CEE 680
Spring 2016
Homework #3
1. Acid/Base Equilibria I (2 POINTS)
Determine the complete solution composition of the following systems (all are added to 1
liter of pure water), based on the governing equations. Make one or more
simplifying assumptions, then solve the system of equations based on those
assumptions, and check the assumptions.
a. 5x10-3 moles HOCl
Approach
• addition of a simple monoprotic acid to water
• try to use simplified solutions developed in class
PBE
[H+] = [OH-] + [OCl-]
Assumptions:
1.
2.
Acidic Solution: [H+] >> [OH-]
Weak Acid: [HOCl] >> [OCl-]
Solve for [H+] based on simplifying assumptions
[ H + ] = k a CT
pH = 0.5(pka + pCT)
pH = 0.5(7.6 + 2.3)
pH = 4.95
Solve for all species
[H+]
[OH-]
= kw/[H+]
[HOCl] = CT
[OCl-]
= ka[HOCl]/[H+]
= 10-4.95
= 10-9.05
= 10-2.30
= 10-4.95
or 1.1x10-5
or 8.9x10-10
or 5x10-3
or 1.1x10-5
Check Assumptions
1.
2.
Acidic Solution: [H+] >> [OH-]; 10-4.95 >> 10-9.05 , OK
Weak Acid: [HOCl] >> [OCl-]; 10-2.30 >> 10-4.95 , OK
1
b. 5x10-3 moles NaCN
Approach
• addition of a simple monoprotic base to water
• try to use simplified solutions developed in class
PBE
[HCN] + [H+] = [OH-]
Assumptions:
1.
2.
Basic Solution: [OH-] >> [H+]
Weak Base: [CN-] >> [HCN]
Solve for [H+] based on simplifying assumptions
[ OH − ] = k b CT
pOH = 0.5(pkb + pCT)
pOH = 0.5(4.8 + 2.3)
pOH = 3.55
pH = 14-pOH = 10.45
Solve for all species
[H+]
[OH-]
= kw/[H+]
[CN-]
= CT
[Na+]
= CT
[HCN]
= kb[CN-]/[OH-]
= 10-10.45
= 10-3.55
= 10-2.30
= 10-2.30
= 10-3.55
or 3.6x10-11
or 2.8x10-4
or 5x10-3
or 5x10-3
or 2.8x10-4
Check Assumptions
1.
2.
Basic Solution: [OH-] >> [H+]; 10-3.55 >> 10-10.45 , OK
Weak Base: [CN-] >> [HCN]; 10-2.30 >> 10-3.55 , OK
c. 5x10-8 moles HCl
Approach
• addition of a simple monoprotic acid to water
• try to use simplified solutions developed in class
Assumptions:
1.
2.
Acidic Solution: [H+] >> [OH-]
Strong Acid: [Cl-] >> [HCl]
PBE
[H+] = [OH-] + [Cl-]
2
Solve for [H+] based on simplifying assumptions
[H+] = CT
pH = pCT
pH = 7.3
Solve for all species
[H+]
[OH-]
= kw/[H+]
[HCl]
= [H+][Cl-]/ka
[Cl-]
= CT
= 10-7.3
= 10-6.7
= 10-17.6
= 10-7.3
or 5x10-8
or 2x10-7
or 2.5x10-18
or 5x10-8
Check Assumptions
1.
2.
Acidic Solution: [H+] >> [OH-]; 10-7.3 >> 10-6.7 , No!
Strong Acid: [Cl-] >> [HCl]; 10-7.3 >> 10-17.6 , OK
Try again, but make only a strong acid assumption:
Assumptions:
1.
Strong Acid: [Cl-] >> [HCl]
Solve for [H+] based on simplifying assumptions
CT + CT2 + 4 Kw
[H ] =
2
[H+] = 10-7.3 + (10-14.6 + 4x10-14)0.5
[H+] = 1.28x10-7
pH = 6.89
+
Solve for all species
[H+]
[OH-]
= kw/[H+]
[HCl]
= [H+][Cl-]/ka
[Cl-]
= CT
= 10-6.89
= 10-7.11
= 10-17.2
= 10-7.3
or 1.28x10-7
or 7.80x10-8
or 6.5x10-18
or 5x10-8
Check Assumptions
1.
Strong Acid: [Cl-] >> [HCl]; 10-7.3 >> 10-17.2 , OK
d. 1x10-3 moles HAc
+
5x10-4 moles NaAc
Approach
• addition of a simple monoprotic acid and its conjugate base to water
• cannot use simplified solutions developed in class
3
•
return to original equations and solve, simplifying wherever possible
Species
H+, OH-, HAc, Ac-, Na+
Equations
Kw = 10-14 = [H+][OH-]
Ka = 10-4.7 = [Ac-][H+]/[HAc]
CT1= 1.5x10-3 = [Ac-] + [HAc]
CT2 = 5x10-4 = [Na+]
[H+] + [Na+] = [OH-] + [Ac-]
Assumptions
addition of both an acid and a base keeps the pH from dropping or rising very much,
this means that neither the hydrogen ion, nor the hydroxide ion will be significant
compared to the added salts (e.g., sodium, acetate).
