The equilibrium for the reaction between (CH,),NH, a weak base, and water is represented by the equation below. The table shows the pH of three solutions of (CH,),NH(aq) at 25°C. (CH3),NH(aq) + H2O(1) 2 (CH,),NH,†(aq)+ OH¯(ag) K, = 5.4 x 10-4 at 25°C %3| [(CH3),NH] pH at 25°C 0.050 11.69 0.10 11.85 0.20 12.01 Based on the information given, which of the following is true? Solutions with a higher concentration of (CH,),NH have a higher pOH because to reach A equilibrium a smaller amount of the conjugate acid (CH,),NH2* is produced. Solutions with a higher concentration of (CH,),NH have a higher pOH because to reach equilibrium more OH is produced. 21 Solutions with a higher concentration of (CH,),NH have a higher pH because to reach equilibrium a smaller amount of the conjugate acid (CH),NH2* is produced. Solutions with a higher concentration of (CH,),NH have a higher pH because to reach equilibrium more OH is produced.
The equilibrium for the reaction between (CH,),NH, a weak base, and water is represented by the equation below. The table shows the pH of three solutions of (CH,),NH(aq) at 25°C. (CH3),NH(aq) + H2O(1) 2 (CH,),NH,†(aq)+ OH¯(ag) K, = 5.4 x 10-4 at 25°C %3| [(CH3),NH] pH at 25°C 0.050 11.69 0.10 11.85 0.20 12.01 Based on the information given, which of the following is true? Solutions with a higher concentration of (CH,),NH have a higher pOH because to reach A equilibrium a smaller amount of the conjugate acid (CH,),NH2* is produced. Solutions with a higher concentration of (CH,),NH have a higher pOH because to reach equilibrium more OH is produced. 21 Solutions with a higher concentration of (CH,),NH have a higher pH because to reach equilibrium a smaller amount of the conjugate acid (CH),NH2* is produced. Solutions with a higher concentration of (CH,),NH have a higher pH because to reach equilibrium more OH is produced.
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter13: Acids And Bases
Section: Chapter Questions
Problem 60QAP: Consider a 0.33 M solution of the diprotic acid H2X. H2X H+(aq)+ HX(aq)Ka1=3.3 10 4 HX H+(aq)+...
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