HW3_17

# HW3_17 - 1 and NH 3 have reacted The resulting solution is...

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Homework due 3/17 Calculate the pH of a solution prepared by mixing 20 mL of 0.20 M NH 3 with 40 mL of 0.10 M HNO 3 . (K b for NH 3 = 1.8x10 -5 ) This is a typical titration type problem in which the strong acid HNO 3 is being added to the weak base NH 3 . To solve this problem one must first account for the acid-base reaction between HNO 3 and NH 3 . Then, based on what is present after the first reaction, the equilibrium present is determined and used to determine the pH of the solution. HNO 3 -----> H 1+ + NO 3 1- (Note: NO £ 1- is a spectator ion) NH 3 + H 1+ -------> NH 4 1+ (this acid base reaction goes to completion. moles NH 3 moles H 1+ moles NH 4 1+ initial (0.02 L)(0.2 M) (0.04 L)(0.1 M) 0 (from HNO 3 ) 0.004 moles 0.004 moles 0 moles change -0.004 moles -0.004 moles +0.004 moles after rxn 0 0 0.004 moles The situation after the reaction in this case represents the endpoint since all of the H
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Unformatted text preview: 1+ and NH 3 have reacted. The resulting solution is a solution of NH 4 1+ ion. To get the pH of this solution one must use the ionization of the weak acid NH 4 1+ . [NH 4 1+ ] = 0.004 moles / 0.60 L = 0.067 M (where 0.60 L is the total of the volumes of the two solutions added.) NH 4 1+ NH 3 + H 1+ K a = K w /K b = [H 1+ ][NH 3 ]/[NH 4 1+ ] [NH 4 1+ ] [H 1+ ] [NH 3 ] initial 0.067 M change- x x x equil. 0.067 - x x x where x = [H 1+ ] produced K w 1.0x10-14 [H 1+ ][NH 3 ] (x)(x) K a = ------ = ------------ = ------------------- = ---------------- K b 1.8x10-5 [NH 4 1+ ] 0.067 -x ignoring the subtracted x x = [H 1+ ] = 6.12x10-6 pH = 5.21...
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