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

Level 1 — Recognition

Recall Solution L1·Q1

(a) Caustic soda $= NaOH$ (b) Washing soda $= Na_{2} CO_{3} \cdot 10 H_{2} O$ (c) Baking soda $= NaHCO_{3}$ (d) Quicklime $= CaO$ (e) Slaked lime $= Ca(OH)_{2}$ (f) Plaster of Paris $= CaSO_{4} \cdot \frac{1}{2} H_{2} O$ , i.e. $(CaSO_{4})_{2} \cdot H_{2} O$

Recall Solution L1·Q2

NaCl (brine / halite) → supplies $Na^{+}$ (and $Cl^{-}$ ) for all sodium chemistry.
CaCO₃ (limestone) → supplies $Ca^{2 +}$ , and on heating gives $CO_{2}$ and $CaO$ .

Level 2 — Application

Intuition Mini-primer: what electrolysis and "overpotential" mean (so the solution is self-contained)

Electrolysis = using a battery to force a chemical split that would not happen on its own. Two metal rods dip in the salt solution: the cathode (minus terminal) pushes electrons in → something gains electrons (reduction); the anode (plus terminal) pulls electrons out → something loses electrons (oxidation). At each rod, the species that discharges most easily wins. "Easiest" is measured by a number called reduction potential (background in Electrolysis and Electrode Potentials): the more positive, the easier to reduce. Overpotential = a real cell often needs more voltage than the textbook number to actually make a gas — this "extra push required" differs for different gases. For $O_{2}$ the extra push is large, for $Cl_{2}$ it is small, so at high $Cl^{-}$ concentration $Cl_{2}$ comes off the anode first even though water "on paper" looks competitive. That gap is drawn in the figure below.

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

Figure — for each anode option, the bar shows the total voltage the cell must supply: the teal part is the "on-paper" discharge voltage and the orange part is the overpotential (the extra push a real cell needs to actually release that gas). Read the two totals: making $O_{2}$ from water needs about $1.83 V$ in total, but making $Cl_{2}$ needs only about $1.46 V$ . The cell always takes the cheaper (smaller total voltage) route, so chlorine is released, not oxygen — this is exactly why the anode gives $Cl_{2}$ and the overall product list contains $Cl_{2}$ , not $O_{2}$ .

Recall Solution L2·Q1

Cathode (reduction): water is reduced (its reduction is far easier than $Na^{+}$ , whose very negative reduction potential means $Na^{+}$ clings to its charge): $2 H_{2} O + 2 e^{-} \to H_{2} + 2 OH^{-}$ Anode (oxidation): concentrated $Cl^{-}$ is discharged. From the figure, making $Cl_{2}$ costs a smaller total voltage ( $\approx 1.46 V$ ) than making $O_{2}$ from water ( $\approx 1.83 V$ ), because $O_{2}$ carries a much larger overpotential — so chlorine wins: $2 Cl^{-} \to Cl_{2} + 2 e^{-}$ Overall (recall $↑$ = "escapes as gas"): $2 NaCl + 2 H_{2} O electrolysis 2 NaOH + Cl_{2} ↑ + H_{2} ↑$ Three products = three industries: NaOH (base), Cl₂ (bleach/PVC), H₂ (see Hydrogen — preparation and uses).

Recall Solution L2·Q2

(a) Excess acidic gas protonates carbonate all the way to bicarbonate: $NaOH + CO_{2} \to NaHCO_{3}$ (b) Carbonate absorbs more $CO_{2}$ to become bicarbonate: $Na_{2} CO_{3} + CO_{2} + H_{2} O \to 2 NaHCO_{3}$ (c) Slaking (exothermic): $CaO + H_{2} O \to Ca(OH)_{2} + heat$

Level 3 — Analysis

Figure — approximate solubilities in cold water (about $10 - 15^{\circ} C$ , the temperature of a Solvay tower). The $y$ -axis is grams of salt per $100 mL$ of water (how much can stay dissolved before the salt drops out); the number is printed on top of each bar. The shortest bar wins the race to precipitate — here that is $NaHCO_{3}$ (about $9.6$ ), far below $NaCl$ and $NH_{4} Cl$ (about $36 - 37$ ). Values are illustrative, chosen to show the ordering, not exact lab constants.

Recall Solution L3·Q1

The tank holds four ions, so four salts could form: $NaCl$ , $NH_{4} Cl$ , $NaHCO_{3}$ , $NH_{4} HCO_{3}$ . Which one leaves the water? The least soluble one — because once its concentration product exceeds its (small) solubility, it can no longer stay dissolved and crystals drop out (in the figure the tallest bar is the most soluble; $NaHCO_{3}$ is the shortest, so it hits its limit first). In the cold, concentrated ammoniacal brine, $NaHCO_{3}$ has the lowest solubility, so it precipitates. Written as a net ionic equation (spectator ions $Cl^{-}$ and $NH_{4}^{+}$ removed, since they stay dissolved and unchanged): $Na^{+} + HCO_{3}^{-} \to NaHCO_{3} ↓$ and the full molecular form (bringing the spectators back in) is: $NH_{4} HCO_{3} + NaCl \to NaHCO_{3} ↓ + NH_{4} Cl$ Removing solid $NaHCO_{3}$ from the solution pulls the equilibrium forward (Le Chatelier's Principle): the system keeps making more to replace what left. The other three salts stay dissolved.

