To find the pH of a 0.22 M ammonia solution, we need to first determine its hydroxide ion concentration, as ammonia (b1 NH3) is a weak base. The ionization of ammonia in water can be represented by the following equilibrium expression:
NH3 + H2O <=> NH4+ + OH–
The equilibrium constant for this reaction is known as Kb, and it can be derived from the ionization constant of water (Kw) and the Ka of NH4+. Given that Ka is 1.8 x 10-5, we can find Kb using:
Kb = Kw / Ka = (1.0 x 10-14) / (1.8 x 10-5) ≈ 5.56 x 10-10
Now, using the Kb value, we can set up the equilibrium expression:
Kb = [NH4+][OH–] / [NH3]
Assuming x is the amount of NH3 that ionizes, we can write:
- [NH4+] = x
- [OH–] = x
- [NH3] = 0.22 – x (approximately 0.22, since x is small)
Substituting these values into the Kb expression:
5.56 x 10-10 = (x)(x) / (0.22)
5.56 x 10-10 = x2 / 0.22
x2 = 5.56 x 10-10 x 0.22
x2 ≈ 1.223 x 10-10
x ≈ √(1.223 x 10-10) ≈ 1.11 x 10-5 M (OH–)
Now that we have the concentration of hydroxide ions, we can find the pOH:
pOH = -log(1.11 x 10-5) ≈ 4.95
To find the pH, we can use the relationship:
pH + pOH = 14
pH = 14 – 4.95 ≈ 9.05
Next, we calculate the percent ionization of ammonia:
Percent Ionization = (Amount ionized / Initial concentration) * 100%
Which is:
Percent Ionization = (1.11 x 10-5 / 0.22) * 100% ≈ 0.00505%
In summary, the pH of a 0.22 M ammonia solution is approximately 9.05, and the percent ionization of ammonia at this concentration is about 0.00505%.