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Module2_Homework_solutions - ENV 6666 Module 2...

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ENV 6666 Module 2 Homework (with solutions) Key equations needed: 𝑝𝐾 𝑊 = 14.00 , thus { 𝑂𝐻 } = 10 −𝑝𝐾 𝑊 { 𝐻 + } and assumption that I = 0 means we will approximate: { 𝐻𝐴 } ( 𝑇𝑂𝑇 A)( 𝛼 0 ) = ( 𝑇𝑂𝑇 A) { 𝐻 + } { 𝐻 + } +𝐾 𝑎1 and { 𝐴 } ( 𝑇𝑂𝑇 A)( 𝛼 1 ) = ( 𝑇𝑂𝑇 A) 𝐾 𝑎1 { 𝐻 + } +𝐾 𝑎1 Build spreadsheet like so, then plot: Note: “TOTA” 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.
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Charge balance may be expressed as negative charges = positive charges: 𝐶 . 𝐵 .: [ 𝐻 + ] = [ 𝑂𝐻 ] + [ 𝐴 ] Since we use the ‘dilute solution approximation, we use this instead: { 𝐻 + } { 𝑂𝐻 } + { 𝐴 } Consider the charge balance equation a “left-hand side” = “right-hand side” (i.e., LHS = RHS) expression, and add a column to spreadsheet for “LHS – RHS” (i.e., program the cells of a column for [ 𝐻 + ] [ 𝑂𝐻 ] [ 𝐴 ] although we approximate with the Activities { 𝐻 + } { 𝑂𝐻 } { 𝐴 } ) and look for the “zero-crossing”, i.e., when LHS – RHS = 0. Zero-crossing occurs at pH 5.5 as shown: The pH closest to the equilibrium pH would be the one whose “LHS – 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 } 3.87 𝐸 -6 { 𝐴 } 6.13 𝐸 -6 { 𝐻 + } 6.31 𝐸 -06 { 𝑂𝐻 } 1.58 𝐸 -09
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This is solved by the same method used in problem 1, but with one more column and more complicated “alpha” expressions: 𝛼 0 = { 𝐻 + } 2 { 𝐻 + } 2 + { 𝐻 + } 𝐾 𝑎1 + 𝐾 𝑎1 𝐾 𝑎2 𝛼 1 = { 𝐻 + } 𝐾
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