When you have ONLY CH3COONa in solution, the only thing that can happen is the hydrolysis of CH3COONa which is
CH3COO^- + HOH ==> CH3COOH + OH^-
When you have a mixture of CH3COOH and CH3COONa, this is a buffered solution and the pH is determined by the Henderson-Hasselbalch equation. The PRIMARY acid is CH3COOH and it will ionize as you have shown. The CH3COONa CAN hydrolyze and produce some CH3COOH + OH^-; however, the excess of CH3COOH in the solution to begin with does two things: (1) first it essentially neutralizes any OH^- that forms AND (2) the CH3COOH is a common ion (see right of the hydrolysis equation) and it forces the equilibrium of the hydrolysis reaction to the left by Le Chatelier's principle. The hydrolysis reaction is not large, anyway, (Kb for CH3COONa is 1 x 10^-14/1.8 x 10^-5 = 5.55 x 10^-10) and the CH3COOH in the CH3COOH/CH3COONa mixture makes the hydrolysis even smaller. In essence we use the hydrolysis of CH3COONa when it is the ONLY chemistry going on but we ignore it otherwise because it is small enough to be negligible.
For a question like “calculate the pH of an aq.solution that is 1.0 M CH3COOH and 1.0 M CH3COONa, how do you know to write the equation like this: CH3COOH + H2O => H3O+ + CH3COO- and not like H3O+ + CH3COO- => CH3COOH + H2O for the ICE chart. This would switch the reactants and products.
Like “calculate pH of soln that .25M CH3COONa how do you know to do CH3COO- + H2O => CH3COOH + OH- and not the other way?
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