Solve an equilibrium problem (using an ice table) to calculate the ph of each solution: part a a solution that is 0.195 m in hc2h3o2 and 0.125 m in kc2h3o2

To determine the pH of a solution which has 0.195 M hc2h3o2 and 0.125 M kc2h3o2, we use the ICE table and the acid dissociation constant of hc2h3o2 to determine the concentration of the hydrogen ion present at equilibrium. We do as follows:

HC2H3OO = H+ + C2H3OO-
KC2H3OO = K+ + C2H3OO-

Therefore, the only source of hydrogen ion would be the acid. We use the ICE table,
                    HC2H3OO           H+        C2H3OO-
I                     0.195                  0              0.125
C                      -x                    +x               +x

E                 0.195-x                x              0.125 + x

Ka = 1.8*10^-5 = (0.125 + x) (x) / 0.195 -x 
x = 2.81x10^-5 M = [H+]

pH = - log [H+]
pH = -log 2.81x10^-5
pH = 4.55

Therefore, the pH of the resulting solution would be 4.55.

Ithink carbon dioxide, i am not completely sure.tell me if i am wrong.