           0                    0

C                -x                                              +x                  +x

E           0.445-x                                           +x                  +x

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

                1.8*10^-5    = x*x/(0.445-x)

               1.8*10^-5 *(0.445-x) = x^2

                    x   = 0.00282

           [OH^-]    = x   = 0.00282M

          POH     = -log[OH^-]

                        = -log0.00282

                        = 2.5497

         PH          = 14-POH

                       = 14-2.5497

                        = 11.45 >>>>>answer


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