Benzoic acid is C6H5COOH. It ionizes as
C6H5COOH + H2O ==> H3O+ + C6H5COO^-
The ICE table runs from left to right in three horizontal rows:
I = initial...then the concns of each are listed.
C = change are listed in the row.
E = equilibrium listed in the row.
Unfortunately these boards don't allow spacing so I will use periods to space. Just ignore the periods.
pH = 3.65 so (H3O^+) = 2.24 x 10^-4 and I'm sure that is an equilibrium value.
C6N5COOH + H2O ==> H3O^+ + C6H5COO^-
I=0.00250...........0........0
C= -2.24E-4........+2.24E-4..+2.24E-4
E= 0.00250-2.24E-4...+2.24E-4...+2.24E-4
Ka = ((H3O^+)(C6H5COO^-)/(C6H6COOH)
Plug in the equilibrium values and solve for Ka.
I don't understand ice table at all..can anyone please guide me in how to solve the following question? :
The pH of a 0.00250 mol/L solution of benzoic acid is 3.65. Calculate the Ka for the benzoic acid. (This question does not state that the given concentration is an equilibrium concentration)
Thank u in advance
3 answers
Thank u sooo much!! =)
Just to confirm, the answer is 2.004748935 x 10^-5, right?
1 answer