C2H5COOH ---> C2H5COO-Â Â +Â Â
H+
0.551Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â (initial)
0.551-x                    Â
x                      Â
x   (at equilibrium)
Ka = [C2H5COO-][H+] / [C2H5COOH ]
1.34*10^-5 = x*x / (0.551-x)
Here Ka is very small, so x will be very small and it
can be ignored as compared to 0.551
So, above expression becomes
1.34*10^-5 = x*x / 0.551
x=2.72*10^-3 M
Answer:
[C2H5COO–] = x = 2.72*10^-3 M
[C2H5COOH ] = 0.551-x= 0.551-2.72*10^-3 = 0.548 M
