Chapter 5 · Carbonated Water · Ten playable problems

Ch 5 · P1 — Pushed or pulled

PROBLEM 41 · CARBONATED WATER — HENRY'S LAW

A gas and an ionic solid dissolve for different reasons. Assign each solute its mechanism, then watch what heat does to the one nothing is holding.

Live model

pressure pushes gases · polarity pulls saltsAssign each solute its mechanism.

Ch 5 · P2 — Who rules at this pH

PROBLEM 42 · CARBONATED WATER — THE CARBONATE EQUILIBRIUM

CO₂(g) ⇌ CO₂(aq) ⇌ H₂CO₃ ⇌ HCO₃⁻ + H⁺ ⇌ CO₃²⁻ + 2H⁺. Walk the pH handle to each flag and stamp which species dominates there.

Live model

pH referees the whole chainWalk the handle to each flag and stamp the ruling species — 0 of 3 stamped.

Ch 5 · P3 — The two-way sponge

PROBLEM 43 · CARBONATED WATER — WHY BICARBONATE BUFFERS

Hit the bicarbonate with acid, then with base. Watch what it becomes each time and how little the pH cares — then say why that makes it a buffer.

Live model

absorb H⁺ one way · donate H⁺ the otherThrow acid AND base at the bicarbonate, then answer.

Ch 5 · P4 — The pH that rose by itself

PROBLEM 44 · CARBONATED WATER — CO₂ STRIPPING

The operator stops feeding acid. Nothing is added, yet pH climbs from 7.5 to 8.8. Click the four events in causal order.

Live model

strip → shift → consume → climbClick the four events in causal order.

Ch 5 · P5 — Alkalinity in equivalents

PROBLEM 45 · CARBONATED WATER — THE CaCO₃ CURRENCY

A cooling tower makeup carries 250 ppm alkalinity as CaCO₃. Drag the handle and read the same water in three currencies.

Live model

50 mg CaCO₃ = 1 milliequivalentDrag the makeup alkalinity to 250 ppm.

Ch 5 · P6 — Reading the regime

PROBLEM 46 · CARBONATED WATER — TWO TITRATIONS

P is 80 and M is 240, both as CaCO₃. Set both titrations and name the species the sample is carrying.

Live model

NO SAMPLE

2P vs M decides the regimeSet P to 80 and M to 240. Then push P past half of M and watch the field rearrange.

Ch 5 · P7 — The alkalinity tax

PROBLEM 47 · CARBONATED WATER — WHAT THE BOILER MAKES OF IT

A boiler's feedwater M-alkalinity is 15 ppm as CaCO₃. Drag to it and read the CO₂ in the steam and condensate, and the amine demand at 2.5 ppm per 1 ppm CO₂.

Live model

0.79 ppm CO₂ per ppm M · 2.5 ppm amine per ppm CO₂Drag M-alkalinity to 15 ppm.

Ch 5 · P8 — Free caustic

PROBLEM 48 · CARBONATED WATER — pH WITHOUT A METER

P is 400 and M is 450, both as CaCO₃. Set both titrations, find the hydroxide alkalinity, and walk it to the boiler's pH.

Live model

OH⁻ is 2P − M · pH is 14 + log₁₀[OH⁻]Set P to 400 and M to 450.

Ch 5 · P9 — Upstream or downstream

PROBLEM 49 · CARBONATED WATER — TWO WAYS TO FIGHT CARBONATE

Neutralize the alkalinity before the boiler, or neutralize the acidity after it. Put each tool on the leg of the circuit where it fights.

Live model

either way, the alkalinity is the targetPut each tool where it fights.

Ch 5 · P10 — Two thousand pounds of permission

PROBLEM 50 · CARBONATED WATER — THE ACID FEED FAILURE

An acid feed fails and 2,000 pounds of scale land on the tower fill. Rebuild the chain of events in order.

Live model

fail → strip → climb → carbonate → rockRebuild the failure: click the five events in order.

