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Kale Eichmann

Feb 20, 2025

NH3 is a weak base (Kb = 1.8 × 10–5) so the salt NH4Cl acts as weak acid. What is the pH of a solution that is 0.043 M in NH4Cl at 25 °C?

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Hydrolysis of NH4Cl can be written as NH4Cl + H2O ==> Cl^- + NH3 + H+ or simply NH4^+ + H2O ==> NH3 + H3O^+ Kw = KaKbKa = Kw/Kb = 1×10^-14/1.8×10^-5 = 5.56×10^-10Ka = 5.56×10^-10 = [NH3][H3O+]/[NH4Cl] = (x)(x)/0.043 -x and assuming x is small relative to 0.043 M, then ignore it x^2 = 2.39×10^-11x = 4.9×10^-6 = [H3O+] (note: this is small relative to 0.043 so assumption was valid, and no need for quadratic)pH = -log 4.9×10^-6pH = 5.3... Show More
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