Calculate the Ka values for each of the following?

Question

The pH of an 0.041 M solution of weak acid A is 2.37. The pH of an 0.024 M solution of weak acid B is 2.90. The pH of an 0.097 M solution of weak acid C is 5.08.

Answer 1

pH = 2.37 => [H+] = 10^-2.37 = 0.00427 M
HA <- -- - - -- - - > H+ + A-
initial concentration
0.041
at equilibrium
0.041-0.00427.....0.00427...0.00427
Ka = ( 0.00427)^2 / 0.0367 = 0.000497


pH = 2.90 => [H +]= 10^-2.90 =0.00126 M
In the same way [ HA] = 0.024 - 0.00126 =0.0227 M at equilibrium
Ka = ( 0.00126)^2 / 0.0227 = 6.99 x 10^-5

pH = 5.08 => [H+] = 10^-5.08 = 8.32 x 10^-6 M
[HA] = 0.097 - 8.32 x 10^-6 = 0.097
Ka = ( 8.32 x 10^-6)^2 / 0.097 = 7.14 x 10^-10

Answer 2

Ka = [H+][A-] / [HA]

Assuming HA is monoprotic then H+ = A-
So by substitution
Ka = [H+]^2 / [HA]

& pH = -log(10)[H+] = 2.37
[H+] = 10^-2.37 = 4.2658 x 10^-3
[H+]^2 = (4.2658 x 10^-3)^2 = 1.8197 x 10^-5
Ka = 1.8197 x 10^-5 / 0.041 = 4.4383 x 10^-4
Ka = 4.4383 x 10^-4

Ka = [H+][A-] / [HA]


Assuming HB is monoprotic then H+ = B-
So by substitution
Ka = [H+]^2 / [HB]

& pH = -log(10)[H+] = 2.90
[H+] = 10^-2.90 = 1.2589 x 10^-3
[H+]^2 = (1.2589 x 10^-3)^2 = 1.5849 x 10^-6
Ka = 1.5849 x 10^-6 / 0.024 = 6.6037 x 10^-5
Ka = 6.6037 x 10^-5

Assuming HC is monoprotic then H+ = C-
So by substitution
Ka = [H+]^2 / [HC]

& pH = -log(10)[H+] = 5.08
[H+] = 10^-5.08 = 8.3176 x 10^-6
[H+]^2 = (8.3176 x 10^-6)^2 = 6.9183 x 10^11
Ka = 6.9183 x 10^-11 / 0.097 = 7.1323 x 10^-10
Ka = 7.1323 x 10^-10

Answer 3

A:
pH= 2.37 => c(H+)=10^(-2.37)=0.00427mol/L

A->H+ +Aalkalineversion

c(H+) =c(Aalkalineversion)

now you know you know you had 0.041 mol/L A originally , of which now 0.00427 mol/l have reacted away.

Ka=[c(H+)]^2/[0.041-0.00427]
Ka= 4.96*10^(-4)

Try the others yourself