ENCE 433 Dr. Alba Torrents

EXAMINATION 2 "SHOW ALL WORK" Fall 1996

Thursday, October 31

No Partial credit for YES/NO; HIGH/LOW without explanation.

State assumption; remember to check them

Clearly indicate intermediate and final results

1. (25) Complete the following table to show how CT and ACIDITY change when NaOH or HCl or CO2 etc. is added to an aqueous solution in a closed system.

­ indicates the term increases.

--- indicates the term remains the same.

¯ indicates the term decreases.

CT

ACIDITY

NaOH

HCl

CO2

NaHCO3

Na2CO3

NaCl

NH3

Recall: NH4+ NH3 + H+ pKa = 9.3

2.(15) Determine the alkalinity of the following systems:

  1. [CO2]T = 10-3.5 M, [NaOH]T = 10-3 M
  2. [CO2]T = 10-3.5 M, [HCl]T = 10-3 M
  3. [CO2]T = 10-3.5 M, [NH4Cl]T = 10-3.5 M, [Na3PO4]T = 10-4.5 M

Assume that NH4Cl is a neutral salt and that HPO4-2 is the dominant species for the phosphate system



3. (30) Wastewater having CT = 2.0 x 10-3 M and alkalinity = 1.0 x 10-3 M is discharged to lakewater having CT = 2.5 x 10-3 M and alkalinity = 2.5 x 10-3 M.

(a) What are the CT and Alk of a 1:10 dilution (1 part wastewater, 9 parts lakewater) if the mixture is "closed" (i.e., not in equilibrium with the atmosphere)? What is the pH?

(b) What concentration of strong acid or base would need to be added to take this sample to its Na2CO3 equivalence point? (assume closed system)

4. (30) Budget cuts have left a scientist with a sparsely stocked laboratory. Besides pure water, sodium bicarbonate (NaHCO3) and sodium carbonate (Na2CO3) are the only chemicals available.

(a) Describe how a solution having pH 9.0 can be prepared if no interaction is allowed to take place with the atmosphere. (i.e., how many moles of each chemical should be added to water to make 1 liter of pH 9.0 solution?)

  1. Describe how a solution having pH 9.0 can be prepared if the solution is allowed to equilibrate with the atmosphere.

ADDITIONAL INFORMATION

R = 1,987 cal /mole oK = 0.082 liter-atm deg-1 mole-1 = 8.31 joules deg-1 mole-1

PV = nRT

gal = 3.785 liters

F = 96,500 coulombs

2.3 RT/F = 0.059 V at 25oC

P(CO2) = 10-3.5 atm; P (O2) = 10-0.7 atm; P (N2) = 10-0.1 atm

H2CO3 pKa1 = 6.3; pKa2 = 10.3

[Alk] = [HCO3-] + 2 [CO3-2] + [OH-] - [H+]

[Acy] = [HCO3-] + 2 [H2CO3] - [OH-] + [H+]