A 0.039 M solution of a weak acid (HA) has a pH of 4.28. What is the Ka of the acid?
pH = -log (H^+)
(H^+) = ?. I estimate 5E-5 but you need a better answer than that.
...........HA ==> H^+ + A^-
I........0.039....0......0
C.........-x......x......x
E......0.039-x....x......x
The pH tells you the value of x; i.e., approx 5E-5 so you can evaluate x and 0.039-x.
Plug those values into the Ka expression and evaluate Ka.
To find the Ka of the weak acid (HA), we first need to determine the concentration of H+ ions in the solution.
Since the pH of the solution is given as 4.28, we can determine the concentration of H+ ions using the formula:
[H+] = 10^(-pH)
[H+] = 10^(-4.28)
[H+] = 4.17 x 10^(-5) M
Now, we can set up the equation for the dissociation of the weak acid:
HA ⇌ H+ + A-
We can assume that initially, the concentration of the weak acid (HA) is equal to the concentration of the acid that dissociates (A-) after equilibrium is reached. Therefore, the initial concentration of HA is also 4.17 x 10^(-5) M.
Using the equilibrium expression for Ka:
Ka = [H+][A-] / [HA]
Substituting the known values:
Ka = (4.17 x 10^(-5))(4.17 x 10^(-5)) / (4.17 x 10^(-5))
Ka = 1.73 x 10^(-9)
To find the Ka of the acid (HA), we need to use the given information about the pH and concentration of the solution.
First, let's recall the relationship between pH and the concentration of hydrogen ions (H+):
pH = -log[H+]
We know that the pH of the solution is 4.28, so we can calculate [H+] using this equation:
[H+] = 10^(-pH)
[H+] = 10^(-4.28)
[H+] ≈ 4.2 x 10^(-5) M
Since HA is a weak acid, it dissociates in water according to the following equation:
HA ⇌ H+ + A-
The concentration of H+ is the same as the concentration of the weak acid, which is 0.039 M.
Now, we can use the formula for Ka to find the value:
Ka = [H+][A-] / [HA]
Substituting the values we calculated:
Ka = (4.2 x 10^(-5) M) * (0.039 M) / (0.039 M)
Ka ≈ 4.2 x 10^(-5)
Therefore, the approximate value of Ka for the weak acid HA is 4.2 x 10^(-5).