If a weak acid has a dissociation constant of 6.4 x 10(-6), what is the pH of a solution of this acid in 0.10 molar concentration?
..........HA ==> H^+ + A^-
initial...0.1....0......0
change....-x....+x......+x
final.....0.1-x...+x.....+x
Ka = (H^+)(A^-)/(HA)
Substitute into Ka expression from ICE chart above and solve for (H^+), then convert to pH.
To find the pH of a solution of a weak acid, you can use the dissociation constant (Ka) of the acid. The dissociation constant is a measure of the extent to which the acid dissociates in water.
The equation for the dissociation of a weak acid, denoted as HA, can be written as follows:
HA ⇌ H+ + A-
The dissociation constant (Ka) expression for this reaction is:
Ka = [H+][A-] / [HA]
In this case, you are given the value of the dissociation constant (Ka) as 6.4 x 10^(-6), and the molar concentration of the acid (HA) as 0.10 M.
To find the concentration of [H+] (hydrogen ion) in the solution, we can assume that the concentration of [A-] (conjugate base) is essentially equal to the concentration of [H+], as the acid is weak.
Let's assume the concentration of [H+] is denoted as 'x'. At equilibrium, the concentration of [A-] would also be 'x'.
Using this assumption, we can rewrite the dissociation constant expression as follows:
Ka = x * x / (0.10 - x)
Since the dissociation constant (Ka) is given, we can plug it in and solve for 'x', which represents the concentration of [H+].
6.4 x 10^(-6) = x^2 / (0.10 - x)
Simplifying the equation, we get:
6.4 x 10^(-6) = x^2 / 0.10
Now, we can solve this equation using the quadratic equation or approximations. Since the dissociation constant is small, we can assume that 'x' is much smaller than 0.10, and thus, 0.10 - x ≈ 0.10.
Simplifying the equation further, we have:
6.4 x 10^(-6) = x^2 / 0.10
Rearranging the equation, we get:
x^2 = 6.4 x 10^(-6) * 0.10
Solving for 'x', we find:
x ≈ 0.008
Therefore, the concentration of [H+] is 0.008 M.
To find the pH, we can use the formula:
pH = -log10[H+]
Plugging in the value of [H+], we get:
pH = -log10(0.008) ≈ 2.10
Hence, the pH of the solution of this weak acid with a concentration of 0.10 M is approximately 2.10.