Calculate the percent ionization of 0.120 M lactic acid (Ka=1.4*10^-4).
To calculate the percent ionization of lactic acid, we need to use the equilibrium expression for the ionization of lactic acid, which is given by:
Ka = [H+][C3H5O3-]/[C3H6O3]
Where Ka is the acid dissociation constant, [H+] is the concentration of hydrogen ions (protons), [C3H5O3-] is the concentration of lactate ions, and [C3H6O3] is the concentration of lactic acid.
Since we know the value of Ka and the initial concentration of lactic acid ([C3H6O3] = 0.120 M), we can set up an equation to find the concentration of lactate ions ([C3H5O3-]) at equilibrium.
Let x be the concentration of lactate ions at equilibrium. Then, the concentration of hydrogen ions at equilibrium will also be x, since the acid dissociates in a 1:1 ratio.
Thus, the equilibrium expression becomes:
1.4 x 10^-4 = x * x / (0.120 - x)
Now, we can solve this equation to find the value of x.
Simplifying the equation:
1.4 x 10^-4 = x^2 / (0.120 - x)
Multiplying both sides by (0.120 - x):
1.4 x 10^-4 * (0.120 - x) = x^2
Rearranging the equation:
x^2 + 1.4 x 10^-4 * x - 1.4 x 10^-4 * 0.120 = 0
Now, we can solve this quadratic equation using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Where a = 1, b = 1.4 x 10^-4, and c = -1.4 x 10^-4 * 0.120.
Plugging in these values into the quadratic formula and solving for x:
x = (-1.4 x 10^-4 ± √((1.4 x 10^-4)^2 - 4 * 1 * (-1.4 x 10^-4 * 0.120))) / (2 * 1)
After calculating the value of x, plug it back into the equation for the percent ionization:
Percent ionization = (x / [C3H6O3]) * 100
This will give you the percent ionization of lactic acid.
pH = pKa + log([A-]/[HA])
Let's call lactic acid HL.
.........HL ==> H^+ + L^-
initial.0.120....0.....0
change...-x......x.....x
equil...0.120-x...x....x
Ka = 1.4E-4 = (H^+)(L^-)/(HL)
Substitute from the ICE chart and solve for x which = (H^+). Then
% ionization = [(H^+)/(0.120)]*100 = ?