HW3_12 - Homework due 3/12 Calculate the pH of a solution...

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Homework due 3/12 Calculate the pH of a solution that is 0.25 M pyridine, C 5 H 5 N and 0.75 M in pyridinium bromide, C 5 H 5 NHBr. (K b for C 5 H 5 N = 1.7x10 -9 ) This is an example of a solution containing a weak base, C 5 H 5 N, and the salt of its conjugate acid C 5 H 5 NHBr, which is a buffer. The salt is 100 % ionized to C 5 H 5 H 1+ and Br 1- . The Br 1- ion is a spectator ion and can be ignored. There are 4 approaches to this problem. Two using the acid equilibrium and 2 using the base equilibrium. Starting with the base equilibrium approach. The equilibria of interest is thus C 5 H 5 N + H 2 O C 5 N 5 NH 1+ + OH 1- . Using the Henderson-Hasselbalch equation pOH = pK b + log ([C 5 H 5 NH 1+ ] / [C 5 H 5 N]) pOH = - log 1.7x10 -9 + log (0.75/0.25) = 8.77 + (0.48) = 9.25 pH = 14 – pOH = 14 – 9.25 = 4.75 Alternatively, using the x’s approach [C 5 H 5 N] [C 5 H 5 NH 1+ ] [OH 1- ] initial 0.25 M 0.75 M 0 (from C 5 H 5 NHBr) change -x +x +x let x = [OH 1- ] formed equil. 0.25 – x
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This note was uploaded on 04/03/2008 for the course CHEM 114 taught by Professor Martin during the Spring '08 term at Moravian.

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HW3_12 - Homework due 3/12 Calculate the pH of a solution...

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