CEE 680

Fall 2011

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

Assumptions:

1. Acidic Solution: [H⁺] >> [OH^-]

2. Weak Acid: [HOCl] >> [OCl^-]

Solve for [H⁺] based on simplifying assumptions

pH = 0.5(pk_a + pC_T)

pH = 0.5(7.6 + 2.3)

pH = 4.95

Solve for all species

[H^+]		= 10^-4.95	or 1.1x10^-5
[OH^-]	= kw/[H^+]	= 10^-9.05	or 8.9x10^-10
[HOCl]	= CT	= 10^-2.30	or 5x10^-3
[OCl^-]	= ka[HOCl]/[H^+]	= 10^-4.95	or 1.1x10^-5

Check Assumptions

1. Acidic Solution: [H⁺] >> [OH^-]; 10^-4.95 >> 10^-9.05 , OK

2. Weak Acid: [HOCl] >> [OCl^-]; 10^-2.30 >> 10^-4.95 , OK

b. 5x10-3 moles NaCN

Approach

· addition of a simple monoprotic base to water

· try to use simplified solutions developed in class

Assumptions:

1. Basic Solution: [OH^-] >> [H⁺]

2. Weak Base: [CN^-] >> [HCN]

Solve for [H⁺] based on simplifying assumptions

pOH = 0.5(pk_b + pC_T)

pOH = 0.5(4.8 + 2.3)

pOH = 3.55

pH = 14-pOH = 10.45

Solve for all species

[H^+]		= 10^-10.45	or 3.6x10^-11
[OH^-]	= kw/[H^+]	= 10^-3.55	or 2.8x10^-4
[CN^-]	= CT	= 10^-2.30	or 5x10^-3
[Na^+]	= CT	= 10^-2.30	or 5x10^-3
[HCN]	= kb[CN^-]/[OH^-]	= 10^-3.55	or 2.8x10^-4

Check Assumptions

1. Basic Solution: [OH^-] >> [H⁺]; 10^-3.55 >> 10^-10.45 , OK

2. 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. Acidic Solution: [H⁺] >> [OH^-]

2. Strong Acid: [Cl^-] >> [HCl]

Solve for [H⁺] based on simplifying assumptions

[H⁺] = C_T

pH = pC_T

pH = 7.3

Solve for all species

[H^+]		= 10^-7.3	or 5x10^-8
[OH^-]	= kw/[H^+]	= 10^-6.7	or 2x10^-7
[HCl]	= [H^+][Cl^-]/ka	= 10^-17.6	or 2.5x10^-18
[Cl^-]	= CT	= 10^-7.3	or 5x10^-8

Check Assumptions

1. Acidic Solution: [H⁺] >> [OH^-]; 10^-7.3 >> 10^-6.7 , No!

2. 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

[H⁺] = 10^-7.3 + (10^-14.6 + 4x10^-14)^0.5

[H⁺] = 1.28x10^-7

pH = 6.89

Solve for all species

[H^+]		= 10^-6.89	or 1.28x10^-7
[OH^-]	= kw/[H^+]	= 10^-7.11	or 7.80x10^-8
[HCl]	= [H^+][Cl^-]/ka	= 10^-17.2	or 6.5x10^-18
[Cl^-]	= CT	= 10^-7.3	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

· return to original equations and solve, simplifying wherever possible

Species

H⁺, OH^-, HAc, Ac^-, Na⁺

Equations

K_w = 10^-14 = [H⁺][OH^-]

K_a = 10^-4.7 = [Ac^-][H⁺]/[HAc]

C_T1= 1.5x10^-3 = [Ac^-] + [HAc]

C_T2 = 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. [Na⁺] >> [H⁺]

2. [Ac^-] >> [OH^-]

Now Solve using the basic equations

Start with the ENE

[H⁺] + [Na⁺] = [OH^-] + [Ac^-]

[Na⁺] = [Ac^-]

And combining with the mass balance equations

[Na⁺] = [Ac^-] = C_T2 = 5x10^-4

[HAc] = C_T1 - [Ac^-] = C_T1 - C_T2 = 1x10^-3

Now use the equilibria to get the remaining species

K_a = 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^-] = k_w/[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.

