Question

- 7.4 x 10 What A chemistry graduate student is given 125. ml. of a 0.50 M trimethylamine (CH N ) solution. Trimethylamine is a weak base with K mass of (CH), NHCI should the student dissolve in the (CH), N solution to turn it into a buffer with pH = 11.05? You may assume that the volume of the solution doesn't change when the (CH), NHCl is dissolved in it. Be sure your answer has a unit symbol, and round it to 2 significant digits.

Answer 1

Given Base = (CH4), N salt = (CH), NHCl (CH), NHCI = molay mass = 95.59/mol Basic Buffes I number of moles of salt pok = pky +log/ ( number of moles of base - For from 1 data Kb = 7.4810lt: Base lasmi . O.SM . concentration, Number of moles of Base: na Molarity x volcevre (mi) 1000 5 na 6.5) (125) 103 - 0.0625 mol we know that pko = -log Ke So pho = -log (7.4X164) = -109(714) - log (104) = - 0.869 - (-4) log 10 = -0.869 +4 CD PK6 = 3.13

Buffer pH is given 11.05 we know that pH + pol - 14 » pol = 14-pH =14-11.05 pol = 2.95 substitute all the values in a 2.95 = 3.13 +log ( her sal humber of moles of salt di 0625 mol a log / salt C :0.06a's mol = 2.95 -13113 nsalt 0.0625 mol - antilog(-0.18) = 10-0.18 = 0.66 Msalt = 0.0625 moi) (0.66) = 0.04125 mol Number of moles in) a mass molay mass mass = (salt) (molar mass) = (0104125 mol) 95.59 = 3.946 Answer: 3.9 g mass of (EM) NHCl = 3.99 mol