Calculate the percent ionization of a
0.76 molar solution of acetic acid. The ioniza-
tion constant of acetic acid is 1.8 × 10−5.
To calculate the percent ionization of a 0.76 molar solution of acetic acid, we need to use the equation for the ionization constant:
Ka = [H+][CH3COO-]/[CH3COOH]
Given:
Ka = 1.8 × 10^(-5) (ionization constant of acetic acid)
Assumptions:
We assume that the initial concentration of acetic acid is equal to its molar concentration.
Let x be the amount of acetic acid that ionizes.
At equilibrium, the concentration of acetic acid is (0.76 - x) and the concentrations of H+ and CH3COO- are both x.
Plugging these values into the equation for Ka:
(1.8 × 10^(-5)) = (x)(x)/(0.76 - x)
Simplifying the equation:
1.8 × 10^(-5) = x^(2)/(0.76 - x)
To solve this equation, we use the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
In this case, the equation becomes:
x = (-0 ± √(0^2 - 4(1)(-1.8 × 10^(-5))(0.76)))/(2(1.8 × 10^(-5)))
Simplifying further:
x = (√(0.6928 × 10^(-5)))/(3.6 × 10^(-5))
x = 0.026
Therefore, the amount of acetic acid that ionizes is 0.026 moles.
To calculate the percent ionization, we divide the amount of acetic acid that ionizes by the initial concentration of acetic acid and multiply by 100:
Percent ionization = (0.026 moles)/(0.76 moles) × 100
Percent ionization ≈ 3.42%
Therefore, the percent ionization of a 0.76 molar solution of acetic acid is approximately 3.42%.
To calculate the percent ionization of a solution, you need to know the ionization constant and the concentration of the acid. In this case, the concentration of acetic acid is given as 0.76 M and the ionization constant is given as 1.8 × 10−5.
The ionization constant (Ka) is a measure of the extent to which a weak acid ionizes in solution. It is defined as the ratio of the concentration of the ionized form of the acid ([H+][A-]) to the concentration of the undissociated acid ([HA]).
The equation for the ionization of acetic acid is:
CH3COOH ⇌ H+ + CH3COO-
To find the percent ionization, follow these steps:
Step 1: Calculate the initial concentration of acetic acid (HA) before ionization:
[HA] = 0.76 M
Step 2: Calculate the initial concentration of the ionized form (A-) before ionization:
[A-] = 0 M (since no ionization has occurred yet)
Step 3: Calculate the initial concentration of the hydrogen ion (H+) before ionization:
[H+] = 0 M (since no ionization has occurred yet)
Step 4: Calculate the change in concentration of the ionized form (A-) at equilibrium:
[A-]change = x (where x is the extent of ionization, which we need to find)
Step 5: Calculate the change in concentration of the hydrogen ion (H+) at equilibrium:
[H+]change = x (since the number of hydrogen ions released is equal to the number of acetate ions formed)
Step 6: Calculate the concentration of the undissociated acid (HA) at equilibrium:
[HA] = 0.76 - x (since some of the initial acid has ionized into acetate ions)
Step 7: Set up the equilibrium expression based on the ionization of acetic acid:
Ka = [H+][A-] / [HA]
Step 8: Substitute the equilibrium concentrations into the equilibrium expression:
1.8 × 10−5 = (x)(x) / (0.76 - x)
Step 9: Solve the quadratic equation for x:
1.8 × 10−5 = x^2 / (0.76 - x)
This equation can be solved using numerical methods or by approximation. Let's assume the extent of ionization, x, is small compared to the initial concentration of acetic acid (0.76 M). This allows us to make the approximation that 0.76 - x ≈ 0.76. With this approximation, the equation becomes:
1.8 × 10−5 ≈ x^2 / 0.76
Simplifying the equation, we get:
x^2 ≈ (1.8 × 10−5)(0.76)
x^2 ≈ 1.368 × 10−5
x ≈ √(1.368 × 10−5)
x ≈ 0.0037
Step 10: Calculate the percent ionization:
Percent ionization = (extent of ionization / initial concentration of acid) x 100
Percent ionization = (0.0037 / 0.76) x 100
Percent ionization ≈ 0.49%
Therefore, the percent ionization of the 0.76 M acetic acid solution is approximately 0.49%.