What is the acid dissociation constant for a 0.010 M solution of the unkown acid HX that has a pH of 2.70 at equilibrium?
I know the answer but not how to get it, please help? Thanks in advance.
pH = 2.70 = -log(H^+)
(H^+) = 0.002M
...........HX ==> H^+ + X^-
I.......0.010.....0......0
C..........-y.....y......y
E........0.010-y...y.....y
Ka = (H^+)(X^-)/(HX)
y = 0.002 from the pH calculation above. Substitute and solve for Ka.
Thanks for the help, but I keep getting 4*10^-4 and not 5*10^-4 which is what I'm told the correct answer is. :S
To find the acid dissociation constant (Ka) for the unknown acid HX, we can start by using the pH value to determine the concentration of H+ ions at equilibrium.
First, we need to convert the pH to a concentration of H+ ions. The pH is given as 2.70, which represents the negative logarithm (base 10) of the H+ concentration. So, we can calculate the H+ concentration using the equation:
[H+] = 10^(-pH)
[H+] = 10^(-2.70)
[H+] = 0.001995 M
Since HX is a weak acid, we can assume that it partially dissociates in water according to the equation:
HX ⇌ H+ + X-
The equilibrium concentration of H+ ions can be considered as the concentration of the acid that has dissociated, whereas the initial concentration of HX is equal to the equilibrium concentration of X- ions.
Therefore, on equilibrium, the concentration of H+ ions is 0.001995 M.
The initial concentration of HX is given as 0.010 M.
Hence, the fraction of acid that has dissociated (α) is given by:
α = [H+]/[HX]
α = 0.001995 M / 0.010 M
α = 0.1995
Now, we can use the definition of the acid dissociation constant (Ka) to find its value:
Ka = [H+][X-] / [HX]
Since the initial concentration of X- ions is equal to the equilibrium concentration of HX (0.010 M), we can rewrite the equation as:
Ka = (0.001995 M) * (0.010 M) / 0.010 M
Ka = 0.001995 M
Therefore, the acid dissociation constant (Ka) for the 0.010 M solution of the unknown acid HX with a pH of 2.70 at equilibrium is 0.001995 M.