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 C_T is -1

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	pC
H^+	2.2e-5	4.65
OH^-	4.5e-10	9.35
H3PO4	2.8e-4	3.55
H2PO4^-	1e-1	1.0
HPO4^-2	2.8e-4	3.55
PO4^-3	8e-12	11.1
Na^+	1e-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	pC
H^+	1.7e-10	9.75
OH^-	5.6e-5	4.25
H3PO4	8e-12	11.1
H2PO4^-	2.8e-4	3.55
HPO4^-2	1e-1	1
PO4^-3	2.8e-4	3.55
Na^+	2e-1	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	pC
H^+	2.5e-13	12.6
OH^-	4e-2	1.4
H3PO4	1e-17	17
H2PO4^-	2e-7	6.7
HPO4^-2	4e-2	1.4
PO4^-3	6.3e-2	1.2
Na^+	3e-1	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 C_T is -1 for carbonate system; -0.7 for ammonia

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

** this is the sum of [CO₃^-2] (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	pC
H^+	2.5e-8	7.6
OH^-	4e-7	6.4
H2CO3	5e-3	2.3
HCO3^-	1e-1	1
CO3^-2	1.8e-4	3.75
NH4^+	2e-1	0.7
NH3	5e-3	2.3
Cl^-	2e-1	0.7
Na^+	1e-1	1

iii) 0.10 F (NH4)2CO3

PBE

[NH₃] + [OH^-] = [H⁺] + 2[H₂CO₃] + [HCO₃^-]

which reduces to:

[NH₃] = [HCO₃^-]

Solution Composition

Species	Graph
	C	pC
H^+	5.6e-10	9.25
OH^-	1.8e-5	4.75
H2CO3	7e-5	4.2
HCO3^-	1e-1	1
CO3^-2	8e-3	2.1
NH4^+	1e-1	1.0
NH3	1e-1	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
CO3^-2	1e-1	1
Na^+	2e-1	0.7

** this is the sum of [CO₃^-2] (2.8e-2) and [NaCO₃^-] (6.9e-2).

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 = pK_a + log{C_A/C_HA}

· pK_a's are 3.75 for formic acid and 9.3 for ammonia

i) 0.10 F NaCOOH + 0.40 F HCOOH

pH = pK_a + log{C_A/C_HA}

pH = 3.75 + log{0.1/0.4}

pH = 3.15

Species	Equation	C	pC
H^+	buffer eq.	7.1e-4	3.15
OH^-	Kw	1.4e-11	10.85
HCOOH	=CHA	0.4	0.40
COOH^-	=CA	0.1	1
Na^+	=CA	0.1	1

ii) 0.20 F NH3 + 0.50 F NH4Cl

pH = pK_a + log{C_A/C_HA}

pH = 9.3 + log{0.2/0.5}

pH = 8.90

Species	Equation	C	pC
H^+	buffer eq.	1.2e-9	8.90
OH^-	Kw	8.0e-5	5.10
NH4^+	=CHA	0.5	0.3
NH3	=CA	0.2	0.7
Cl^-	=CHA	0.5	0.3

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 C_T, and known pH, find pK_a

· 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 C_T is the pK_a (about 3.8)

· then write PBE for base addition (i.e., NaUr) and solve

PBE for Base Addition (NaUr)

[HUr] + [H⁺] = [OH^-]

