A 1.0 x 10-2 M solution of a weak acid is found to be 3.7% dissociated. Calculate Ka of this acid.
I found the amount dissociated by (amount dissociated/initial conc)*100=3.7
Where do I go from here?
I think you need to back up before going forward.
(HA) = 0.01
(H^+) = (A^-) = 0.01 x 0.037 = 3.7E-4
............HA ==> H^+ + A^-
I.........0.01.....0.....0
C........-3.7E-4..3.7E-4..3.7E-4
E.......you finish
Substitute the E line into the Ka expression and solve for Ka. Your answer should be approx 1E-5 if I plugged in the right numbers.
To calculate the Ka of the weak acid, you need to use the information about the percentage of dissociation and initial concentration.
The percentage of dissociation is given as 3.7%. This means that 3.7% of the initial concentration of the acid has dissociated into its ions.
To find the amount dissociated (in moles), you can use the equation:
(amount dissociated/initial concentration) = percentage dissociation/100
Let's assume the initial concentration of the acid is represented by [HA]. Therefore, the amount dissociated is:
(amount dissociated) = ([HA] * (percentage dissociation/100))
Now, since the acid is weak, we can assume that the concentration of [HA] after dissociation is approximately equal to the initial concentration (since only a small fraction dissociates). Therefore, we can say that:
([HA] - amount dissociated) ≈ [HA]
Using this approximation, we can express the equilibrium expression for the dissociation of the acid as:
Ka = ([H+][A-]) / [HA]
Substituting the values we know:
Ka = ([H+][A-]) / ([HA] * (percentage dissociation/100))
Now, we need to determine the concentrations of [H+] and [A-]. Since the acid is weak, we can assume they are both equal to the amount dissociated, which we calculated earlier:
[H+] = amount dissociated
[A-] = amount dissociated
Therefore, we can rewrite the equation as:
Ka = (amount dissociated * amount dissociated) / ([HA] * (percentage dissociation/100))
Now, you can substitute the values you calculated for amount dissociated and the given initial concentration [HA]. Using those values, you can solve for Ka.