[mandy9008]

HC₂H₃O₂(aq) + H₂O(l) H₃O⁺(aq) + C₂H₃O₂^-(aq)

K_c=1.8*10^-5

solution initially contains 0.180M HC₂H₃O₂, what is equilibrium concentration of H₃O?

i initially thought that since there were no gasses it would be 0, but that is not the case, then i thought that it would be equal(0.180M), but that is also not the case.

the answer is 1.80*10^-3M

how do i arrive there?

[Astrokel]

ICE Table

[sm2345]

Here as HC2H3O2 is a weak acid and its degree of dissociation (α) must be taken into account.

Initially, the conc. of HC2H3O2 is c = .18 M.

Now,

           HC2H3O2(aq) + H2O(l)  H₃O⁺ +C2H3O2^-(aq)

At t=0:     c                                     0                                             0

At t=t_eq:    c(1-α)                 cα                                            cα

Now,   k_c = ([C2H3O2^-]*[H₃O⁺])/[HC2H3O2]

=   (cα*cα)/c(1-α)

=  cα² (Assuming α<<1)

Now k_c & c are known. Put these values and get α = .01

Hence [H3O⁺] = cα = 1.8x10^-3

Also pH of the solution can be calculated now,