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... Question 9 Homework. Unanswered Determine the pH of a solution of aniline of 0.025 aniliniumion PK, 46 0 A pH=5,5 O BPH 1
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T YNH₂ + H₂O - 0Hz toh 0.025 m (initial) 0.025€ 0.025(1-x) lat equilibrium) Kb = (0.0256) (0.0254) = 0.0254? 0.025(1-4) l-ana0-0250² = 3.9881070 *2= 3.98 x 10 10 = 159.2 x 10 10 0.625 1159.2 x 10-105 % 1262x104 [OH-] = 0.025 * = 0.025*1.262 x 10WM =

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