CEE 680 Lecture #10
9/12/2013
Print version
Updated: 12 September 2013
Lecture #10
Acids & Bases: Analytical Solutions with simplifying assumptions III
(Stumm & Morgan, Chapt.3 )
(Benjamin, Chapt. 3)
David Reckhow
CEE 680 #10
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Hydrochloric Acid Example
 1. List all species present

H+, OH‐, HCl, Cl‐
Four total
 2. List all independent equations
 equilibria
 Ka = [H+][Cl‐]/[HCl] = 10+3
 Kw = [H+][OH‐] = 10‐14
2
 mass balances
 C = [HCl]+[Cl‐] = 10‐3
1
3
 proton balance: (proton rich species) = (proton poor species)
HCl
H2O

David Reckhow
[H+] = [OH‐] + [Cl‐]
CEE 680 #10
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CEE 680 Lecture #10
9/12/2013
HCl: Exact Solution
[H+]3 + Ka[H+]2 - {Kw + KaC}[H+] - KWKa = 0
 Exact solution: pH = 3.0000004
 [H+] = 1.00 x 10‐3
[OH-] = Kw/[H+]
 [OH‐] = 1.00 x 10‐11
[Cl-]=KaC/{Ka+[H+]}
 [Cl‐] = 1.00 x 10‐3
[HCl] = C-[Cl-]
 [HCl] = 1.00 x 10‐11
David Reckhow
CEE 680 #10
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HCl example (cont.)
 Can we simplify?
[H+]3
1.000E-9
+ Ka[H+]2 - Kw[H+] -KaC[H+] - KWKa
0.001000 1.000E-17
 What about the PBE?
 [H+] = [OH‐] + [Ac‐]
0.001000 1.000E-11
0
 And the MBE too?

~0
David Reckhow
=0
CEE 680 #10
C = [HCl] + [Cl‐]
~0
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CEE 680 Lecture #10
9/12/2013
Simplified HCl Example
K = [H+][OH-]
w
2 [OH
-] = K /[H+]
w
 3. Use simplified PBE & MBE
4
 [H+] = [OH‐] + [Cl‐]
 [H+]  [Cl‐]
Assumes [H+]>>[OH-]
3 C [HCl]+[Cl-]
C  [Cl-]
[Cl-]  C
3+4
 [H+] = C
Assumes [HCl]<<[Cl-]
 [H+] = C
1 Ka = [H+][Cl-]/[HCl]
 4. Solve for other species
1+3
David Reckhow
Ka = [H+] C / [HCl]
[HCl] = [H+] C /Ka
CEE 680 #10
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Assumptions
 Use both & Compare answers
 Exact:
 Simplified:
pH = 3.0000004 pH = 3.0000000  Use simplified equation, and check assumptions!
 [OH‐] << [H+]
 1.00 x 10‐11 << 1.00 x 10‐3 yes!
 [Cl‐] >> [HCl]
 1.00 x 10‐3 >> 1.00 x 10‐11 yes!
David Reckhow
CEE 680 #10
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CEE 680 Lecture #10
9/12/2013
Simplified HCl Example for low C
 3. Use simplified PBE & MBE
2+3+4
4
K = [H+][OH-]
w
2 [OH
-] = K /[H+]
w
 [H+] = [OH‐] + [Cl‐]
3 C= [HCl]+[Cl-]
C  [Cl-]
[Cl-]  C
 [H+] = KW/ [H+] + [Cl‐]
 [H+] = KW/ [H+] + C
 [H+]2 ‐ C[H+] ‐ Kw = 0
 [H+] = {C(C2 + 4Kw)0.5}/2
Assumes [HCl]<<[Cl-]
1 Ka = [H+][Cl-]/[HCl]
 4. Solve for other species
1+3
David Reckhow
Ka = [H+] C / [HCl]
[HCl] = [H+] C /Ka
CEE 680 #10
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Calculation for 2nd HCl example
 For a 10‐7 solution of HCl
 pH = 6.79
[H  ] 
 Check Assumptions
 [Cl‐]  C = 10‐7
 [HCl] = [H+] C /Ka