1.
2.
[Na+] >> [H+]
[Ac-] >> [OH-]
Now Solve using the basic equations
Start with the ENE (cannot use PBE for mixtures of acids and conjugate bases)
[H+] + [Na+] = [OH-] + [Ac-]
[Na+] = [Ac-]
And combining with the mass balance equations
[Na+] = [Ac-] = CT2 = 5x10-4
[HAc] = CT1 - [Ac-] = CT1 - CT2 = 1x10-3
Now use the equilibria to get the remaining species
Ka = 10-4.7 = [Ac-][H+]/[HAc]
[H+] = 10-4.7 [HAc]/[Ac-]
[H+] = 10-4.7 1x10-3/5x10-4
[H+] = 4x10-5 = 10-4.4
[OH-] = kw/[H+] = 10-9.6
2. Acid/Base Equilibria II: graphical method (2 POINTS)
Solve the following problems (A. and B.) graphically. Later in question #4, I will ask
you to solve them exactly using MINEQL. Show the graphs and circle your solution
point. Then present the approximate concentrations in a table.
4
A). Construct a log C vs pH diagram for a 0.10 F phosphate (H3PO4, H2PO4-, HPO4-2,
PO4-3) system. Using it, calculate the pH and the concentration of all species in the
following solutions:
i) 0.10 F NaH2PO4
ii) 0.10 F Na2HPO4
iii) 0.10 F Na3PO4
Approach
• prepare Log C vs pH diagram
• write PBE for each solution
• locate pHs for each solution
• read off concentrations for each species
Log C vs pH Diagram
• pKs are 2.1, 7.2, and 12.35
• Log CT is -1
5
Log C vs pH Diagram
0
H2PO4-
H3PO4
-1
HPO4-2
PO4-3
PBE Solutions
iii
-2
-3
-
H+
OH
i and ii
-4
-5
Log C
-6
-7
-8
-9
-10
-11
-12
-13
-14
0
1
2
3
4
5
6
7
8
pH
6
9
10
11
12
13
14
i) 0.10 F NaH2PO4
PBE
[HPO4-2] + 2[PO4-3] + [OH-] = [H+] + [H3PO4]
which reduces to:
[HPO4-2] = [H3PO4]
Solution Composition
Species
Graph
C
H+
OHH3PO4
H2PO4HPO4-2
PO4-3
Na+
2.2e-5
4.5e-10
2.8e-4
1e-1
2.8e-4
8e-12
1e-1
pC
4.65
9.35
3.55
1.0
3.55
11.1
1
ii) 0.10 F Na2HPO4
PBE
[PO4-3] + [OH-] = [H+] + 2[H3PO4] + [H2PO4-]
which reduces to:
[PO4-3] = [H2PO4-]
Solution Composition
Species
Graph
C
H+
OHH3PO4
H2PO4HPO4-2
PO4-3
Na+
1.7e-10
5.6e-5
8e-12
2.8e-4
1e-1
2.8e-4
2e-1
7
pC
9.75
4.25
11.1
3.55
1
3.55
0.7
iii) 0.10 F Na3PO4
PBE
[OH-] = [H+] + 3[H3PO4] + 2[H2PO4-] + [HPO4-2]
which reduces to:
[OH-] = [HPO4-2]
Solution Composition
Species
Graph
C
+
H
OHH3PO4
H2PO4HPO4-2
PO4-3
Na+
2.5e-13
4e-2
1e-17
2e-7
4e-2
6.3e-2
3e-1
8
pC
12.6
1.4
17
6.7
1.4
1.2
0.5
B) Construct similar log C vs pH diagrams for 0.10 F carbonate system (H2CO3,
HCO3-, CO3-2) and 0.20 F ammonia system (NH4+, NH3), and use this to calculate
pH and composition of the following systems:
i) 0.10 F NaHCO3
ii) 0.10 F NaHCO3 + 0.20 F NH4Cl
iii) 0.10 F (NH4)2CO3
iv) 0.10 F Na2CO3
Approach
• prepare Log C vs pH diagram
• write PBE for each solution
• locate pHs for each solution
• read off concentrations for each species
Log C vs pH Diagram
• pKs are 6.3 and 10.3 for carbonate system; 9.3 for ammonia
• Log CT is -1 for carbonate system; -0.7 for ammonia
9
Log C vs pH Diagram
0
NH4
-1
+
PBE Solutions
NH3
H2CO3
HCO3
-2
-
CO3
iii
-2
ii and iv
i
H+
-3
-
OH
-4
-5
Log C
-6
-7
-8
-9
-10
-11
-12
-13
-14
0
1
2
3
4
5
6
7
8
pH
10
9
10
11
12
13
14
i) 0.10 F NaHCO3
PBE
[CO3-2] + [OH-] = [H+] + [H2CO3]
which reduces to:
[CO3-2] = [H2CO3]
