Module2_Homework_solutions

Module2_Homework_sol - ENV 6666 Module 2 Homework(with solutions Key equations needed 𝑝𝐾 π‘Š = 14.00 thus π‘‚π βˆ’ = 10 βˆ’Β‘Β’ Β£ π

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Unformatted text preview: ENV 6666 Module 2 Homework (with solutions) Key equations needed: 𝑝𝐾 π‘Š = 14.00 , thus { 𝑂𝐻 βˆ’ } = 10 βˆ’Β‘Β’ Β£ { 𝐻 + } and assumption that I = 0 means we will approximate: { 𝐻𝐴 } β‰ˆ ( 𝑇𝑂𝑇 A)( 𝛼 ) = ( 𝑇𝑂𝑇 A) { Β€ Β₯ } { Β€ Β₯ } +Β’ π‘Ž1 and { 𝐴 βˆ’ } β‰ˆ ( 𝑇𝑂𝑇 A)( 𝛼 Β¦ ) = ( 𝑇𝑂𝑇 A) Β’ π‘Ž1 { Β€ Β₯ } +Β’ π‘Ž1 Build spreadsheet like so, then plot: Note: β€œTOTA” column simply adds {HA}+{A-}. It is a good practice to include this column because it can alert you to a programming error, since you know it should always be 1.00E-05. Charge balance may be expressed as negative charges = positive charges: 𝐢 . 𝐡 .: [ 𝐻 + ] = [ 𝑂𝐻 βˆ’ ] + [ 𝐴 βˆ’ ] Since we use the β€˜dilute solution approximation, we use this instead: { 𝐻 + } β‰ˆ { 𝑂𝐻 βˆ’ } + { 𝐴 βˆ’ } Consider the charge balance equation a β€œleft-hand side” = β€œright-hand side” (i.e., LHS = RHS) expression, and add a column to spreadsheet for β€œLHS – RHS” (i.e., program the cells of a column for [ 𝐻 + ] Β‘ [ 𝑂𝐻 βˆ’ ] Β‘ [ 𝐴 βˆ’ ] although we approximate with the Activities { 𝐻 + } Β‘ { 𝑂𝐻 βˆ’ } Β‘ { 𝐴 βˆ’ } ) and look for the β€œzero-crossing”, i.e., when LHS – RHS = 0. Zero-crossing occurs at pH β‰ˆ 5.5 as shown: The pH closest to the equilibrium pH would be the one whose β€œLHS – RHS” is closest to zero (-4.44E-06 is slightly closer to zero than is +5.00E-06). The approximate equilibrium pH is 5.5, but we could insert more spreadsheet rows between pH = 5.0 and pH = 5.5 to get an answer more precise than the nearest 0.5 on the pH scale. This is easy to do, with the result: The value closest to zero is 1.77E-07, so the equilibrium pH = 5.2 (to the nearest 0.1 on pH scale). The values can be read from the pH = 5.2 row. They are: { 𝐻𝐴 } β‰ˆ 3.87 𝐸-6 { 𝐴 βˆ’ } β‰ˆ 6.13 𝐸-6 { 𝐻 + } β‰ˆ 6.31 𝐸-06 { 𝑂𝐻 βˆ’ } β‰ˆ 1.58 𝐸-09 This is solved by the same method used in problem 1, but with one more column and more complicated β€œalpha” expressions: 𝛼 = { 𝐻 + } 2 { 𝐻 + } 2 + { 𝐻 + } 𝐾 π‘Ž1 + 𝐾 π‘Ž1 𝐾 π‘Ž2 𝛼 1 = { 𝐻 + } 𝐾 π‘Ž1 { 𝐻 +...
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This note was uploaded on 03/06/2011 for the course ENV 6666 taught by Professor Fuss during the Spring '11 term at Uni San Francisco de Quito.

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Module2_Homework_sol - ENV 6666 Module 2 Homework(with solutions Key equations needed 𝑝𝐾 π‘Š = 14.00 thus π‘‚π βˆ’ = 10 βˆ’Β‘Β’ Β£ π

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