[alex021095]

The acid dissociation constant of ethanoic acid is 1.80x10^-5M.
Calculate the pH of a mixture:

a)25.0cm^3 of 0.100M ethanoic acid and 50.0cm^3 of 0.100M sodium ethanoate  acid
      My ans: CH3COOH + H2O-->H3O^+ + CH3COO^-
                  [CH3COO^-] = 50.0 x 0.100 / (50.0+25.0) = 0.0667M
                  [CH3COOH]   = 25.0 x 0.100 / (50.0+25.0) = 0.0333M
                  pH = -log1.80x10^-5 + log0.0667 / 0.0333 = 5.05
                 
             Could anyone help me check my ans whether right or wrong^^"?

[TWOFOUSAND]

yup thats right XD   by my calculations anyway