Propanoic acid has a Ka of 1.3 x 10^-5. What is the % ionization in a 3.0M solution?
thank you so much!
To determine the percent ionization of propanoic acid in a 3.0M solution, we need to understand that the ionization of a weak acid is represented by its dissociation constant (Ka). The Ka is the ratio of the concentration of the products to the concentration of the reactants in a chemical equilibrium expression.
The chemical equation for the ionization of propanoic acid (CH3CH2COOH) can be written as:
CH3CH2COOH ⇌ CH3CH2COO- + H+
The equilibrium expression for this reaction can be written as:
Ka = [CH3CH2COO-][H+] / [CH3CH2COOH]
Given the value of Ka for propanoic acid (1.3 x 10^-5), we can use this information to determine the percent ionization.
Step 1: Write the expression for Ka:
1.3 x 10^-5 = [CH3CH2COO-][H+] / [CH3CH2COOH]
Step 2: Assuming that x represents the initial concentration of [H+], and considering that propanoic acid is a weak acid with a low ionization extent, we can approximate the concentration of undissociated propanoic acid ([CH3CH2COOH]) as (3.0 - x) M.
Step 3: Since the concentration of CH3CH2COO- will be approximately equal to x M and the concentration of [H+] will be approximately equal to x M, we can substitute these values into the equilibrium expression.
1.3 x 10^-5 = (x)(x) / (3.0 - x)
Step 4: Rearrange the equation and solve for x.
1.3 x 10^-5 * (3.0 - x) = x^2
3.9 x 10^-5 - 1.3 x 10^-5x = x^2
x^2 + 1.3 x 10^-5x - 3.9 x 10^-5 = 0
Step 5: Solve this quadratic equation using the quadratic formula, or any appropriate method. In this case, solving the quadratic equation gives us:
x ≈ 0.00243 M
Step 6: Calculate the percent ionization:
Percent ionization = (x / initial concentration) * 100
Percent ionization = (0.00243 M / 3.0 M) * 100
Percent ionization ≈ 0.08%
Therefore, the percent ionization of propanoic acid in a 3.0M solution is approximately 0.08%.
Let's call propanoic acid HPr. Then
...........HPr ==> H^+ + Pr^-
Initial....3.0......0.....0
Change......-x......x......x
Equil.....3.0-x.....x.......x
Ka = (H^+)(Pr^-)/(HPr)
Substitute from the ICE chart into the Ka expression and solve for x = (H^+).
Then %ionization = [(H^+)/3.0)]*100 = ?