Bases: Sodium Acetate Example 10 -3 M in 1 L 1. List all species - - PDF document

SLIDE 1

CEE 680 Lecture #10 2/5/2020 1

Lecture #10 Acids & Bases: Analytical Solutions with simplifying assumptions IV

(Stumm & Morgan, Chapt.3 )

David Reckhow CEE 680 #10 1

Updated: 5 February 2020

(Benjamin, Chapt. 4)

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Bases: Sodium Acetate Example

 1. List all species present

 H+, OH‐, HAc, Ac‐, Na+

 2. List all independent equations

 equilibria

 Ka = [H+][Ac‐]/[HAc] = 10‐4.77  Kw = [H+][OH‐] = 10‐14

 mass balances

 C = [HAc]+[Ac‐] = 10‐3

 proton balance: (proton rich species) = (proton poor species)

 [HAc] + [H+] = [OH‐]

David Reckhow CEE 680 #8 2

1 2 3 4 Five total

H2O Ac- CNa = [Na+] = 10-3

5 10-3 M in 1 L

SLIDE 2

CEE 680 Lecture #10 2/5/2020 2

NaAc Example (cont.)

 3. Combine equations and solve for H+

 [H+] = [OH‐] ‐ [HAc]

 [H+] = KW/ [H+] ‐ [HAc]  [H+] = KW/ [H+] ‐ [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+] ‐ KWKa = 0

 4. Solve for other species

David Reckhow CEE 680 #8 3

4 2+4

Kw = [H+][OH-] [OH-] = Kw/[H+]

2

C = [HAc]+[Ac-] [Ac-] = C-[HAc]

3 1 Ka = [H+][Ac-]/[HAc]

Ka = [H+]{C-[HAc]} /[HAc] Ka[HAc]= [H+]C-[H+][HAc] [HAc] = C[H+]/{Ka+[H+]}

1+3 1+2+3+4

CNa = [Na+] = 10-3

5

Answer

 [OH‐] = 7.67 10‐7  pOH = 6.115  pH = 7.885

David Reckhow CEE 680 #8 4

SLIDE 3

CEE 680 Lecture #10 2/5/2020 3

In‐class Practice

10‐4M Sodium Acetate

 UMass: Alvin & Cielo, Chris  UNISA: Sikelelwa

10‐3M Sodium Cyanide

 UMass: Ian &JQ  UNISA:

10‐3M Calcium Oxalate

 UMass:  UNISA: Alfred

10‐4M Sodium Bicarbonate



UMass: Laura, Bridgette, Isaac  UNISA:

David Reckhow CEE 680 #9 5

For Oxalic acid, H2OxHOx-Ox-2 pKa1=1.25 pKa2=4.27

Writing PBE’s

 Monoprotic Acid

 same as ENE

 Diprotic Acid

 same as ENE

 Diprotic: Ampholyte

 Not ENE

 e.g., NaHCO3

 Diprotic Base

 Not ENE



[H+] = [OH‐] + [Ac‐]



[H+] = [OH‐] + [HCO3

‐] + 2[CO3 ‐2]  [H2CO3] + [H+] = [OH‐] + [CO3 ‐2]  2[H2CO3] + [HCO3 ‐] + [H+] = [OH‐] David Reckhow CEE 680 #10 6

H2O HAc H2O H2CO3 H2O HCO3

• H2O

CO3

• 2

Note: Alkali metals such as Na and K do not act as acids, so they are ignored here

SLIDE 4

CEE 680 Lecture #10 2/5/2020 4

Guide to Simplified Acid/Base Solutions #1

 Neutral

 if C<10‐8

 Acid Addition, C>10‐6.5

 Acidic

 if KaC > 10‐13

 Strong Acid

 if Ka > 10C

 Weak Acid

 if C > 100Ka

David Reckhow CEE 680 #10 7

2 4 ] [

2 w

K C C H   



w a

K C K H  

]

[

2 4 ] [

2

C K K K H

a a a

   



C H 

]

[ C K H

a



]

[

w

K H 

]

[

Guide to Simplified Acid/Base Solutions #2

Base Addition, C>10‐6.5

 Basic

 if KbC > 10‐13

 Strong Base

 if Kb > 10C

 Weak Base

 if C > 100Kb

Very Dilute Systems

 if 10‐8 < C < 10‐6.5

 try strong acid/base or weak acid/base assumption  otherwise may need to use general solution

David Reckhow CEE 680 #10 8

2 4 ] [

2 w

K C C OH   



w b

K C K OH  

]

[

2 4 ] [

2

C K K K OH

b b b

   



C OH 

]

[ C K OH

b



]

[

SLIDE 5

CEE 680 Lecture #10 2/5/2020 5

Exact Solutions: Summary

 Monoprotic

 Acids: [H+]3 + Ka[H+]2 ‐ {Kw + KaC}[H+] ‐ KWKa = 0  Bases: [H+]3 + {C+Ka}[H+]2 ‐ Kw[H+] ‐ KWKa = 0

 Diprotic

 Acids:



[H+]4 + K1[H+]3 + {K1K2‐Kw‐ K1C}[H+]2 ‐ K1{2CK2+Kw}[H+] ‐ KWK1K2 = 0

 Ampholytes:



[H+]4 + {C+K1}[H+]3 + {K1K2‐Kw}[H+]2 ‐ K1{CK2+Kw}[H+] ‐ KWK1K2 = 0

 Bases:



[H+]4 + {2C+K1}[H+]3 + {CK1+K1K2‐Kw}[H+]2 ‐ K1Kw[H+] ‐ KWK1K2 = 0

David Reckhow CEE 680 #10 9

To next lecture

David Reckhow CEE 680 #10 10