What is the PH of a 0.075M solution of the acid HA if the Ka for the acid is 4.8×10raise to power -8?
To find the pH of a solution, we need to calculate the concentration of hydrogen ions (H+ ions) in the solution.
In this case, you're given a 0.075 M solution of the acid HA, with a Ka (acid dissociation constant) value of 4.8 × 10^-8.
The Ka value indicates the extent of the ionization of the acid in water. For a weak acid like HA, it can be assumed that only a small portion of HA will dissociate into H+ ions and the conjugate base A-.
The equilibrium expression for the dissociation of the acid can be written as:
HA ⇌ H+ + A-
The Ka expression related to this dissociation is:
Ka = [H+][A-] / [HA]
Since the concentration of H+ ions is the same as the concentration of A- ions (as they are formed in a 1:1 ratio), we can assume that [H+] = [A-] and express Ka as:
Ka = [H+]^2 / [HA]
Rearranging the equation, we get:
[H+]^2 = Ka × [HA]
Plugging in the values, we have:
[H+]^2 = (4.8 × 10^-8) × (0.075)
[H+]^2 = 3.6 × 10^-9
Now, we can solve for [H+] by taking the square root of both sides:
[H+] ≈ √(3.6 × 10^-9)
[H+] ≈ 6 × 10^-5
The concentration of hydrogen ions in the solution is approximately 6 × 10^-5 M.
To find the pH, which measures the acidity or alkalinity of a solution, we can use the equation:
pH = -log[H+]
Plugging in the value of [H+], we get:
pH = -log(6 × 10^-5)
pH ≈ -(-4.22)
pH ≈ 4.22
Therefore, the pH of the 0.075 M solution of the acid HA, with a Ka of 4.8 × 10^-8, is approximately 4.22.