NH3 is a weak base (Kb = 1.8 × 10–5) and so the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.050 M in NH4Cl?

.........NH4^+ + H2O ==> H3O^+ + NH3

I......0.05...............0........0
C........-x...............x........x
E......0.05-x..............x.......x

ka(for NH4^+) = (Kw/Kb for NH3) =
(H3O^+)(NH3)/(NH4^+)
Substitute and solve for x = (H3O^+), convert to pH.

1254.23

To find the pH of a solution of NH4Cl, we need to consider the hydrolysis reaction of NH4Cl. NH4Cl dissociates in water to form NH4+ and Cl- ions. NH4+ acts as a weak acid by accepting water molecules to form hydronium ions (H3O+), while Cl- ions remain as spectator ions and do not participate in the pH determination.

The hydrolysis reaction of NH4+ can be represented as follows:
NH4+ + H2O ↔ NH3 + H3O+

The equilibrium constant for this reaction is given by the base ionization constant, Kb, for NH3, which is 1.8 × 10^–5.

Using the equilibrium expression for Kb:
Kb = [NH3][H3O+]/[NH4+]

Since we are given the concentration of NH4Cl (0.050 M), the concentration of NH4+ and Cl- ions is also equal to 0.050 M.

Let's assume that x mol/L of NH4+ reacts with water to form NH3 and H3O+.

At equilibrium, the concentration of NH4+ will be (0.050 - x) M, the concentration of NH3 will be x M, and the concentration of H3O+ will also be x M.

Substituting these values into the Kb expression, we have:
1.8 × 10^–5 = (x)(x)/(0.050 - x)

Solving this quadratic equation, we find that x ≈ 0.0044 M.

Since x represents the concentration of H3O+, which is also the concentration of H+ ions, and pH is defined as -log[H+], we can calculate the pH by taking the negative logarithm of x:
pH = -log(0.0044) ≈ 2.36