a) Calculate the percent ionization of 0.125M lactic acid (Ka=1.4x10^-4) b) Calculate the percent ionization of 0.125M lactic acid in a solution containing 0.00075 M sodium lactate.
a) To calculate the percent ionization of lactic acid, we can use the Ka (acid dissociation constant) equation:
Ka = [H+][A-] / [HA]
Where:
- [H+] is the concentration of the hydronium ion (H+)
- [A-] is the concentration of the conjugate base (lactate ion, C3H5O3-)
- [HA] is the initial concentration of the acid (lactic acid, C3H6O3)
The percent ionization can be calculated as follows:
% Ionization = ([H+] / [HA]) × 100
Given:
- [HA] (initial concentration of lactic acid) = 0.125 M
- Ka (acid dissociation constant) = 1.4 x 10^-4
First, we need to calculate the concentration of H+ and A- using the Ka equation.
Let x be the concentration of the hydronium ion (H+), and since lactic acid is monoprotic, it will be equal to the concentration of the conjugate base (lactate ion, C3H5O3-).
Ka = x^2 / (0.125 - x)
Since x is small compared to 0.125, we can assume that 0.125 - x ≈ 0.125.
Ka = x^2 / 0.125
Rearranging the equation and solving for x:
x^2 = Ka * 0.125
x = √(Ka * 0.125)
Substituting the given values:
x = √(1.4 x 10^-4 * 0.125)
x ≈ 6.64 x 10^-3 M
Now, we can calculate the percent ionization:
% Ionization = (6.64 x 10^-3 M / 0.125 M) × 100
% Ionization ≈ 5.3%
Therefore, the percent ionization of 0.125 M lactic acid is approximately 5.3%.
b) To calculate the percent ionization of lactic acid in a solution containing both lactic acid and sodium lactate, we need to consider the common-ion effect.
First, we can calculate the concentration of the acetate ion (A-) provided by sodium lactate:
- Sodium lactate is a strong electrolyte, so it will fully dissociate in solution.
- 0.00075 M sodium lactate will provide 0.00075 M acetate ions (A-).
Now, we can calculate the new concentration of lactic acid in the presence of the acetate ions.
Let's assume the new concentration of lactic acid is x.
We can set up an equilibrium expression using the new concentrations:
Ka = [H+][A-] / [HA]
1.4 x 10^-4 = (6.64 x 10^-3 + 0.00075) * (0.00075) / x
Simplifying the equation:
1.4 x 10^-4 = (7.39 x 10^-3 * 0.00075) / x
x = (7.39 x 10^-3 * 0.00075) / (1.4 x 10^-4)
x ≈ 3.975 x 10^-2 M
Now, we can calculate the percent ionization in the new solution:
% Ionization = (6.64 x 10^-3 M / 3.975 x 10^-2 M) × 100
% Ionization ≈ 16.7%
Therefore, the percent ionization of 0.125 M lactic acid in a solution containing 0.00075 M sodium lactate is approximately 16.7%.
To calculate the percent ionization of lactic acid in these scenarios, we need to use the concept of equilibrium and the acid dissociation constant (Ka).
a) In the first scenario, we have 0.125M lactic acid and no other relevant species.
The chemical equation for the ionization of lactic acid can be written as follows:
CH3CH(OH)COOH ⇌ CH3CH(OH)COO- + H+
The equilibrium expression for this reaction is:
Ka = [CH3CH(OH)COO-][H+] / [CH3CH(OH)COOH]
Given that Ka = 1.4x10^-4, we can set up the following equation:
1.4x10^-4 = ([CH3CH(OH)COO-][H+]) / (0.125)
Since lactic acid is a weak acid, its initial concentration will remain nearly unchanged after ionization, so we can approximate it as 0.125.
Next, let's assign the degree of ionization as x. This means x is the concentration of CH3CH(OH)COO- and H+ ions formed, and we can assume the initial concentration of lactic acid to be (0.125 - x).
The equation now becomes:
1.4x10^-4 = (x)(x) / (0.125 - x)
Simplifying this equation gives us a quadratic equation, which we can solve to find the value of x. Once we have the value of x, we can calculate the percent ionization.
b) In the second scenario, we have 0.125M lactic acid and 0.00075M sodium lactate. Sodium lactate is the conjugate base of lactic acid. Its dissociation in water can be written as follows:
CH3CH(OH)COO- (from sodium lactate) ⇌ CH3CH(OH)COO- + H+
To calculate the percent ionization in this case, we need to consider the common ion effect. The presence of the conjugate base (CH3CH(OH)COO-) will lower the ionization of lactic acid.
We can again set up an equation using the Ka value:
1.4x10^-4 = ([CH3CH(OH)COO-][H+]) / (0.125 + 0.00075)
Since sodium lactate is a strong electrolyte, it will completely dissociate into its constituent ions, resulting in an increase in the concentration of CH3CH(OH)COO- and H+ ions.
Let's assume the degree of ionization for lactic acid is y. This means the concentration of CH3CH(OH)COO- and H+ ions formed is y, and the initial concentration of lactic acid is (0.125 - y).
The equation now becomes:
1.4x10^-4 = (y)(y) / (0.125 + 0.00075)
Again, simplifying this equation will give us a quadratic equation, which we can solve to find the value of y. Then, we can calculate the percent ionization using the obtained value of y.
Lactic acid = HL, sodium lactate = NaL.
..........HL ==> H^+ + L^-
I...... 0.125....0.....0
C.........-x.....x.....x
E......0.125-x...x.....x
Substitute the E line into Ka expression and solve for x = (H^+), the
% ion = [((H^+)/(HL)]*100 = ?
for b part, plug into Ka expression the following concns:
Remember NaL is 100% ionized in solution.
.........NaL ==> Na^+ + L^-
I......7.5E-4.....0......0
C.....-7.5E-4....7.5E-4..7.5E-4
so final concns are
(H^+) = x
(L^-) = x from HL + 7.5E-4 from NaL to make x + 7.5E-4
(HL) = 0.125-x
These are plugged into the Ka expression and solve for x = (H^+), then
%ion = [(H^+)/(HL)]*100 = ?