What is the pH of a 0.208 M NH4Cl solution at 25°C, given that the Kb of NH3 is 1.8x10^-5?

• A. 4.98
• B. 7.00
• C. 9.02
• D. 11.22

Answer: (A)

Verified step by step guidance

1. Identify that NH4Cl is a salt formed from the weak base NH3 and the strong acid HCl. In solution, NH4+ will act as a weak acid.
2. Write the hydrolysis reaction for NH4+: NH4+ + H2O ⇌ NH3 + H3O+. This shows NH4+ donating a proton to water, forming NH3 and H3O+.
3. Use the relationship between Ka and Kb for conjugate acid-base pairs: Ka * Kb = Kw, where Kw is the ion-product constant of water (1.0 x 10^-14 at 25°C). Calculate Ka for NH4+ using Ka = Kw / Kb.
4. Set up the expression for the acid dissociation constant (Ka) of NH4+: Ka = [NH3][H3O+] / [NH4+]. Assume [NH3] = [H3O+] = x and [NH4+] = 0.208 - x, but since x is small, approximate [NH4+] ≈ 0.208.
5. Solve for x, which represents [H3O+], using the equation Ka = x^2 / 0.208. Once x is found, calculate the pH using pH = -log[H3O+].