What is the pH of a 0.1 M solution of ammonia (NH3), a weak base with a Kb of 1.8 x 10^-5?

• A. 11.13
• B. 9.25
• C. 12.75
• D. 7.00

Answer: (A)

Verified step by step guidance

1. Identify that ammonia (NH3) is a weak base and will partially ionize in water to form NH4+ and OH-. The equilibrium expression for this reaction is: NH3 + H2O ⇌ NH4+ + OH-.
2. Write the expression for the base dissociation constant (Kb) for ammonia: Kb = [NH4+][OH-] / [NH3].
3. Assume that the initial concentration of NH3 is 0.1 M and that the change in concentration due to ionization is 'x'. Therefore, at equilibrium, [NH4+] = x, [OH-] = x, and [NH3] = 0.1 - x.
4. Substitute these equilibrium concentrations into the Kb expression: 1.8 x 10^-5 = (x)(x) / (0.1 - x). Since Kb is small, assume x << 0.1, simplifying the expression to 1.8 x 10^-5 ≈ x^2 / 0.1.
5. Solve for x to find the concentration of OH- ions, then calculate the pOH using the formula pOH = -log[OH-]. Finally, use the relationship pH + pOH = 14 to find the pH of the solution.