Important compounds — NaOH, NaCl, Na₂CO₃ (Solvay), NaHCO₃; CaO, CaCO₃, gypsum, plaster of Paris: Question bank

Before the traps, three small toolboxes build the jargon these questions lean on, each anchored to a picture — so you reinforce the ideas here instead of hunting external notes.

Toolbox 1 — Reading the electrode-potential ladder

Look at the ladder below. Each rung is a half-reaction placed at its potential. At the cathode (where reduction happens) the competition is between reducing water and reducing Na⁺.

Figure — Important compounds — NaOH, NaCl, Na₂CO₃ (Solvay), NaHCO₃; CaO, CaCO₃, gypsum, plaster of Paris

Toolbox 2 — Le Chatelier, common-ion effect & why NaHCO₃ drops out

The graph below shows why cold matters: NaHCO₃'s solubility stays low, while its rivals (NaCl, NH₄Cl) stay high, so only NaHCO₃ crosses its precipitation line in the cold, concentrated mixture.

Toolbox 3 — Amphoteric metals & the real zinc species

Toolbox 4 — Why heating gypsum to different temperatures gives different solids

The hydration state (how many waters ride with each CaSO₄) is set by temperature. The staircase figure shows the three landings: gypsum (2 waters), plaster of Paris (½ water), and dead-burnt anhydrite (0 water) — and marks which landing still sets.

True or false — justify

In electrolysis of aqueous NaCl, sodium metal is deposited at the cathode.

False. Water is reduced instead —

2 H_{2} O + 2 e^{-} \to H_{2} + 2 O H^{-}

— because Na⁺'s reduction potential (

\approx - 2.71 V

) sits far below water's, so water out-bids Na⁺ for electrons (Toolbox 1).

Chlorine is released at the anode only because Cl⁻ is easier to oxidise than water.

False (subtle). By raw potentials water is actually easier to oxidise, but the large overpotential for O₂ (the "extra-voltage toll," Toolbox 1) makes O₂ so slow that Cl₂ wins in practice — kinetics beats thermodynamics.

The Solvay overall reaction

2 N a C l + C a C O_{3} \to N a_{2} C O_{3} + C a C l_{2}

happens directly by mixing the two solids.

False. That net equation is a bookkeeping sum; it never occurs directly. The real driver is precipitating insoluble NaHCO₃ and recycling NH₃ (Le Chatelier, Toolbox 2).

Washing soda and baking soda are the same compound in different amounts.

False. Washing soda is the carbonate

N a_{2} C O_{3} \cdot 10 H_{2} O

; baking soda is the bicarbonate

N a H C O_{3}

. They differ in one extra H and one C–O per formula unit, giving very different basicity and edibility.

An aqueous solution of Na₂CO₃ is neutral because it is a sodium salt.

False. It is alkaline:

C O_{3}^{2 -} + H_{2} O ⇌ H C O_{3}^{-} + O H^{-}

. The anion of a weak acid (carbonic acid) hydrolyses and releases OH⁻. See Salt Hydrolysis and pH.

Heating gypsum more strongly always makes better plaster of Paris.

False. Above ~473 K you get dead-burnt anhydrous CaSO₄ (anhydrite) which will not set with water (Toolbox 4). PoP needs the controlled ~393 K that leaves exactly the ½-water hemihydrate.

Slaked lime and quicklime are chemically identical.

False. Quicklime is CaO; slaked lime is Ca(OH)₂, formed when CaO reacts with water exothermically. One is an oxide, the other a hydroxide.

The setting of plaster of Paris is a purely physical drying process.

False. It is a chemical reaction — the hemihydrate re-absorbs water to reform gypsum

C a S O_{4} \cdot 2 H_{2} O

, growing an interlocking crystal mesh that hardens. Drying alone would not set it.

K₂CO₃ can be made by the Solvay process just like Na₂CO₃.

False. KHCO₃ is too soluble to precipitate out of the cold ammoniacal brine, so the equilibrium-driving crystallisation step never happens (Toolbox 2 curve).

Spot the error

"NaOH + excess CO₂ → Na₂CO₃ + H₂O."

Error: with excess CO₂ the product is the bicarbonate,

N a O H + C O_{2} \to N a H C O_{3}

. Na₂CO₃ forms only with deficient CO₂ where there is enough base to fully neutralise.

"In the Solvay process, NH₃ is consumed and lost, making it wasteful."

Error: NH₃ is recovered and recycled —

2 N H_{4} C l + C a (O H)_{2} \to C a C l_{2} + 2 N H_{3} + 2 H_{2} O

. The only real by-product waste is CaCl₂; the ammonia loop is the whole point of the economy.

"Calcination of limestone

C a C O_{3} \to C a O + C O_{2}

needs no help because it is exothermic."

