The playable drawings need JavaScript. Every prompt and authored answer remains available below.
Problem 1
What is the difference between suspended solids and dissolved solids? Why can’t dissolved solids be removed by conventional filtration?
Authored answer
Suspended solids are particles physically carried by water and defined as the material retained by a TSS filter. Dissolved solids pass through that filter and remain as residue after the water is evaporated. Conventional granular or cartridge filters cannot remove dissolved ions because the ions exist at molecular scale and travel with the water. Removing them requires processes such as ion exchange, reverse osmosis, electrodialysis, or distillation. Colloids occupy the gray zone and may require coagulation or membrane filtration.
Read the connected sectionProblem 2
The chapter describes dissolution as a “transaction.” What are the three energetic terms involved, and which one represents the cost, the payoff, and the wildcard?
Authored answer
Dissolution involves three energetic terms. Lattice energy is the cost of pulling a crystal apart. Hydration energy is the payoff released when water stabilizes the separated species. Entropy is the wildcard: dispersion tends to favor dissolution, while solvent ordering can oppose it. Gibbs free energy combines these effects and tells you whether dissolution is favored.
Read the connected sectionProblem 3
Sodium chloride dissolves readily in water while calcium carbonate dissolves reluctantly. Using the language of lattice energy and hydration energy, explain why.
Authored answer
Sodium chloride has a manageable lattice cost, sufficient hydration payoff, and often favorable entropy. Dissolution is favored under normal conditions. Calcium carbonate has a much higher lattice cost because calcium and carbonate bind strongly in a stable crystal structure. Water can hydrate the separated ions, but the payoff is often conditional and not enough to overcome the strength of the lattice, especially as temperature, pH, calcium, alkalinity, and carbon dioxide shift the balance toward solid scale.
Read the connected sectionProblem 4
Why does conductivity provide a useful estimate of TDS, and what is its most important limitation?
Authored answer
Dissolved ions are charge carriers. More ions means more current and higher conductivity, making it a fast, continuous proxy for TDS. Its limitation: conductivity measures total ionic load but not which ions are present. Two waters at the same conductivity can behave very differently. One may scale, the other may corrode.
Read the connected sectionProblem 5
Your conductivity meter reads 1,450 µS/cm. Estimate the TDS in ppm.
Authored answer
TDS ≈ 0.7 × 1,450 = 1,015 ppm.
Problem 6
You add 5 pounds of NaCl to a 1,000-gallon cooling loop. Assuming complete dissolution, estimate the increase in concentration (ppm).
Authored answer
Water mass: 1,000 gal × 8.34 lb/gal = 8,340 lb. Weight of resulting solution = 8,340 lb + 5 lb = 8,345 lb. ppm increase: (5 lb / 8,345 lb) × (1,000,000) ≈ 599 ppm.
Problem 7
A 12,000-gallon cooling loop needs to be dosed to 150 ppm of a corrosion inhibitor that is 35% active by mass. How many pounds of product must be added? Show your dimensional analysis.
Authored answer
Water mass: 12,000 gal × 8.34 lb/gal = 100,080 lb. Active ingredient required: 100,080 lb × (150 / 1,000,000) = 15.0 lb active. Product required: 15.0 lb ÷ 0.35 = 42.9 lb of product. The percent-active conversion matters. Dosing 15 lb of product instead of 15 lb of active would leave the system at roughly 53 ppm, about a third of target.
Problem 8
A deaerator heats boiler feedwater and vents a small amount of steam to atmosphere. Using Henry's Law and the solubility behavior described in this chapter, explain why both the heating and the venting reduce dissolved oxygen. Then explain why the same equipment does nothing to remove dissolved calcium.
Authored answer
Oxygen is held by physical equilibrium, not hydration. Heating lowers its solubility, so the water can hold less; venting steam lowers the partial pressure of oxygen above the water. Both actions push the equilibrium toward the gas phase. Calcium is not governed by equilibrium with the atmosphere. It was dismantled from a solid and is held in solution by hydration shells, so no amount of heating or venting will drive it out as a gas.
Problem 9
Two cooling towers both read 2,000 µS/cm conductivity. Tower A’s makeup is high in sodium chloride. Tower B’s makeup is high in calcium bicarbonate. Which tower faces a greater scaling risk, and why does identical conductivity not mean identical behavior?
Authored answer
Identical conductivity means equal electrical conductance under the measurement conditions, not identical total ionic concentration or composition. Ion charge and mobility both affect the reading. Tower B faces the greater scaling risk because calcium and bicarbonate can concentrate and shift toward calcium carbonate precipitation, while sodium chloride generally remains soluble.
Read the connected sectionProblem 10
A cooling tower's makeup water reads 400 µS/cm. The circulating water reads 1,600 µS/cm. Estimate the cycles of concentration and the circulating TDS. If the makeup contains 80 mg/L calcium, what is the approximate circulating calcium, and why does the answer matter more at the heat exchanger surface than in the basin?
Authored answer
Cycles = 1,600 ÷ 400 = 4 cycles. Circulating TDS ≈ 0.7 × 1,600 = 1,120 ppm. Calcium ≈ 80 × 4 = 320 mg/L. Calcium carbonate has inverse solubility, so the hottest surface in the system is where ΔG turns over first. The basin may stay clear while the exchanger tubes scale.
Read the connected section