Solution Composition
Species
Graph
C
pC
H+
5e-9
8.3
OH
2e-6
5.7
H2CO3
1e-3
3
HCO3
1e-1
1
CO3-2
1e-3
3
+
Na
1e-1
1
-2
** this is the sum of [CO3 ] (5.69e-4) and [NaCO3 ] (9.93e-4).
ii) 0.10 F NaHCO3 + 0.20 F NH4Cl
PBE
[CO3-2] + [NH3] + [OH-] = [H+] + [H2CO3]
which reduces to:
[NH3] = [H2CO3]
Solution Composition
Species
Graph
C
+
H
OHH2CO3
HCO3CO3-2
NH4+
NH3
ClNa+
2.5e-8
4e-7
5e-3
1e-1
1.8e-4
2e-1
5e-3
2e-1
1e-1
11
pC
7.6
6.4
2.3
1
3.75
0.7
2.3
0.7
1
iii) 0.10 F (NH4)2CO3
PBE
[NH3] + [OH-] = [H+] + 2[H2CO3] + [HCO3-]
which reduces to:
[NH3] = [HCO3-]
Solution Composition
Species
Graph
C
H+
OHH2CO3
HCO3CO3-2
NH4+
NH3
5.6e-10
1.8e-5
7e-5
1e-1
8e-3
1e-1
1e-1
pC
9.25
4.75
4.2
1
2.1
1.0
1.0
iv) 0.10 F Na2CO3
PBE
[OH-] = [H+] + 2[H2CO3] + [HCO3-]
which reduces to:
[OH-] = [HCO3-]
Solution Composition
Species
Graph
C
+
pC
H
2.2e-12
11.65
OH
4.5e-3
2.35
H2CO3
2e-8
7.7
HCO3
4.5e-3
2.35
-2
CO3
1e-1
1
Na+
2e-1
0.7
-2
** this is the sum of [CO3 ] (2.8e-2) and [NaCO3 ] (6.9e-2).
12
3. Acid/Base Equilibria III: Acids & Conjugate Bases (1 POINT)
A. Calculate the composition and pH of the following solutions:
i) 0.10 F NaCOOH + 0.40 F HCOOH
ii) 0.20 F NH3 + 0.50 F NH4Cl
General Approach
• these are solutions of acids and conjugate bases
• use the buffer equation, and its simplifying assumptions:
pH = pKa + log{CA/CHA}
• pKa's are 3.75 for formic acid and 9.3 for ammonia
i) 0.10 F NaCOOH + 0.40 F HCOOH
pH = pKa + log{CA/CHA}
pH = 3.75 + log{0.1/0.4}
pH = 3.15
Species
+
H
OHHCOOH
COOHNa+
Equation
C
pC
buffer eq.
Kw
=CHA
=CA
=CA
7.1e-4
1.4e-11
0.4
0.1
0.1
3.15
10.85
0.40
1
1
ii) 0.20 F NH3 + 0.50 F NH4Cl
pH = pKa + log{CA/CHA}
pH = 9.3 + log{0.2/0.5}
pH = 8.90
Species
Equation
C
pC
H+
OHNH4+
NH3
Cl-
buffer eq.
Kw
=CHA
=CA
=CHA
1.2e-9
8.0e-5
0.5
0.2
0.5
8.90
5.10
0.3
0.7
0.3
13
B.
A 3.16x10-3 F solution of uric acid has a pH of 3.2. What is the pH of
an equimolar solution (i.e., 3.16x10-3 F) of the Na+ salt of its conjugate
base (Na-urate)?
Approach
• prepare a Log C vs pH diagram, but working backwards
• known CT, and known pH, find pKa
• PBE suggests that pH lies at intersection of urate (Ur-) line and the H+ line
• draw line with +1 slope passing through H+ line at pH=3.2
• where it intersects CT is the pKa (about 3.8)
• then write PBE for base addition (i.e., NaUr) and solve
Log C vs pH Diagram
0
-1
HUr
-2
-
H+
Ur
OH
-
Acid
PBE Solution
-3
-4
Conjugate Base
-5
PBE Solution
Log C
-6
-7
-8
-9
-10
-11
-12
-13
-14
0
1
2
3
4
5
6
7
8
9
10
pH
PBE for Base Addition (NaUr)
[HUr] + [H+] = [OH-]
14
11
12
13
14
which reduces to:
[HUr] = [OH-]
Read pH from Graph
pH = 7.65
4.
Acid/Base Equilibria II: MINEQL method (1 POINT)
Solve the problems from question #2 (2A. and 2B.) exactly using MINEQL. Present
the MINEQL-based concentrations in a table. Compare your MINEQL results with
the approximate solutions you obtained from your graphs in problem 2 (note: when
solving problems with the carbonate system, you will have to send aqueous CO2 to
the Type VI category just as you did for H+. When we work with open carbonate
systems, you won't have to do this.)
A). Calculate the pH and the concentration of all species in the following solutions:
i) 0.10 F NaH2PO4
ii) 0.10 F Na2HPO4
iii) 0.10 F Na3PO4
due next
15