To find the pH of a 0.5 M ammonium chloride (NH4Cl) solution, you need to consider the hydrolysis reaction of the ammonium ion (NH4+) with water. The hydrolysis of NH4+ produces H3O+ ions, which affects the pH of the solution.
The hydrolysis reaction of NH4+ with water can be represented as follows:
NH4+ + H2O ⇌ NH3 + H3O+
The equilibrium constant for this reaction is the base dissociation constant (Kb) of ammonium ion, which is given as 1.8 × 10^-5.
Now, let's set up an ICE (Initial, Change, Equilibrium) table to calculate the concentrations of the species involved:
NH4+ + H2O ⇌ NH3 + H3O+ I 0.5 M 0 M 0 M C -x +x +x E 0.5-x x x
Since the initial concentration of NH4+ is 0.5 M and the change is -x, the equilibrium concentration is (0.5 - x) M.
Since Kb is given, we can write the Kb expression:
Kb = [NH3][H3O+] / [NH4+]
Kb = x * x / (0.5 - x)
As Kb is a small value, we can assume that x is negligible compared to 0.5. Therefore, we can approximate 0.5 - x as 0.5.
Kb = x * x / (0.5)
1.8 × 10^-5 = x^2 / (0.5)
x^2 = (1.8 × 10^-5) * (0.5) x^2 = 9 × 10^-6
Taking the square root of both sides:
x ≈ 3 × 10^-3
Since we approximated 0.5 - x as 0.5, the concentration of H3O+ is approximately 3 × 10^-3 M.
To find the pH, we can use the formula:
pH = -log10[H3O+]
pH = -log10(3 × 10^-3) pH ≈ -log10(3) + log10(10^-3) pH ≈ -0.4771 + 3 pH ≈ 2.5229
Therefore, the pH of a 0.5 M ammonium chloride (NH4Cl) solution is approximately 2.52.