WebMay 16, 2024 · [OH −] = 0.006928 M pH = 11.85 Explanation: You are given some solution of ethylamine, an organic molecule, which gives off a basic solution in pure water. How do I know the solution will be basic and not acidic? Well, look at the Kb provided. The Kb is called the base dissociation constant. WebJun 8, 2024 · The pH is at the lower end of this range, pH = p Ka – 1, when the weak acid’s concentration is 10 × greater than that of its conjugate weak base. The buffer reaches its upper pH limit, pH = p Ka + 1, when the weak acid’s concentration is 10 × smaller than that of its conjugate weak base.
chemistry chapter 16.3 Flashcards Quizlet
WebApr 6, 2024 · Explanation: EtN H 2(aq) +H 2O(l) ⇌ EtN H + 3 + H O−. Kb = [EtN H + 3][−OH] [EtN H 2] We could solve this equation had we a value for Kb for ethylamine, or Ka for ethyl ammonium cation; such values are available, and it should have been supplied with the question. Answer link. WebA: The pH of any solution is given by pH = 14 + log[OH- ] where [OH- ] = concentration of OH- ions question_answer Q: The base-dissociation constant of ethylamine (C2H5NH2) is 5.6x10 at 25.0°C. how do polar bears travel
Explained: A 0.100 M solution of ethylamine (C2H5NH2) …
WebMinor pH increase disclosed for some n-alkylamine titanates (pH 1 < pH 2) should be due to partial amine deintercalation into the reaction solution. In the course of preparation of photocatalytic suspensions, it was established that the sample dispersibility is strongly dependent on the polarity of the interlayer organic modifier. WebApr 19, 2024 · pH = 11.7 Explanation: We address the equilibrium... H 2O(l) + N (CH 2CH 3)3(aq) ⇌ H O− +H + N (CH 2CH 3)3 For which Kb = [H O−][H + N (CH 2CH 3)3] [N (CH 2CH 3)3(aq)] ... And now we simply put in some numbers, and NOTE that [H O−] = [H + N (CH 2CH 3)3] = x ...so... Kb = x2 0.050 − x = 5.3 ×10−4 ...and if 0.050>>x ...then... WebAug 2, 2024 · pH = 11.87 so H+ = 1.35×10^-12 [OH-] = Kw/ [H+] = 7.41×10^-3 [BH+]= [OH-]=7.41×10^-3 [B] = 0.100 – 0.007 = 0.093 Kb = 5.9×10^-4 First, a reaction: Et-NH2 + H2O-> Et-NH3+ + OH- So: [Et-NH3+] [OH-]/ [Et-NH2] = Kb Let x be the moles/liter of Et-NH2 reacting. Since you know the pH, the pOH = 2.13. how do police assess vulnerability