David Reckhow
=10‐6.7910‐7/10+3=10‐16.79
[Cl‐]>>[HCl], yes!
CEE 680 #10
C  C 2  4K w
2
10 7  10 14  4 x10 14
2
1  5 7

10
2
 1.62 x10 7

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CEE 680 Lecture #10
9/12/2013
Hypochlorous Acid Example
 1. List all species present

H+, OH‐, HOCl, OCl‐ Four total
 2. List all independent equations
 equilibria
 Ka = [H+][OCl‐]/[HOCl] = 10‐7.6
 Kw = [H+][OH‐] = 10‐14
2
 mass balances
 C = [HOCl]+[OCl‐] = 10‐6
1
3
 proton balance: (proton rich species) = (proton poor species)
HOCl
H2O

[H+] = [OH‐] + [OCl‐] 4
David Reckhow
CEE 680 #10
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HOCl Example for low C

3. Combine equations and solve for H+
4
1+2+3+4
2+4
K = [H+][OH-]
w
2 [OH
-] = K /[H+]
w
 [H+] = [OH‐] + [OCl‐]
 [H+] = KW/ [H+] + [OCl‐]
 [H+] = KW/ [H+] + KaC/[H+]

[H+]2 = KW + KaC
 [H+] = (KaC + KW)0.5
3 C = [HOCl]+[OCl-]
[HOCl]  C
Assumes [HOCl]>>[OCl-]
1 Ka = [H+][OCl-]/[HOCl]
Ka = [H+][OCl-]/ C
 4. Solve for other species
1+3
[OCl-]=KaC/[H+]
David Reckhow
CEE 680 #10
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CEE 680 Lecture #10
9/12/2013
Calculation for HOCl example
 For a 10‐6 solution of HOCl
[ H  ]  K aC  K w
 pH = 6.73
 10 7.610 6  10 14
 3.51x10 14
 Check Assumptions
 [HOCl]  C = 10‐6
 [OCl‐]=KaC/[H+]

 1.87 x10 7
=10‐7.610‐6/10‐6.73=10‐6.87
[HOC]>>[OCl‐], OK

David Reckhow
CEE 680 #10
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Bases: Sodium Acetate Example
 1. List all species present

H+, OH‐, HAc, Ac‐, Na+
10-3 M in 1 L
Five total
 2. List all independent equations
 equilibria
 Ka = [H+][Ac‐]/[HAc] = 10‐4.77 1
 Kw = [H+][OH‐] = 10‐14
2
 mass balances
 C = [HAc]+[Ac‐] = 10‐3
3
CNa = [Na+] = 10-3
5
 proton balance: (proton rich species) = (proton poor species)
AcH2O

David Reckhow
[HAc] + [H+] = [OH‐] 4
CEE 680 #10
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CEE 680 Lecture #10
9/12/2013
CNa = [Na+] = 10-3
NaAc Example (cont.)

1+2+3+4
2+4
K = [H+][OH-]
3. Combine equations and solve for H+
4
5
w
2 [OH
-] = K /[H+]
w
 [H+] = [OH‐] ‐ [HAc]
 [H+] = KW/ [H+] ‐
 [H+] = KW/ [H+] ‐



[HAc]
[H+]C/{Ka+[H+]}
[H+]2 = KW ‐ C[H+]2/{Ka+[H+]}
Ka[H+]2 + [H+]3 = KWKa + Kw[H+] ‐ C[H+]2
[H+]3
+ {C+Ka}[H+]2
‐ Kw
[H+] ‐
3 C = [HAc]+[Ac-]
[Ac-] = C-[HAc]
KWKa = 0
1 Ka = [H+][Ac-]/[HAc]
 4. Solve for other species
1+3
David Reckhow
CEE 680 #10
Ka = [H+]{C-[HAc]} /[HAc]
Ka[HAc]= [H+]C-[H+][HAc]
[HAc] = C[H+]/{Ka+[H+]}
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Answer
 [OH‐] = 7.67 10‐7
 pOH = 6.115
 pH = 7.885
David Reckhow
CEE 680 #10
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CEE 680 Lecture #10
9/12/2013
 To next lecture
DAR
David Reckhow
CEE 680 #10
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