What is the pH of a buffer solution that has CH3CO2H=0.68 M and CH3CO2- =0.58 M, and has 0.01 M concentration of H2SO4 added to it? Ka of acetic acid is 1.8x10-5
Ac is acetate and HAc is acetic acid
.......Ac + H^+ ==> HAc
I....0.58...0........0.68
add........0.02...........
C...-0.02..-0.02......+0.02
E.....0.56...0.........0.70
Plug the E line into the HH equation and solve for pH.
To find the pH of the buffer solution, we need to consider the equilibrium between the weak acid (CH3CO2H) and its conjugate base (CH3CO2-).
First, calculate the initial concentrations of CH3CO2H and CH3CO2- in the buffer solution with the given concentrations:
CH3CO2H: 0.68 M
CH3CO2-: 0.58 M
Next, determine the amount of CH3CO2H that will ionize due to the addition of H2SO4:
H2SO4 reacts with CH3CO2H to form HSO4- and H3O+.
The concentration of H2SO4 added is 0.01 M, but since acetic acid reacts as a 1:1 ratio with H2SO4, the concentration of CH3CO2H that will react is also 0.01 M.
Now, calculate the new concentrations of CH3CO2H and CH3CO2- after the reaction:
CH3CO2H: 0.68 M - 0.01 M = 0.67 M
CH3CO2-: 0.58 M + 0.01 M = 0.59 M
The equilibrium between CH3CO2H and CH3CO2- can be described by the dissociation constant (Ka) as:
Ka = [H3O+][CH3CO2-] / [CH3CO2H]
Rearranging the equation, we have:
[H3O+] = (Ka * [CH3CO2H]) / [CH3CO2-]
Substituting the provided values:
Ka = 1.8x10^-5
[CH3CO2H] = 0.67 M
[CH3CO2-] = 0.59 M
[H3O+] = (1.8x10^-5 * 0.67) / 0.59
[H3O+] ≈ 2.04x10^-5 M
To find the pH, we use the equation:
pH = -log[H3O+]
Substituting the value:
pH = -log(2.04x10^-5)
pH ≈ 4.69
Therefore, the pH of the buffer solution, after adding 0.01 M H2SO4, is approximately 4.69.