Question

You make a solution of 0.035 M NaCH3CO2 (sodium acetate) and 0.035 M CH3CO2H (acetic acid). Solve for the pH of this solution if you add 0.1 M LiClO4.

Answer 1

pH of acidic buffer = pka+log( H3C-COONa/H3C-COOH)

           pka of H3C-COOH = -log(1.8*10^-5) = 4.74

concentration of H3C-COOna( sodium acetate ) = 0.035 M

concentration of H3C-COOH(acetic acid) = 0.035 M

pH = 4.74 +log(0.035/0.035)

     = 4.74

after addition of LiClO4 , i think it is concentration = 0.01 M,

   LiClO4 - acidic salt

pH of acidic buffer = pka+log( H3C-COONa/H3C-COOH + LiClo4)

pH = 4.74 +log(0.035/(0.035+0.01))

       = 4.63