Question

The pOH of an aqueous solution of 0.481 M trimethylamine (a weak base with the formula (CH3)3N) is ?.

Answer 1

To find the pOH of an aqueous solution of 0.481 M trimethyl aminek(CH3), N) Given that Kof trimethylamine (CH), N) is 6.3x10-5 It will dissociates (CHỊ), N+H,» CHÀ NH*+OH- - [(CHỊ) NH][OH] ° [(CH, N] and if x moles of the initial concentration dissociates, then usingicetable Then (CH), N+H,0-(CH3), NH+ + OH- I 0.481 0 0 C -X +x +x E 0.481- x K7 = 6.3x10-4 5 - [(CH), NH*][OH-] |(CHỊ), N| 6.3x10^-0.481- x 6.3x10-4- 0.481- Since x is small we have 0.481- XN 0.481 Hence 6.3x10-4 6.3x10-0481 x2 = 6.3x10-*x0.481 x= 6.3x10-*x0.481 x = 5.50x10 PM Since x represents equilibrium concentration of [OH-Jions x= [OH-] = 5.50x10²M To find the pOH POH=-log[OH-] =-10g[5.50x102] = 2.26 Hence the pOH of an aqueous solution of 0.481 M trimethylamine (CH), N) = 2.26