Question

In: Chemistry

What are [HCOOH], [H3O+], [HCOO–], [OH–] and pH of a) 0.300 M HCOOH, b) 0.005 M...

What are [HCOOH], [H3O+], [HCOO–], [OH–] and pH of

a) 0.300 M HCOOH,

b) 0.005 M HCOOH solutions? (Ka = 1.8·10–4)

HCOOH(aq) + H2O(l) H3O+(aq) + HCOO–(aq)

Solutions

Expert Solution

           HCOOH(aq) + H2O(l)--------------> H3O+(aq) + HCOO–(aq)

I         0.3                                                  0                 0

C        -x                                                     +x               +x

E        0.3-x                                                  +x               +x

              Ka = [H3O^+][HCOO^-]/[HCOOH]

             1.8*10^-4   = x*x/0.3-x

            1.8*10^-4 *(0.3-x) = x^2

              x= 0.00726

         [HCOOH] = 0.3-x = 0.3-0.00726 = 0.29274M

          [H3O^+]   = x   = 0.00726M

          [HCOO^-] = x   = 0.00726M

         [OH-]    = Kw/[H3O^+]

                     = 1*10^-14/0.00726   = 1.37*10^-12M

        PH = -log[H3O+]

                = -log1.37*10^-12    = 11.86 >>>>answer

b.

HCOOH(aq) + H2O(l)--------------> H3O+(aq) + HCOO–(aq)

I        0.005                                            0                 0

C           -x                                             +x               +x

E         0.005-x                                             +x               +x

              Ka = [H3O^+][HCOO^-]/[HCOOH]

             1.8*10^-4   = x*x/0.005-x

            1.8*10^-4 *(0.005-x) = x^2

              x= 0.000863

     [HCOOH] = 0.005-x = 0.005-0.000863 = 0.004137M

          [H3O^+]   = x   = 0.000863M

          [HCOO^-] = x   = 0.000863M

         [OH-]    = Kw/[H3O^+]

                      = 1*10^-14/0.000863   = 1.158*10^-11M

      PH         = -log[H3O+]       

                    = -log0.000863 = 3.0639 >>>>answer

        


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