Question

Starting with a solution at equilibrium containing 0.5 M acetate and 0.5 M acetic acid, then adding 0.1 sodium acetate, what would the pH of the original solution and the solution after the addition be? Acetic acid Ka 1x10-6 (use this cant use calculator on test) PLEASE EXPLAIN

Answer 1

[CH3COO^-]   = 0.5M

[CH3COOH]   = 0.5M

Ka   = 1*10^-6

PKa   = -logKa

          = -log1*10^-6

           = 6

PH   = PKa + log[CH3COO^-]/[CH3COOH]

         = 6 + log0.5/0.5

          = 6 + log1

          = 6 + 0

           = 6

after addition of 0.1M sodium acetate(CH3COONa)

[CH3COO^-]   = 0.5 + 0.1 = 0.6M

[CH3COOH]   = 0.5M

Ka   = 1*10^-6

PKa   = -logKa

          = -log1*10^-6

           = 6

PH   = PKa + log[CH3COO^-]/[CH3COOH]

       = 6 + log(0.6/0.5)

       = 6 + 0.07918

       = 6.0792>>>answer