Error: it is strongly endothermic. It goes forward because the open kiln lets CO₂ escape, removing a product and pulling the equilibrium right (Le Chatelier, Toolbox 2).

"Mortar hardens by simply drying out the water."

Error: it hardens by re-carbonation —

C a (O H)_{2} + C O_{2} \to C a C O_{3} + H_{2} O

— slaked lime reabsorbing atmospheric CO₂ and turning back into stone. Drying alone leaves a soft paste.

"PoP is

C a S O_{4} \cdot 2 H_{2} O

Error: that is gypsum. PoP is the hemihydrate

C a S O_{4} \cdot \frac{1}{2} H_{2} O

, i.e.

(C a S O_{4})_{2} \cdot H_{2} O

, with only half a water per CaSO₄ (Toolbox 4).

"The mercury cathode in Castner–Kellner is used because mercury is a good conductor."

Error: its real job is to hold discharged sodium as an amalgam, physically separating Na from the OH⁻ so they can't back-react; the amalgam is later hydrolysed to give pure NaOH.

"Zinc dissolves in NaOH as

2 N a O H + Z n \to N a_{2} Z n O_{2} + H_{2}

with nothing else needed."

Error: water is missing and the real dissolved species is tetrahydroxozincate:

Z n + 2 N a O H + 2 H_{2} O \to N a_{2} [Z n (O H)_{4}] + H_{2}

(Toolbox 3).

N a_{2} Z n O_{2}

is just the dehydrated shorthand.

Why questions

Why does NaHCO₃, not Na₂CO₃, precipitate first in the Solvay tower?

Because NaHCO₃ is the least soluble species in the cold, concentrated ammoniacal brine; the common-ion effect from high Na⁺/HCO₃⁻ pushes it out of solution, dragging the equilibrium forward (Toolbox 2).

Why does baking soda make cake rise?

On heating it decomposes,

2 N a H C O_{3} Δ N a_{2} C O_{3} + H_{2} O + C O_{2}

; the released CO₂ gas inflates the dough into a spongy structure.

Why is exactly ½ a water molecule retained in PoP?

Two gypsum units carry 4 H₂O total; on gentle heating they lose 3 H₂O, leaving 1 H₂O shared between 2 CaSO₄ — that averages to ½ H₂O per CaSO₄ (Toolbox 4).

Why does washing soda crumble to a powder on standing in air?

Efflorescence —

N a_{2} C O_{3} \cdot 10 H_{2} O

loses most of its crystal water to the atmosphere, collapsing to the monohydrate powder.

Why can NaOH dissolve zinc but a weak base cannot?

Zn is amphoteric (Toolbox 3); only a strong base is basic enough to attack it, giving soluble tetrahydroxozincate —

Z n + 2 N a O H + 2 H_{2} O \to N a_{2} [Z n (O H)_{4}] + H_{2}

Why is the lime cycle called a "cycle"?

CaCO₃ → CaO (bake) → Ca(OH)₂ (slake) → back to CaCO₃ (re-carbonate). The final product is the same stone you started with, so the chemistry closes a loop.

Edge cases

What happens if you bubble CO₂ through Na₂CO₃ solution instead of adding it to NaOH?

You get the bicarbonate:

N a_{2} C O_{3} + C O_{2} + H_{2} O \to 2 N a H C O_{3}

. Excess CO₂ protonates carbonate down to bicarbonate — an alternative industrial route to baking soda.

What is the limiting product if gypsum is heated past 473 K?

All water leaves, giving anhydrous CaSO₄ (dead-burnt anhydrite) — the top landing in Toolbox 4. It will not re-absorb water, so it is useless as plaster.

At the cathode, why doesn't the OH⁻ concentration eventually stop H₂ evolution?

Cations (Na⁺) migrate to keep the region near-neutral in charge and the electrode keeps supplying electrons; OH⁻ builds up in bulk as NaOH but H₂O reduction continues as long as current flows.

What would happen to the Solvay equilibrium if the brine were warm instead of cold?

NaHCO₃ solubility rises with temperature (see the upward curve in Toolbox 2), so it would fail to precipitate — the driving crystallisation stops and Na₂CO₃ yield collapses. The cold step is essential.

If CaO were slaked with a huge excess of water, what changes?

You still form Ca(OH)₂, but the strongly exothermic reaction can boil the water and, in excess, gives a dilute suspension (lime water/milk of lime) rather than a dry hydroxide — the chemistry is the same, only the physical state differs.

Recall One-line self-test

Cover every answer above and re-derive the reason, not the verdict. If you can state the driver — potential ladder, overpotential, precipitation + common-ion, recycling, endothermic + open kiln, re-carbonation, controlled 393 K — you have mastered every trap in this topic.

Related prerequisite ideas: Hydrogen — preparation and uses, Hardness of Water, s-Block — anomalous properties of Li and Be.