I performed an experiment to find the pH of strong acid, weak acid, salt, and bu
ID: 514031 • Letter: I
Question
I performed an experiment to find the pH of strong acid, weak acid, salt, and buffer solutions. I used HCl, HC2H3O2 solutions, and buffer solutions of HC2H3O2/NaC2H3O2. I also added HCl and NaOH solutions to separate samples of buffer solutions.
2.) Preparing HC2H3O2 Solutions and determining pH
Find the theoretical pH from the concentration of HC2H3O2 and the measured pH
1.0x10^-1 .....1.87....????
1.0x10^-2 ..... 2.54....????
1.0x10^-3 ...... 3.01 .... ????
1.0x10^-4 ...... 4.52.....????
3.)Given the concentrations of HC2H3O2, find the *calculated Ka* of HC2H3O2 and the *literature Ka* based on the pH data
Concentration:
1.0x10^-1
1.0x10^-2
1.0x10^-3
1.0x10^-4
Explanation / Answer
2) theoretical pH
[HC2H3O2] = 1 x 10^-1 M
HC2H3O2 <==> H+ + C2H3O2-
let x amount dissociated
Ka = 1.8 x 10^-5 = x^2/1 x 10^-1
x = [H+] = 1.34 x 10^-3 M
theoretical pH = -log[H+] = 2.87
Literature Ka = (1.34 x 10^-3)^2/(1 x 10^-1 - 1.34 x 10^-3) = 1.82 x 10^-5
measured pH = -log[H+] = 1.87
[H+] = 0.0135 M
Calculated Ka = (0.0135)^2/(1 x 10^-1 - 0.0135) = 2.11 x 10^-3
[HC2H3O2] = 1 x 10^-2 M
HC2H3O2 <==> H+ + C2H3O2-
let x amount dissociated
Ka = 1.8 x 10^-5 = x^2/1 x 10^-2
x = [H+] = 4.24 x 10^-4 M
theoretical pH = -log[H+] = 3.37
Literature Ka = (4.24 x 10^-4)^2/(1 x 10^-2 - 4.24 x 10^-4) = 1.88 x 10^-5
measured pH = -log[H+] = 2.54
[H+] = 2.88 x 10^-3 M
Calculated Ka = (2.88 x 10^-3)^2/(1 x 10^-2 - 2.88 x 10^-3) = 1.17 x 10^-3
[HC2H3O2] = 1 x 10^-3 M
HC2H3O2 <==> H+ + C2H3O2-
let x amount dissociated
Ka = 1.8 x 10^-5 = x^2/1 x 10^-3
x = [H+] = 1.34 x 10^-4 M
theoretical pH = -log[H+] = 3.87
Literature Ka = (1.34 x 10^-4)^2/(1 x 10^-3 - 1.34 x 10^-4) = 2.07 x 10^-5
measured pH = -log[H+] = 3.01
[H+] = 9.77 x 10^-4 M
Calculated Ka = (9.77 x 10^-4)^2/(1 x 10^-3 - 9.77 x 10^-4) = 0.0415
[HC2H3O2] = 1 x 10^-4 M
HC2H3O2 <==> H+ + C2H3O2-
let x amount dissociated
Ka = 1.8 x 10^-5 = x^2/1 x 10^-4
x = [H+] = 4.24 x 10^-5 M
theoretical pH = -log[H+] = 4.37
Literature Ka = (4.24 x 10^-5)^2/(1 x 10^-4 - 4.24 x 10^-5) = 3.12 x 10^-5
measured pH = -log[H+] = 4.52
[H+] = 3.02 x 10^-5 M
Calculated Ka = (3.02 x 10^-5)^2/(1 x 10^-4 - 3.02 x 10^-5) = 1.31 x 10^-5
3) Theoretical and calculate Ka
2) theoretical pH
[HC2H3O2] = 1 x 10^-1 M
theoretical pH = -log[H+] = 2.87
Literature Ka = (1.34 x 10^-3)^2/(1 x 10^-1 - 1.34 x 10^-3) = 1.82 x 10^-5
measured pH = -log[H+] = 1.87
[H+] = 0.0135 M
Calculated Ka = (0.0135)^2/(1 x 10^-1 - 0.0135) = 2.11 x 10^-3
[HC2H3O2] = 1 x 10^-2 M
theoretical pH = -log[H+] = 3.37
Literature Ka = (4.24 x 10^-4)^2/(1 x 10^-2 - 4.24 x 10^-4) = 1.88 x 10^-5
measured pH = -log[H+] = 2.54
[H+] = 2.88 x 10^-3 M
Calculated Ka = (2.88 x 10^-3)^2/(1 x 10^-2 - 2.88 x 10^-3) = 1.17 x 10^-3
[HC2H3O2] = 1 x 10^-3 M
theoretical pH = -log[H+] = 3.87
Literature Ka = (1.34 x 10^-4)^2/(1 x 10^-3 - 1.34 x 10^-4) = 2.07 x 10^-5
measured pH = -log[H+] = 3.01
[H+] = 9.77 x 10^-4 M
Calculated Ka = (9.77 x 10^-4)^2/(1 x 10^-3 - 9.77 x 10^-4) = 0.0415
[HC2H3O2] = 1 x 10^-4 M
theoretical pH = -log[H+] = 4.37
Literature Ka = (4.24 x 10^-5)^2/(1 x 10^-4 - 4.24 x 10^-5) = 3.12 x 10^-5
measured pH = -log[H+] = 4.52
[H+] = 3.02 x 10^-5 M
Calculated Ka = (3.02 x 10^-5)^2/(1 x 10^-4 - 3.02 x 10^-5) = 1.31 x 10^-5