When asked to name all of the ions and molecules present in a 0.1 mol L-1 solution of CH3COOH, why are there OH- ions present if the equation is
CH3COOH(aq) + H2O(l) --> CH3COO-(aq) + H3O+(aq) ?
The water ionizes in such a manner that the product of the concentrations of H3O+ ions and OH- ions is constant (at 25C, the product is equal to 1*10^-14.
To understand why there are OH- ions present in a solution of CH3COOH, let's break down the dissociation of water and the ionization of acetic acid (CH3COOH).
First, let's look at the dissociation of water:
H2O(l) → H+(aq) + OH-(aq)
In water, a small fraction of water molecules dissociate into hydronium ions (H+) and hydroxide ions (OH-). The concentration of H+ ions and OH- ions in pure water at 25°C is 1x10^-7 mol/L each. This is due to the autoprotolysis of water, where water acts as both an acid and a base, transferring a proton from one water molecule to another.
Now, let's focus on the ionization of acetic acid (CH3COOH):
CH3COOH(aq) + H2O(l) → CH3COO-(aq) + H3O+(aq)
Acetic acid, CH3COOH, is a weak acid meaning it does not fully dissociate into ions in solution. When acetic acid dissolves in water, it donates a proton (H+) to water, forming acetate ions (CH3COO-) and hydronium ions (H3O+). The concentration of acetate ions (CH3COO-) in the solution will depend on the concentration of the acetic acid and how much it ionizes.
However, due to the autoprotolysis of water mentioned earlier, some water molecules can also act as bases and accept protons, forming hydroxide ions (OH-). Since water is constantly undergoing autoprotolysis, there will always be a small concentration of OH- ions present in a solution.
The product of H3O+ ions and OH- ions concentrations is constant at 1x10^-14 at 25°C, as determined by the autoprotolysis of water. So, even though the equation for the ionization of acetic acid does not explicitly show OH- ions, they are present in the solution at a concentration determined by the product of H3O+ and OH- ions.