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Example · Example 17

Q.Calculate the pH\text{pH} of a 0.1 M0.1\ \text{M} solution of NH4Cl\text{NH}_4\text{Cl}, given Kb(NH3)=1.8×10−5K_b(\text{NH}_3) = 1.8 \times 10^{-5} and Kw=1.0×10−14K_w = 1.0 \times 10^{-14}.

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NH4Cl\text{NH}_4\text{Cl} is the salt of the strong acid HCl\text{HCl} and the weak base NH4OH\text{NH}_4\text{OH} (NH3\text{NH}_3). Cl−\text{Cl}^- does not hydrolyze, but NH4+\text{NH}_4^+ does: NH4++H2O⇌NH4OH+H+\text{NH}_4^+ + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4\text{OH} + \text{H}^+, with Ka(NH4+)=Kw/Kb=(1.0×10−14)/(1.8×10−5)≈5.56×10−10K_a(\text{NH}_4^+) = K_w/K_b = (1.0\times10^{-14})/(1.8\times10^{-5}) \approx 5.56\times10^{-10}. Treating this exactly like a weak acid ionization, $[\text{H}^+] = \sqrt{K_a C} = \sqrt{(5.56\times10^{-10})(0.1)} = \sqrt{5.56\times10^{-11}} \ …

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