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Section A: Long Answer Questions

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5 questions
1long10 marks

Define a galvanic (voltaic) cell and explain its working with the help of the Daniell cell. State and explain the Nernst equation for a single electrode. A galvanic cell is set up as:

ZnZn2+(0.10 M)Cu2+(0.50 M)Cu\text{Zn} \mid \text{Zn}^{2+}(0.10\ \text{M}) \parallel \text{Cu}^{2+}(0.50\ \text{M}) \mid \text{Cu}

Given the standard electrode potentials EZn2+/Zn=0.76 VE^{\circ}_{\text{Zn}^{2+}/\text{Zn}} = -0.76\ \text{V} and ECu2+/Cu=+0.34 VE^{\circ}_{\text{Cu}^{2+}/\text{Cu}} = +0.34\ \text{V}, calculate the EMF of the cell at 25C25\,^{\circ}\text{C}.

Galvanic cell: A galvanic (voltaic) cell is an electrochemical device that converts chemical energy released by a spontaneous redox reaction into electrical energy. It consists of two half-cells joined by a salt bridge, each containing an electrode dipped in an electrolyte.

Working — Daniell cell:

  • Anode (oxidation): Zinc rod in ZnSO4\text{ZnSO}_4 solution.   ZnZn2++2e\;\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-
  • Cathode (reduction): Copper rod in CuSO4\text{CuSO}_4 solution.   Cu2++2eCu\;\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}
  • Overall: Zn+Cu2+Zn2++Cu\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}

Electrons flow externally from Zn (negative terminal) to Cu (positive terminal); the salt bridge (e.g. KCl in agar) maintains electrical neutrality by allowing ion migration.

  e- -->            e- -->
 [Zn]----wire----[Cu]
   |   salt bridge  |
  ZnSO4 ~~~~~~~~~ CuSO4
 (anode -)        (cathode +)

Nernst equation for a single electrode: For the reduction Mn++neMM^{n+} + ne^- \rightarrow M,

E=ERTnFln1[Mn+]=E+0.0591nlog[Mn+](at 25C)E = E^{\circ} - \frac{RT}{nF}\ln\frac{1}{[M^{n+}]} = E^{\circ} + \frac{0.0591}{n}\log[M^{n+}] \quad (\text{at } 25\,^{\circ}\text{C})

where EE^{\circ} is the standard electrode potential, nn the number of electrons, F=96500 C/molF = 96500\ \text{C/mol}.

Calculation: Cell reaction transfers n=2n = 2 electrons.

Ecell=EcathodeEanode=0.34(0.76)=1.10 VE^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = 0.34 - (-0.76) = 1.10\ \text{V}

Reaction quotient:

Q=[Zn2+][Cu2+]=0.100.50=0.20Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{0.10}{0.50} = 0.20

Nernst equation for the cell:

Ecell=Ecell0.0591nlogQ=1.100.05912log(0.20)E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n}\log Q = 1.10 - \frac{0.0591}{2}\log(0.20) log(0.20)=0.699\log(0.20) = -0.699 Ecell=1.10(0.02955)(0.699)=1.10+0.02066=1.1207 VE_{\text{cell}} = 1.10 - (0.02955)(-0.699) = 1.10 + 0.02066 = 1.1207\ \text{V}

EMF of the cell 1.121 V\approx \mathbf{1.121\ V}.

electrochemistrygalvanic-cellnernst-equation
2long10 marks

What is corrosion? Explain the electrochemical (wet) theory of corrosion of iron, writing the anodic and cathodic reactions for both oxygen-absorption and hydrogen-evolution mechanisms. Discuss any two factors that influence the rate of corrosion and describe two methods used to protect iron structures from corrosion.

Corrosion: Corrosion is the slow, spontaneous destruction of a metal by chemical or electrochemical reaction with its environment, converting the metal into more stable compounds such as oxides, hydroxides or carbonates. For iron the visible product is hydrated ferric oxide (rust), Fe2O3xH2O\text{Fe}_2\text{O}_3\cdot x\text{H}_2\text{O}.

Electrochemical (wet) theory: When iron is exposed to moisture, the heterogeneous surface develops tiny anodic and cathodic regions forming countless short-circuited galvanic cells, with the thin water film acting as electrolyte.

