Graphical Solution of Acid/Base Problems. Graphical Solution of Acid/Base Problems. Graphical Solution of Acid/Base Problems

Transcription

1 H 3.45 RHS Ac OH Example: Speciation of M H +0.3 M NaAc :,OH,H,HSO,SO 3,,Ac, Na + Equilibrium constants (a) K w =0 4.0 (b) K a,s = 0.86, K a,s = , K a,ac = (c) N/A (d) None Mass balances: TOT =0 4.0, TOTNa=TOTAc=0.3 Charge balance: [Na + ] +[ ] = [OH ]+[HSO ]+[SO 3 ]+[Ac ] Na + Ac H OH H RHS Ac OH

2 Modify CB to eliminate the big number on each side of the equation, by subtracting a value equal to that big number. [Na + ]+[ ] = [OH ]+[HSO ]+[ ]+[Ac ] ([Na + ] = [] + [Ac ]) [] + [ ] = [OH ]+[HSO ]+[SO 3 ] - - H Ac RHS * OH Summary: Graphical Solution of Acid/Base Problems Making Graphical Solution of Acid/Base Problems More Efficient Graphical analysis is a relatively fast, convenient way to estimate the speciation of acid/base systems. It is most useful when: The TOTc values are known, but the is not We wish to consider the speciation in several different systems, all with similar TOTc values How can we write a modified CB containing only nondominant species directly? Two steps:. Identify anticipated dominant species.. Determine how to write modified CB that doesn t contain those species (even if they have charge). The species in the CB are all non-dominant species

3 Predicting Dominant As the drops from pk a + to pk a, almost all of each H n A/H n A conjugate pair changes from deprotonated to protonated, thereby consuming TOTA available Stronger bases become protonated at higher than weaker bases All available, exchangeable ions are used to protonate available bases, from strongest (highest pk a ) to weakest. This process continues until no available, exchangeable remains. - - H Ac OH Example: Speciation of M H +0.3 M NaAc Input in terms of bases and exchangeable : SO 3 (pk a =7., pk a =.08) plus 0.3 (i.e., 5x0 3 )Ac (pk a =4.76) plus *0 4 exchangeable to convert all SO 3 to HSO ;0 4 remains to convert % of Ac to No exchangeable remains Conclusion: HSO 3 and Ac likely to be dominant M H +0.3 M NaAc pk a Conversion Available Main Ac Main -- Initial Condition.0x0 4 Ac x0 4 Ac 4.76 Ac 0.0 Ac Example: 4x0 4 Na CO NaHCO NaH PO 4 + 3*0 3 HCl Input in terms of bases and exchangeable : 5x0 4 CO 3 (pk a =0.33, pk a =6.35) plus 0 3 PO 4 (pk a3 =.35, pk a =7.0, pk a =.6) plus 5.*0 3 exchangeable.86 HSO H 3

4 4x0 4 Na CO NaHCO NaH PO 4 +3*0 3 HCl pk a Conversion Available Main CO 3 Main PO 4 -- Initial Condition 5.x0 3 CO 3 PO 4.35 PO 4 4.x0 3 CO CO 3 HCO 3 3.6x0 3 HCO H PO 4.6x0 3 HCO 3 H PO 4 Making Graphical Solution of Acid/Base Problems More Efficient How can we write a modified CB containing only nondominant species directly? Two steps:. Identify anticipated dominant species.. Determine how to write modified CB that doesn t contain those species (even if they have charge). Approach: Write a modified mass balance on the TOTH equation 6.35 HCO H CO 3.x0 3 H CO 3 H PO 4.6 H PO 4 H 3 PO 4.x0 3 H CO 3 H 3 PO 4 Components,, and the System Tableau Comp + Comp + Comp3 + Spec Comp Comp Comp3 3 Example: 3x0 4 H 3 PO 4 Reaction Written Conventionally Reaction Forming from Components (H O,, PO 4 ) H 3 PO 4 + H PO 4 K a 3 + PO 4 H 3 PO 4 (K a K a K a3 ) H PO 4 + K a + PO 4 H PO 4 (K a K a3 ) + PO 4 K a3 + PO 4 (K a3 ) PO 4 PO 4 K= PO 4 PO 4 K= K= K= Spec Spec Spec3 Spec4 H O + OH K w H O OH K w H O H O K= H O H O K= 4

5 log(h 3 PO 4 ) = log(h O) 0 + log( ) ) + log(k a K a K a3 ) log(h PO 4 ) = log(h O) 0 + log( ) + 4 ) + log(k a K a3 ) 4 ) = log(h O) 0 + log( ) + 4 ) + log(k a3 ) 4 ) = log(h O) 0 + log( ) ) + log() log( ) = log(h O) 0 + log( ) + 4 ) 0 + log() log(oh ) = log(h O) + log( ) + log(k w ) log(h O) = log(h O) + log( ) 0 + log() log(h 3 PO 4 ) = log(h O) 0 + log( ) ) + log(k a K a K a3 ) log(h PO 4 ) = log(h O) 0 + log( ) + 4 ) + log(k a K a3 ) 4 ) = log(h O) 0 + log( ) + 4 ) + log(k a3 ) 4 ) = log(h O) 0 + log( ) ) + log() log( ) = log(h O) 0 + log( ) + 4 ) 0 + log() log(oh ) = log(h O) + log( ) + 4 ) 0 + log(k w ) log(h O) = log(h O) + log( ) ) 0 + log() log(h 3 PO 4 ) = 0 log(h O) + 3 log( ) + 4 ) + log(k a K a K a3 ) log(h PO 4 ) = 0 log(h O) + log( ) + 4 ) + log(k a K a3 ) 4 ) = 0 log(h O) + log( ) + 4 ) + log(k a3 ) 4 ) = 0 log(h O) + 0 log( ) + 4 ) + log() log( ) = 0 log(h O) + log( ) ) + log() log(oh ) = log(h O) log( ) ) + log(k w ) log(h O) = log(h O) + 0 log( ) ) + log() Example: 3x0 4 H 3 PO 4 H O PO 4 log K H 3 PO log (K a K a K a3 ) H PO 4 0 log (K a K a3 ) 0 log (K a3 ) PO log () = log () = 0 OH 0 log K w H O 0 0 log () = 0 This table contains all the information in the equilibrium constant relationships. Next, add information in the mass balances. Equilibrium Tableau Example: 3x0 4 H 3 PO 4 H O PO 4 log K Conc n H 3 PO [H 3 PO 4 ] eq H PO [H PO 4 ] eq HPO [HPO 4 ] eq PO [PO 4 ] eq [ ] eq OH [OH ] eq H O [H O] eq TOTH O eq TOTH eq TOTPO 4, eq TOTHO [ HO] = + OH eq eq eq TOTH [ ] = 3 H PO + H PO + HPO + H OH + eq 3 4 eq 4 eq 4 eq eq eq TOTPO 4 3 [ HPO] HPO HPO PO = + + +, eq 3 4 eq 4 eq 4 eq 4 eq 5