At equilibrium, it was found that #CH_3COOH# = 0.8537 M, #CH_3COO^-# = 0.0040 M, and #H_3O^+# = 0.0040 M . How do you calculate the acidity constant of #CH_3COOH# and the pH value of the solution?

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Answer 1 · 2016-10-30T20:58:51

#pH=2.40;pK_a=4.73#.

#pH=-log_10[H_3O^+]# #=# #-log_10(0.004)=2.40#

Acetic acid undergoes the following reaction:

#H_3C-CO_2H + H_2O rightleftharpoons H_3C-CO_2^(-) + H_3O^(+)#

And we write the equilibrium expression in the usual way:

#([H_3O^+][""^(-)OAc])/([HOAc])# #=# #K_a#

The concentration of water does not appear in this expression, because #[H_2O]# is effectively constant.

#=# #(0.004*mol*L^-1xx0.004*mol*L^-1)/(0.8537*mol*L^-1)# #=# #1.87xx10^-5#

#pK_a=-log_(10)K_a=-log_10(1.87xx10^-5)# #=# #4.73#.