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Erythrosin17
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.
pstnonsonjoku

The pH of the solution is 4.55.

We have to note that the H2C2H3O2/KC2H3O2 is a buffer solution. So we set up an ICE table from the equations;

HC2H3O2(aq) ⇄ H+(aq) + C2H3O2-(aq)

KC2H3O2(aq) ⇄ K+(aq) + C2H3O2-(aq)

Note that H^+ comes from the acid

                   HC2H3O2(aq) ⇄       H+(aq)        C2H3O2-(aq)

I                     0.195                   0              0.125

C                      -x                    +x               +x

E                 0.195-x               x             0.125 + x

Given that the Ka of HC2H3O2 = 1.8*10^-5

So; Ka = [H+] [C2H3O2-]/HC2H3O2          

therefore;

 1.8*10^-5 = [0.195-x] [x]/[0.125 + x]

Solving the equation yields x = 2.81x10^-5 M

Hence;  [H+] = x = 2.81x10^-5 M

pH = - log[ 2.81x10^-5 M]

pH = 4.55

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