Handout6-PolyproticACID-BASE

Handout6-PolyproticACID-BASE - CEM 834 ACID–BASE...

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CEM 834 ACID–BASE EQUILIBRIA FALL 2005 II. Polyprotic Acid–Base Systems A. Simultaneous Equation Approach To illustrate the approach for the simultaneous solution of the stoichiometric equations for polyprotic acid–base systems, we will calculate the pH and concentration of all species in a 0.100 M KH PO 2 4 solution. In order to determine the six species H 3 PO 4 , H 2 PO 4 ‾, HPO 4 2 ‾, PO 4 3 ‾, H + , and OH‾, we must have six independent equations that are obtained in the following manner: 1. Write down all equilibria and formation constants H PO HPO K 2.11x10 4 4 a1 f 12 + - - + = 3 2 H HPO H PO K 1.61x10 4 2 4 a2 f 7 + - - + = 2 H H PO H PO K 1.45x10 2 4 3 4 a3 f 2 + - + = H OH H O K x10 2 w f 14 + - + = 100 . Also calculate the product of the formation constants, where β an f ai f i 1 n K = = For the phosphoric acid system, β a1 f a1 f 12 K 2.11 x10 = = β a2 f a1 f a2 f 19 K K 3.40 x10 = = β a3 f a1 f a2 f a3 f 21 K K K 4.93 x10 = = 2. Mass balance equation ] PO [H ] PO [H ] [HPO ] [PO M 0.100 C 4 3 4 2 2 4 3 4 T + + + = = - - - 3. Proton balance equation for H PO 2 4 - as the starting material [H ] H PO ] [HPO ] 2[PO ] [OH ] 3 4 4 2 4 + - + = + + - [ 3-
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2 4. Simplifying assumptions With H PO 2 4 - as the starting material , we have the following possibilities a. The solution is acidic because of the reaction H PO H HPO K 1 K 6.21x10 2 4 4 eq a2 f 8 - + - - + = = 2 b. The solution is basic because of the reaction
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Handout6-PolyproticACID-BASE - CEM 834 ACID–BASE...

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