What is the pH of a 0.100 M NH₃ solution that has K_b = 1.8 × 10^-5? The equation for the dissociation of NH₃ is NH₃(aq)+ H₂O(l)⇌ NH₄⁺(aq)+ OH^-(aq). A)1.87 B)2.87 C)11.13 D)12.13

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Quizplus · Accepted Answer

The dissociation of NH4+ in water is NH4+ + H2O ⇌ NH3 + H3O+. The given $K_b$ for NH3 is $1.8 \times 10^{-5}$. To find the pH, we first calculate the $pOH$ using the formula $pOH = \frac{1}{2}(pK_w - \log K_b - \log [NH_4^+])$, where $pK_w = 14$ at 25°C. Substituting the given values, $pOH = \frac{1}{2}(14 - \log(1.8 \times 10^{-5}) - \log(0.100))$. This calculation gives a $pOH$ around 2.87. Since $pH + pOH = 14$, the $pH = 14 - pOH = 14 - 2.87 = 11.13$.

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