Respuesta :
Answer:
Ka = 2.76 * 10^-6
Explanation:
The question asks to calculate the acid ionization constant value(Ka) for an acid, given the molarity and pH of the acid in question.
Firstly, we write the dissociation equation for the acid: This can be written as follows:
HA --->H+ + A-
From here, we can write the equilibrium expression of this dissociation equation as:
Ka = [H+][A-]/[HA]
Hence, to calculate the value of Ka, we need to know the value of the concentrations.
We start with the hydrogen ion
Mathematically, pH = -log[H+]
3.26 = -log[H+]
[H+] = -Antilog(3.26)
[H+] = 5.494 * 10^-4 M
We can use the initial, change and equilibrium table I.e ICE to calculate the equilibrium concentration as follows:
HA. [H+]. [A-]
I. 0.11. 0. 0.
C. -x. +x. +x
E. 0.11-0.0005494 0.0005494. 0.0005494
Since 0 + x = 0.0005494, x = 0.0005494
We now calculate the value for the Ka according to the equation below:
Ka = [H+][A-]/[HA]
Ka = (0.0005494)^2/(0.11-0.0005494)
Ka = 0.00000030184/0.1094506
Ka = 2.76 * 10^-6
