What is the pH of a 0.253 M solution of ammonium chloride? Kb (NH₃) = 1.2 × 10^-5

Question

Quizplus · Accepted Answer

Ammonium chloride is a salt of a weak base (NH3) and a strong acid (HCl). It undergoes hydrolysis in water to produce H3O+ and NH3:

NH4Cl + H2O → NH4+ + Cl- + H3O+

Since NH3 is a weak base, it will partially deprotonate H3O+ to form NH4+ and H2O:

NH3 + H3O+ → NH4+ + H2O

Therefore, the solution will be acidic. The pH can be calculated using the Kb and the concentration of NH4Cl:

Kb = Kw/Ka(NH4+) = 1.0 x 10^-14 / 5.6 x 10^-10 = 1.8 x 10^-5

Kb = [NH3][OH-]/[NH4+]

1.8 x 10^-5 = [x][x]/[0.253 - x]

Assuming x<<0.253, we can simplify the denominator to 0.253 and solve for x:

x = [OH-] = 1.2 x 10^-3 M

pOH = -log[OH-] = 2.92

pH = 14 - pOH = 11.08

Therefore, the pH of the solution is approximately 4.8. The correct answer is B.

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