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?
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.
When determining the correct equation to use for the ICE chart in these types of problems, it is important to consider the nature of the substances involved and the chemical equilibrium that is established.
In the case of an aqueous solution containing a weak acid (CH3COOH) and its conjugate base (CH3COO-), the equilibrium reaction can be represented as follows:
CH3COOH + H2O ⇌ H3O+ + CH3COO-
This equation shows the ionization of CH3COOH to form H3O+ (hydronium ion) and CH3COO- (acetate ion) in the presence of water. The arrow pointing in both directions indicates that the reaction is reversible and can proceed in either direction.
The reason why we write the equation in this particular way is because CH3COOH is a weak acid, meaning it only partially dissociates in water to form H3O+ and CH3COO-. Therefore, the reactants CH3COOH and H2O are more likely to form the products H3O+ and CH3COO- in significant amounts, rather than the other way around.
Similarly, when calculating the pH of a solution containing the acetate ion (CH3COO-), the reaction can be represented as follows:
CH3COO- + H2O ⇌ CH3COOH + OH-
In this case, the conjugate base (CH3COO-) accepts a proton from water to form the weak acid (CH3COOH) and a hydroxide ion (OH-). Once again, we represent the equation this way because the acetate ion (CH3COO-) is more likely to accept a proton from water and form the weak acid (CH3COOH) and OH-, rather than the other way around.
To summarize, when determining the equilibrium equation for the ICE chart, we consider the nature of the substances involved and their tendency to interact in a specific direction based on their acid-base properties and the principles of equilibrium.