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You need to prepare 100.0 mL of a pH=4.00 buffer solution using 0.100 M benzoic acid...

You need to prepare 100.0 mL of a pH=4.00 buffer solution using 0.100 M benzoic acid (pKa = 4.20) and 0.220 M sodium benzoate. How much of each solution should be mixed to prepare this buffer?

a) mL of benzoic acid

b) mL of sodium benzoate

Solutions

Expert Solution

Let volume of sodium benzoate =x

      volume of benzoic acid      = 100-x

      no of moles of sodium benzoate = molarity * volume

                                                     = 0.22*x

     no of moles of bezoic acid            = molarity * volume

                                                       = 0.1*(100-x)

       PH = PKa +log[sodium benzoate]/[benzoic acid]

       4      = 4.2 + log0.22*x/0.1*(100-x)

       4-4.2   = log0.22*x/0.1*(100-x)

       -0.2     = log0.22*x/0.1*(100-x)

log0.22*x/0.1*(100-x) = -0.2

0.22*x/0.1*(100-x)       = 10-0.2 = 0.63

0.22*x/0.1*(100-x)   = 0.63

0.22*x               = 0.63*0.1(100-x)

                             = 0.063(100-x)

0.22x                    =6.3-0.063x

0.22x-6.3+0.063x =0

0.263x-6.3 =0

   0.263x = 6.3

         x   = 6.3/0.263   = 23.95

volume of benzoic acid   = 23.95ml

volume of benzoate       = 100-x = 100-23.95 = 76.05ml

   

      

queen_honey_blossom answered 1 year ago
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