ecprob1solution

# ecprob1solution - Extra Credit Problem 1: Calculation of...

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Extra Credit Problem 1: Calculation of the distribution of different forms of phosphate in a 0.01 M solution at pH 7.0. Use the values 2.2, 7.2 and 12.4 for pK 1 , pK 2 and pK 3 respectively. pH p K 7. 0 2. 2 4. 84 24 34 4 2 4 2 p H p K 7. 0 7. 2 0.2 4 2 4 3 4 Youhavethefollowingfourequationsinfourunknowns [H P O] 1 . 1 0 1 0 1 0 6.31x10 [H PO] [H P [H PO]x6.31x10 [HP 2 . 1 0 1 0 1 0 0.631 [H P [HP [H P O ]x0.631 [PO 3. - -- - - - - == = = p K 7. 0 12. 4 5. 46 2 4 3 26 44 23 3 4 2 4 4 2 4 ] 1 0 1 0 1 0 3.98x10 [HP [P [HP O ]x3.98x10 4 . [H PO ] [H P [HP 0.01M Substitute1for[H P O ]in2: 2 . [HP [H PO]x6.31x1 0 x0.631 Substitute2for[HP O ]in3: - - - - - - - - - = ++ += = 3 4 4 4 [H PO]x 6.31x1 0 x0.631x3.98x10 Substitute1,2,and3into4toget4asafunction of[HPO] 4 . [H PO] [H PO ]x6.31x10 0 x0.631 0 x0.631x3.98x1 0 Factor: [H PO ]( 1 6.31x1 - = + + + 4 4 5 5 8 8 43 9.71654x1 0M 0 0 x0.63 1 0) [H PO ](1.02917x10 ) 1.02917x10 Thenplugthisvalueintoequations1,2,and3: 0 Mx 6.31x1 0 6.1311 0. 1 00 x0 - - - + = = 2 8 4 3 4 8 8 6 [HP 0 Mx6.31x1 0 x0.63 1 3.86875x10 Check theanswersbypluggingintoequat 61311M 0.00386875M 1.53976x1
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## This note was uploaded on 11/27/2011 for the course BSC 5936 taught by Professor Staff during the Spring '08 term at FSU.

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