CEE 680

Fall 2011

Homework #5

11 points total

Problem #1 (2 points)

1. Determine the pH and alkalinity for the following solutions. Assume the systems are closed to the atmosphere.

a. 5x10-4F NaOH

b. 5x10-5F Na2CO3

c. 5x10-5F KHCO3

d. 1x10-3F HCl

Solution to #1

a. 5x10-4F NaOH

This is a strong base,

[OH^-] = [Na⁺] = 5 x 10^-4

pH = 10.7

Alk = 5 x 10^-4 eq/L = 25 mg/L as CaCO₃

b. 5x10-5F Na2CO3

This is a weak base. The problem can be solved with a log C vs pH diagram and the following proton balance equation:

2[H₂CO₃] + [HCO₃^-] + [H⁺] =[OH^-]

which reduces to:

[HCO₃^-] =[OH^-]

pH = 9.7

Alk = 10^-4 eq/L = 5 mg/L as CaCO₃

c. 5x10-5F KHCO3

This is a weak base. The problem can be solved with a log C vs pH diagram and the following proton balance equation:

[H₂CO₃] + [H⁺] =[OH^-]+ [CO₃^-2]

which reduces to:

[H₂CO₃] =[OH^-]

pH = 8.0

Alk = 5 x 10^-5 eq/L = 2.5 mg/L as CaCO₃

d. 1x10-3F HCl

This is a strong acid,

[Cl^-] = [H⁺] = 1 x 10^-3

pH = 3

Alk = -1 x 10^-3 eq/L = -50 mg/L as CaCO₃

Problem #2 (1 point)

2. What partial pressure of carbon dioxide are the solutions from question #1 in equilibrium with?

Solution to #2

a. 5x10-4F NaOH

C_T = 0, so:

p_CO2 = 0

b. 5x10-5F Na2CO3

C_T = 5x10-5, and pH = 9.7, so:

alpha₀ ~ [H⁺]/k₁ = 10^-9.7/10^-6.3 = 10^-3.4

p_CO2 = 5 x 10^-5 (10^-3.4)/10^-1.5

p_CO2 = 6.3x10^-7 = 10^-6.2

c. 5x10-5F KHCO3

C_T = 5x10-5, and pH = 8.0, so:

alpha₀ ~ [H⁺]/k₁ = 10^-8.0/10^-6.3 = 10^-1.7

p_CO2 = 5 x 10^-5 (10^-1.7)/10^-1.5

p_CO2 = 3.2x10^-5 = 10^-4.5

d. 1x10-3F HCl

C_T = 0, so:

p_CO2 = 0

Problem #3

3. Repeat question #1, but assume the systems are open to the atmosphere.

Solution to #3 (1 point)

Draw an open system, log C vs pH diagram, and solve using the electroneutrality equation.

a. 5x10-4F NaOH

ENE:

[Na⁺] + [H⁺] = [OH^-] + [HCO₃^-] + 2[CO₃^-2]

which reduces to:

[Na⁺] = [HCO₃^-]

pH = 8.0

Alk = 5 x 10^-4 eq/L = 25 mg/L as CaCO₃

b. 5x10-5F Na2CO3

ENE:

[Na⁺] + [H⁺] = [OH^-] + [HCO₃^-] + 2[CO₃^-2]

which reduces to:

[Na⁺] = [HCO₃^-]

pH = 7.3

Alk = 1 x 10^-4 eq/L = 5 mg/L as CaCO₃

c. 5x10-5F KHCO3

ENE:

[K⁺] + [H⁺] = [OH^-] + [HCO₃^-] + 2[CO₃^-2]

which reduces to:

[K⁺] = [HCO₃^-]

pH = 7.0

Alk = 5 x 10^-5 eq/L = 2.5 mg/L as CaCO₃

d. 1x10-3F HCl

ENE:

[H⁺] = [OH^-] + [Cl^-] + [HCO₃^-] + 2[CO₃^-2]

which reduces to:

[H⁺] = [Cl^-]

pH = 3

Alk = -1 x 10^-3 eq/L = -50 mg/L as CaCO₃

Problem #4 (1 point)

4. A groundwater was found to have the following composition:

ANC = 3x10-2 equ/L, pH = 7.5

What is the partial pressure of CO2 that is in equilibrium with this water?

Solution to #4

Assume all ANC is due to carbonate species; which at pH 7.5 is almost entirely bicarbonate.

Therefore:

[HCO₃^-] = 3 x 10^-2

and since

[H₂CO₃^*] = [HCO₃^-][H⁺]/k₁

so:

[H₂CO₃^*] = (3 x 10^-2)(10^-7.5)/10^-6.3

[H₂CO₃^*] = 10^-2.7

now:

[H₂CO₃^*] = k_H x p_CO2

p_CO2 = 10^-2.7/10^-1.5 = 10^-1.2 = 0.063 atm

Problem #5 (1 point)

5. A surface water and a groundwater are to be used as a raw water supply for the town of Springfield, FL. They will be mixed as follows:

	Surface Water	Groundwater
Flow Rate	3.5 MGD	0.8 MGD
Alkalinity	10 mg/L as CaCO3	250 mg/L as CaCO3
pH	6.9	8.3

What will the pH of the combined waters be?

Solution to #5

Recognize that this must be a closed system. The mixing of waters within an engineered treatment system that does not specifically include aeration or multi-day open-air storage will not result in substantial transfer of CO₂ across the air water interface.

First Calculate C_T for both waters:

Surface Water: C_T = ((10/50,000) - 10^-7.1 + 10^-6.9)/0.8 = 2.5 x 10^-4 M = 10^-3.6

Groundwater: C_T = ((250/50,000) - 10^-5.7 + 10^-8.3)/1 = 5 x 10^-3 M = 10^-2.3

Next Determine the Blended C_T and Alk

C_T(blend) = (3.5(2.5 x 10^-4) + 0.8(5 x 10^-3))/(3.5 + 0.8) = 1.134 x 10^-3 M

Alk_(blend) = (3.5(10/50,000) + 0.8(250/50,000))/(3.5 + 0.8) = 1.093 x 10^-3 eq/L

Lastly, Determine pH of the Blended Water

since the pH of both waters is well below the pk₂, carbonate is likely to be insignificant as compared to bicarbonate. This means that the alkalinity is composed almost entirely of bicarbonate, so:

now insert Alk and C_T and solve for H⁺, and we get:

[H⁺] = 1.87 x 10^-8

pH = 7.73

Summary:

Prob 1

	pH	Alkalinity
		(mg/L)	(equ/L)
a	10.7	25	5x10^-4 or 10^-3.3
b	9.7	5	10^-4
c	8.0	2.5	5x10^-5 or 10^-4.3
d	3.0	-50	-10^-3

Prob 2

	Pco2
a	0
b	6.3 x 10^-7
c	3.2 x 10^-5
d	0

Prob 3

	pH	Alkalinity
		(mg/L)	(equ/L)
a	8.0	25	5x10^-4 or 10^-3.3
b	7.3	5	10^-4
c	7.0	2.5	5x10^-5 or 10^-4.3
d	3.0	-50	-10^-3

Prob 4

Pco₂

10^-1.2 or 0.063 atm

Prob 5

pH

7.73