Recall Solution L3·Q2

$Na_{2} CO_{3}$ is the salt of a strong base (NaOH) and a weak acid (carbonic acid). Its anion is the leftover of a weak acid, so it grabs a proton back from water — this is anion hydrolysis (see Salt Hydrolysis and pH): $CO_{3}^{2 -} + H_{2} O ⇌ HCO_{3}^{-} + OH^{-}$ The freed $OH^{-}$ makes the solution basic → litmus blue. $NaCl$ comes from strong base (NaOH) and strong acid (HCl); neither ion pulls on water, so no extra $OH^{-}$ or $H^{+}$ → neutral.

Level 4 — Synthesis

Recall Solution L4·Q1

Step 1 — kiln (limestone → lime + gas): supplies both $CO_{2}$ and (later) $CaO$ . $CaCO_{3} Δ CaO + CO_{2}$ Step 2 — carbonation of ammoniacal brine: $NH_{3} + H_{2} O + CO_{2} \to NH_{4} HCO_{3}$ $NH_{4} HCO_{3} + NaCl \to NaHCO_{3} ↓ + NH_{4} Cl$ Step 3 — calcine the bicarbonate: returns half the $CO_{2}$ (recyclable) and gives soda ash. $2 NaHCO_{3} Δ Na_{2} CO_{3} + H_{2} O + CO_{2}$ Step 4 — recover ammonia using the Step-1 lime (slaked to $Ca(OH)_{2}$ ): $2 NH_{4} Cl + Ca(OH)_{2} \to CaCl_{2} + 2 NH_{3} + 2 H_{2} O$ The recovered $NH_{3}$ re-enters Step 2. Adding it all and cancelling the recycled $NH_{3} / CO_{2}$ : $2 NaCl + CaCO_{3} \to Na_{2} CO_{3} + CaCl_{2} ✓$

Recall Solution L4·Q2

Calcination: $CaCO_{3} Δ CaO + CO_{2}$ (limestone → quicklime; here $Δ$ = heat, about $1100 K$ ) Slaking: $CaO + H_{2} O \to Ca(OH)_{2}$ (quicklime → slaked lime, exothermic) Mortar = slaked lime + sand + water. Setting / re-carbonation: $Ca(OH)_{2} + CO_{2} \to CaCO_{3} + H_{2} O$ The mortar reabsorbs atmospheric $CO_{2}$ and turns back into $CaCO_{3}$ (stone) — that is why old walls harden over time. The carbon (as $CO_{2}$ ) is the element that leaves at the kiln and returns from the air, closing the lime cycle.

Level 5 — Mastery (quantitative)

Recall Solution L5·Q1

Moles of $NaHCO_{3} = \frac{16.8}{84} = 0.20 mol$ . Ratio $2 NaHCO_{3} : 1 Na_{2} CO_{3} : 1 CO_{2}$ .

$Na_{2} CO_{3}$ : $0.20/2 = 0.10 mol \Rightarrow 0.10 \times 106 = 10.6 g$
$CO_{2}$ : $0.10 mol \Rightarrow 0.10 \times 44 = 4.4 g$ Check: $H_{2} O = 0.10 \times 18 = 1.8 g$ ; total out $10.6 + 4.4 + 1.8 = 16.8 g$ = mass in. ✔ (mass conserved)

Recall Solution L5·Q2

$CaCO_{3} Δ CaO + CO_{2}$ , ratio $1 : 1 : 1$ . Moles of $CaO$ : $\frac{560 000 g}{56} = 10000 mol$ . Pure $CaCO_{3}$ needed: same $1 : 1$ ratio, so $10000 mol \times 100 = 1000000 g = 1000 kg$ . Sample mass (only $80%$ pure, so we need more rock): $\frac{1000}{0.80} = 1250 kg$ . Volume of $CO_{2}$ at STP: the ratio also gives $10000 mol$ of $CO_{2}$ ; at $22.4 L\cdotmol^{- 1}$ , $V = 10000 mol \times 22.4 \frac{L}{mol} = 224000 L = 224 m^{3} (1 m^{3} = 1000 L) .$

Recall Solution L5·Q3

$2 (CaSO_{4} \cdot 2 H_{2} O) 393 K (CaSO_{4})_{2} \cdot H_{2} O + 3 H_{2} O$ Moles gypsum $= \frac{34.4}{172} = 0.20 mol$ . Ratio $2 gypsum \to 1 PoP + 3 H_{2} O$ :

PoP: $0.20/2 = 0.10 mol \Rightarrow 0.10 \times 290 = 29.0 g$
Water off: $0.20 \times \frac{3}{2} = 0.30 mol \Rightarrow 0.30 \times 18 = 5.4 g$ Check: $29.0 + 5.4 = 34.4 g$ ✔. Above 473 K: all water leaves, giving dead-burnt anhydrous $CaSO_{4}$ (anhydrite) which will not set with water. The $\frac{1}{2}$ -water of PoP is essential — it's the "seed" of water the crystals rebuild from when you add water back and it re-forms gypsum.

Recall Solution L5·Q4

Carbonate ion precipitates the calcium out of solution as insoluble chalk (see Hardness of Water): $Ca^{2 +} + CO_{3}^{2 -} \to CaCO_{3} ↓$ Removing $Ca^{2 +}$ softens the water. Baking soda supplies $HCO_{3}^{-}$ , not $CO_{3}^{2 -}$ ; $Ca(HCO_{3})_{2}$ is soluble (that's actually a cause of temporary hardness), so it does not remove the calcium — poor choice.

Recall One-line self-test (cloze)

The Solvay driver is precipitating NaHCO₃ (least soluble) and recycling NH₃. PoP forms at 393 K; above 473 K you get non-setting dead-burnt anhydrite. Washing soda softens water by precipitating CaCO₃.