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

i) 0.10 F NaH2PO4

Solution Composition

Species	Graph		MINEQL
	C	pC	C	pC
H^+	2.2e-5	4.65	2.01e-5	4.70
OH^-	4.5e-10	9.35	4.98e-10	9.30
H3PO4	2.8e-4	3.55	2.87e-4	3.54
H2PO4^-	1e-1	1.0	9.94e-2	1.00
HPO4^-2	2.8e-4	3.55	3.07e-4	3.51
PO4^-3	8e-12	11.1	6.88e-12	11.16
Na^+	1e-1	1	1e-1	1

ii) 0.10 F Na2HPO4

Solution Composition

Species	Graph		MINEQL
	C	pC	C	pC
H^+	1.7e-10	9.75	1.85e-10	9.73
OH^-	5.6e-5	4.25	5.43e-5	4.27
H3PO4	8e-12	11.1	7.88e-12	11.10
H2PO4^-	2.8e-4	3.55	2.97e-4	3.53
HPO4^-2	1e-1	1	9.95e-2	1.00
PO4^-3	2.8e-4	3.55	2.42e-4	3.62
Na^+	2e-1	0.7	2e-1	0.70

iii) 0.10 F Na3PO4

Solution Composition

Species	Graph		MINEQL
	C	pC	C	pC
H^+	2.5e-13	12.6	2.69e-13	12.57
OH^-	4e-2	1.4	3.74e-2	1.43
H3PO4	1e-17	17	6.25e-18	17.20
H2PO4^-	2e-7	6.7	1.62e-7	6.79
HPO4^-2	4e-2	1.4	3.74e-2	1.43
PO4^-3	6.3e-2	1.2	6.26e-2	1.20
Na^+	3e-1	0.5	3e-1	0.52

B) 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

i) 0.10 F NaHCO3

Solution Composition

Species	Graph		MINEQL
	C	pC	C	pC
H^+	5e-9	8.3	7.57e-9	8.12
OH^-	2e-6	5.7	1.33e-6	5.88
H2CO3	1e-3	3	1.56e-3	2.81
HCO3^-	1e-1	1	9.1e-2	1.04
CO3^-2	1e-3	3	1.56e-3 **	3.25
Na^+	1e-1	1	1e-1	1.00

** this is the sum of [CO₃^-2] (5.69e-4) and [NaCO₃^-] (9.93e-4).

ii) 0.10 F NaHCO3 + 0.20 F NH4Cl

Solution Composition

Species	Graph		MINEQL
	C	pC	C	pC
H^+	2.5e-8	7.6	2.45e-8	7.61
OH^-	4e-7	6.4	4.08e-7	6.39
H2CO3	5e-3	2.3	4.94e-3	2.31
HCO3^-	1e-1	1	8.98e-1	1.05
CO3^-2	1.8e-4	3.75	1.71e-4	3.77
NH4^+	2e-1	0.7	1.96e-1	0.71
NH3	5e-3	2.3	4.47e-3	2.35
Cl^-	2e-1	0.7	2e-1	0.70
Na^+	1e-1	1	1e-1	1.00

iii) 0.10 F (NH4)2CO3

Solution Composition

Species	Graph		MINEQL
	C	pC	C	pC
H^+	5.6e-10	9.25	6.4e-10	9.19
OH^-	1.8e-5	4.75	1.57e-5	4.81
H2CO3	7e-5	4.2	1.34e-4	3.87
HCO3^-	1e-1	1	9.31e-2	1.03
CO3^-2	8e-3	2.1	6.8e-3	2.17
NH4^+	1e-1	1.0	1.07e-1	0.97
NH3	1e-1	1.0	9.33e-2	1.03

iv) 0.10 F Na2CO3

Solution Composition

Species	Graph		MINEQL
	C	pC	C	pC
H^+	2.2e-12	11.65	4.0e-12	11.4
OH^-	4.5e-3	2.35	2.5e-3	2.6
H2CO3	2e-8	7.7	2.1e-8	7.68
HCO3^-	4.5e-3	2.35	2.4e-3	2.62
CO3^-2	1e-1	1	9.7e-2 **	1.01
Na^+	2e-1	0.7	2e-1	0.70

** this is the sum of [CO₃^-2] (2.8e-2) and [NaCO₃^-] (6.9e-2).