Anodic reaction (oxidation, same for both mechanisms):

FeFe2++2e\text{Fe} \rightarrow \text{Fe}^{2+} + 2e^-

Cathodic reaction depends on conditions:

  1. Oxygen-absorption (neutral/aerated medium):
12O2+H2O+2e2OH\tfrac{1}{2}\text{O}_2 + \text{H}_2\text{O} + 2e^- \rightarrow 2\text{OH}^-
  1. Hydrogen-evolution (acidic medium, poorly aerated):
2H++2eH22\text{H}^+ + 2e^- \rightarrow \text{H}_2 \uparrow

Rust formation: Fe2+\text{Fe}^{2+} migrates and meets OH\text{OH}^-:

Fe2++2OHFe(OH)2\text{Fe}^{2+} + 2\text{OH}^- \rightarrow \text{Fe(OH)}_2

Further oxidation gives Fe(OH)3\text{Fe(OH)}_3, which dehydrates to rust Fe2O3xH2O\text{Fe}_2\text{O}_3\cdot x\text{H}_2\text{O}.

Two factors affecting rate of corrosion:

  1. Nature of the metal / position in galvanic series — more active (anodic) metals and electrode-potential differences between contacting metals increase corrosion. A small anode with a large cathode causes intense localized attack.
  2. Nature of environment (pH, humidity, dissolved O₂, salts): Lower pH and higher dissolved oxygen, humidity and salt content (e.g. coastal/industrial air) all accelerate corrosion by improving electrolyte conductivity and depolarization at the cathode.

Two protection methods:

  1. Cathodic protection (sacrificial anode): A more active metal such as Mg or Zn is connected to the iron; it becomes the anode and corrodes preferentially, keeping the iron cathodic and protected (used for buried pipelines, ship hulls).
  2. Protective coatings: Barrier coatings (paints, enamel, electroplating) or galvanizing (a Zn coating that also gives sacrificial protection) isolate the metal from the corrosive environment.
corrosionelectrochemical-corrosioncorrosion-control
3long10 marks

Define hardness of water and distinguish between temporary and permanent hardness. A sample of hard water contains the following dissolved salts per litre: Ca(HCO3)2=16.2 mg\text{Ca(HCO}_3)_2 = 16.2\ \text{mg}, Mg(HCO3)2=14.6 mg\text{Mg(HCO}_3)_2 = 14.6\ \text{mg}, CaSO4=13.6 mg\text{CaSO}_4 = 13.6\ \text{mg} and MgCl2=19.0 mg\text{MgCl}_2 = 19.0\ \text{mg}. Calculate the temporary, permanent and total hardness of the water in terms of CaCO3\text{CaCO}_3 equivalent (ppm). (Molar masses: Ca(HCO3)2=162\text{Ca(HCO}_3)_2 = 162, Mg(HCO3)2=146\text{Mg(HCO}_3)_2 = 146, CaSO4=136\text{CaSO}_4 = 136, MgCl2=95\text{MgCl}_2 = 95, CaCO3=100 g/mol\text{CaCO}_3 = 100\ \text{g/mol}.)

Hardness of water: Hardness is the property of water that prevents it from forming a lather (foam) readily with soap; it is caused mainly by dissolved salts of calcium and magnesium (bicarbonates, sulphates, chlorides).

Temporary (carbonate) hardness: due to dissolved bicarbonates of Ca and Mg, Ca(HCO3)2\text{Ca(HCO}_3)_2 and Mg(HCO3)2\text{Mg(HCO}_3)_2. It is removed by boiling (decomposes to insoluble carbonate/hydroxide).

Permanent (non-carbonate) hardness: due to chlorides and sulphates of Ca and Mg (CaCl2,MgCl2,CaSO4,MgSO4\text{CaCl}_2, \text{MgCl}_2, \text{CaSO}_4, \text{MgSO}_4). It is not removed by boiling; requires chemical methods (lime-soda, ion exchange).

