Acetic acid (CH_3COOH) is a weak acid.
a) Write the symbolic equation showing the ionisation of acetic acid.
b)Given that an acetic acid solution of pH 4.8 has a hydrogen ion concentration of 1.58 * 10^-5 mol L^-1, what is the concentration of the acetate ion at equilibrium?
c) Calculate the equilibrium concentration of the undissociated CH_3COOH if the equilibrium constant for this dissociation is 1.8 * 10^-5
a. acetic acid is HAc to save typing.
HAc + H2O ==> H3O^+ + Ac^-
b.
Convert pH = 4.8 to H3O^+. Since the answer to a shows 1 Ac to 1 H3O, then H3O = Ac
c.
Write Ka expression for HAc. Substitute H3O^+ and Ac^- and solve for (HAc).
Post your work if you get stuck.
a) The symbolic equation for the ionization of acetic acid (CH3COOH) is:
CH3COOH ⇌ CH3COO- + H+
b) To find the concentration of the acetate ion (CH3COO-) at equilibrium, we need to use the equation for the dissociation of acetic acid and the pH value given. The pH is a measure of the hydrogen ion concentration in a solution.
We can use the equation:
pH = -log[H+]
To find the concentration of the hydrogen ions, we can rearrange the equation:
[H+] = 10^(-pH)
Plugging in the given pH value of 4.8:
[H+] = 10^(-4.8)
[H+] = 1.58 * 10^(-5) mol L^(-1)
Since acetic acid is a weak acid, we can assume that the concentration of the acetate ion at equilibrium is equal to the concentration of the hydrogen ion.
Therefore, the concentration of the acetate ion at equilibrium is 1.58 * 10^(-5) mol L^(-1).
c) To calculate the equilibrium concentration of the undissociated CH3COOH, we can use the equilibrium constant (Ka) for the dissociation of acetic acid.
The dissociation equation for acetic acid is:
CH3COOH ⇌ CH3COO- + H+
The equilibrium constant expression can be written as:
Ka = [CH3COO-][H+] / [CH3COOH]
We are given the equilibrium constant (Ka) value of 1.8 * 10^(-5), and we have already determined the concentration of the hydrogen ion ([H+]) to be 1.58 * 10^(-5) mol L^(-1).
Let's assume the equilibrium concentration of the undissociated acetic acid is x mol L^(-1).
Using the equilibrium constant expression, we have:
1.8 * 10^(-5) = (1.58 * 10^(-5))(x) / (x)
Simplifying, we get:
1.8 * 10^(-5) = 1.58 * 10^(-5)
Since the equilibrium concentration of the undissociated acetic acid (x) cancels out, we can conclude that the equilibrium concentration of undissociated acetic acid is equal to 1.58 * 10^(-5) mol L^(-1).