If the Ka of a monoprotic weak acid is 4.3*10^-6, what is the pH of a 0.16 M solution of this acid?
To find the pH of a solution of a weak acid, we can use the expression for the Ka of the acid and solve for the concentration of hydronium ions ([H3O+]). The [H3O+] concentration can then be used to calculate the pH using the formula: pH = -log[H3O+].
Given:
Ka = 4.3*10^-6
Concentration of the acid ([HA]) = 0.16 M
The dissociation reaction of the weak acid can be written as:
HA ⇌ H+ + A-
The equilibrium constant expression (Ka) for this reaction is:
Ka = [H+][A-]/[HA]
Since the acid is monoprotic, the concentration of [H+] is equal to the concentration of [A-] because one mole of acid (HA) produces one mole of hydronium ions (H+). Therefore, we can assume [H+] = [A-] = x (let's say).
Substituting the given values into the Ka expression, we have:
4.3*10^-6 = (x)(x)/0.16
Simplifying the expression, we get:
x^2 = 4.3*10^-6 * 0.16
Solving for x, the concentration of [H+] (or [A-]), we find:
x ≈ 0.002071
Now that we know the concentration of [H+] (or [A-]), we can calculate the pH using the formula:
pH = -log[H+]
pH = -log(0.002071)
Calculating this using a calculator, we find:
pH ≈ 2.70
Therefore, the pH of a 0.16 M solution of this acid is approximately 2.70.