Problem set

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Problem 1
Henry’s Law governs gas solubility. How does it differ from the way ionic solids dissolve? Use the language of “pushed” versus “pulled.” (Refer to Chapter 3 if needed).
Authored answer
Ionic solids are pulled into solution. Water’s polarity dismantles the crystal lattice and hydration shells stabilize the freed ions. Gases are pushed into solution by pressure. They occupy gaps in the Hydrogen-Bond Network but are not electrostatically stabilized. When pressure drops or temperature rises, gases leave because nothing is holding them in place.
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Problem 2
Write the full carbonate equilibrium expression. In one sentence, which species dominates at low, neutral, and high pH.
Authored answer
CO₂(g) ⇌ CO₂(aq) ⇌ H₂CO₃ ⇌ HCO₃⁻ + H⁺ ⇌ CO₃²⁻ + 2H⁺. pH determines which species dominates: low pH favors CO₂/carbonic acid, neutral pH favors bicarbonate, and high pH favors carbonate.
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Problem 3
Why is bicarbonate considered a buffer? What dual role does it play?
Authored answer
Bicarbonate can absorb hydrogen (becoming carbonic acid) when acid is added, or donate hydrogen (becoming carbonate) when base is added. This dual role allows it to resist pH change in both directions, which is the definition of a buffer.
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Problem 4
A cooling tower operator stops feeding acid. No chemicals are added, yet the pH rises from 7.5 to 8.8. Explain why, using the concepts of CO₂ stripping and equilibrium shift.
Authored answer
The tower continuously strips CO₂ from the water through warm temperature, massive airflow, and enormous surface area. As CO₂ leaves, equilibrium shifts to replace it: carbonic acid decomposes, bicarbonate consumes H⁺. The decrease in hydrogen increases the pH.
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Problem 5
A cooling tower makeup has 250 ppm alkalinity as CaCO₃. Convert to (a) meq/L and (b) eq/L.
Authored answer
(a) 250 / 50 = 5.0 meq/L. (b) 250 / 50,000 = 0.0050 eq/L.
Problem 6
A water sample shows P-alkalinity = 80 ppm and M-alkalinity = 240 ppm (both as CaCO₃). Which alkalinity species are present? Calculate the bicarbonate and carbonate alkalinity.
Authored answer
½M = 120. Since P (80) < ½M (120), bicarbonate + carbonate are present. Carbonate alk = 2P = 160 ppm as CaCO₃. Bicarbonate alk = M − 2P = 240 − 160 = 80 ppm as CaCO₃.
Problem 7
A boiler’s feedwater M-alkalinity is 15 ppm as CaCO₃. Estimate the CO₂ concentration in the steam/condensate and the neutralizing amine demand at 2.5 ppm amine (active) per 1 ppm CO₂.
Authored answer
CO₂ ≈ 15 × 0.79 = 11.85 ppm. Amine demand = 2.5 × 11.85 = 29.6 ppm active amine.
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Problem 8
A boiler sample shows P-alkalinity = 400 ppm and M-alkalinity = 450 ppm (both as CaCO₃). Identify the species present, calculate the OH-alkalinity, and estimate the pH. Remember that for hydroxide, eq/L = mol/L.
Authored answer
½M = 225. Since P (400) > ½M (225), hydroxide + carbonate are present. OH– alk = 2P − M = 800 − 450 = 350 ppm as CaCO₃. [OH⁻] = 350 / 50,000 = 0.0070 eq/L = mol/L. pOH = −log10(0.0070) = 2.15. pH = 14 − 2.15 = 11.85.
Problem 9
The chapter describes two strategies for managing carbonate in boilers: neutralize alkalinity upstream or neutralize acidity downstream. In 2–3 sentences, explain what each approach targets and why alkalinity must be controlled either way.
Authored answer
Upstream treatment, such as reverse osmosis or dealkalization, removes bicarbonate before it enters the boiler. This greatly reduces alkalinity-derived CO₂ formation and helps control the condensate corrosion pathway. Downstream treatment uses neutralizing amines that volatilize with steam and condense with water, raising condensate pH and neutralizing acidity before carbonic acid can attack metal. Either way, alkalinity must be controlled because bicarbonate can thermally decompose into CO₂ inside the boiler, and that CO₂ can travel with steam, re-dissolve in condensate, lower pH, and create a corrosive environment.
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Problem 10
The chapter opens with a story about 2,000 pounds of scale depositing on cooling tower fill after an acid feed failure. Using carbonate equilibrium, explain the chain of events: what left the water, what shifted, what formed, and why pH was the gatekeeper.
Authored answer
When acid feed stopped, CO₂ stripping continued unchecked. As CO₂ left, equilibrium shifted: carbonic acid decomposed, bicarbonate consumed H⁺, and pH rose. As pH climbed through the 8s, the carbonate fraction increased sharply, making calcium carbonate precipitation increasingly favorable. With calcium already present and 60,000 sq ft of wetted surface area, CaCO₃ precipitated naturally and relentlessly. pH did not cause the scale, but it gave carbonate permission.
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