Quiz-3-07F-Solutions

# Quiz-3-07F-Solutions - I have neither given nor received...

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.CHEM 119 H Quiz 3 12/04/07 USE PENCIL & ERASE CLEANLY! I have neither given nor received aid while t a k i n g t h i s q u i z . H 3 PO 4 : pK A1 = 2.148 pK A2 = 7.198 pK A3 = 12.375 1. (35) Both KH 2 PO 4 {136.09 g/mol} and K 2 HPO 4 {174.18 g/mol} are readily available (from Fisher) in high purity, > 99.0 %. a. (10) Calculate the pH of a solution containing 1.70 g of KH 2 PO 4 in 100.0 mL. pH = 4.67 (4.69) The simplest solution to the problem is the standard equation: pH(ampholyte) = (pK Ai + pK Aj )/2. The following reactions occur: H 2 PO 4 - + H 2 O HPO 4 2- + H 3 O + and H 2 PO 4 - + H 2 O H 3 PO 4 + HO - The reactions involve K A1 and K A2 ; so pH = (2.148 + 7.198)/2 = 4.673 – independent of concentration. You may use the extended equation if you wish: 24 1 mol KH PO (1.70 g KH PO ) 136.09 g F{KH PO }= = 0.125 M .100 L () 2.148 7.198 2.148 14 + 5 12 1W 3 2.148 1 10 *10 *.125 10 *10 K*K*F + K*K H O = 2.066*10 F + K .125 10 pH 4.685 M −− + ⎡⎤ == ⎣⎦ + = b. (5) Assume that you have each of the following pure reagents, which would you use to prepare a buffer at pH = 2.00 and briefly, why? H 3 PO 4 KH 2 PO 4 K 2 HPO 4 K 3 PO 4 At pH = 2.148, [H 3 PO 4 ] = [H 2 PO 4 - ] and these are the two dominant species in solution. To decrease pH to 2.00, [H 3 PO 4 ] > [H 2 PO 4 - ] The solution must contain H 3 PO 4 and KH 2 PO 4 .

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## This note was uploaded on 04/13/2008 for the course CHEM 119H taught by Professor Munson during the Fall '07 term at University of Delaware.

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Quiz-3-07F-Solutions - I have neither given nor received...

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