Calculation — convert each salt to CaCO₃ equivalent:

CaCO3 equiv (mg/L)=amount of salt (mg/L)×100molar mass of salt\text{CaCO}_3\ \text{equiv (mg/L)} = \text{amount of salt (mg/L)} \times \frac{100}{\text{molar mass of salt}}
SaltAmount (mg/L)Factor (100/M)CaCO₃ equiv (ppm)Type
Ca(HCO₃)₂16.2100/16216.2×0.6173=10.016.2 \times 0.6173 = 10.0Temporary
Mg(HCO₃)₂14.6100/14614.6×0.6849=10.014.6 \times 0.6849 = 10.0Temporary
CaSO₄13.6100/13613.6×0.7353=10.013.6 \times 0.7353 = 10.0Permanent
MgCl₂19.0100/9519.0×1.0526=20.019.0 \times 1.0526 = 20.0Permanent

Temporary hardness =10.0+10.0=20 ppm= 10.0 + 10.0 = \mathbf{20\ ppm}

Permanent hardness =10.0+20.0=30 ppm= 10.0 + 20.0 = \mathbf{30\ ppm}

Total hardness =20+30=50 ppm= 20 + 30 = \mathbf{50\ ppm} (as CaCO₃).

water-treatmenthardnessedta-titration
4long10 marks

Define polymerization. Distinguish between addition (chain) polymerization and condensation (step) polymerization with one example of each. Differentiate between thermoplastic and thermosetting polymers, giving two examples of each. Write the repeating unit of polyvinyl chloride (PVC) and state one engineering use of it. Define degree of polymerization and calculate it for a PVC sample of molar mass 125000 g/mol125000\ \text{g/mol}.

Polymerization: The chemical process in which a large number of small molecules (monomers) combine through covalent bonds to form a large macromolecule (polymer) of high molecular mass.

Addition vs condensation polymerization:

FeatureAddition (chain) polymerizationCondensation (step) polymerization
MonomersUnsaturated (C=C) moleculesBi-/poly-functional molecules
MechanismChain growth via free radicals/ions; no by-productStep growth; eliminates small molecule (H₂O, HCl)
Molecular massSame empirical formula as monomerRepeat unit lighter than sum of monomers
ExamplePolyethylene from ethylene nCH2=CH2(CH2CH2)nn\,\text{CH}_2{=}\text{CH}_2 \rightarrow {-}(\text{CH}_2{-}\text{CH}_2){-}_nNylon-6,6 from hexamethylenediamine + adipic acid (releases H₂O)

Thermoplastic vs thermosetting:

FeatureThermoplasticThermosetting
StructureLinear/branched, weak van der Waals forcesHeavily cross-linked 3-D network
On heatingSoften, can be remoulded repeatedlySet permanently; do not soften (char on strong heating)
RecyclableYesNo
ExamplesPolyethylene, PVCBakelite (phenol-formaldehyde), epoxy resin

PVC repeating unit:

 ⁣(CH2 ⁣ ⁣CHCl) ⁣n-\!\left(\text{CH}_2\!-\!\text{CHCl}\right)\!-_n

(formed by addition polymerization of vinyl chloride CH2=CHCl\text{CH}_2{=}\text{CHCl}).

Engineering use of PVC: Manufacture of water-supply and drainage pipes, electrical cable insulation and roofing sheets (rigid, chemically resistant, durable).

Degree of polymerization (DP): It is the number of repeating monomer units (nn) in a polymer chain.

DP=molar mass of polymermolar mass of repeating unit\text{DP} = \frac{\text{molar mass of polymer}}{\text{molar mass of repeating unit}}

For example, a PVC sample of molar mass 125000 g/mol125000\ \text{g/mol} (repeat unit C2H3Cl=62.5 g/mol\text{C}_2\text{H}_3\text{Cl} = 62.5\ \text{g/mol}) has DP=125000/62.5=2000\text{DP} = 125000/62.5 = \mathbf{2000}. A higher DP generally gives greater tensile strength, higher softening point and better chemical resistance, which is why high-molecular-weight polymers are preferred for load-bearing engineering applications.

polymerspolymerizationthermoplastics-thermosetting
5long10 marks

Define calorific value of a fuel and distinguish between gross (higher) and net (lower) calorific value. A sample of coal has the composition by mass: C = 80%, H = 5%, O = 8%, S = 1%, and the rest ash. Using Dulong's formula, calculate the gross calorific value (GCV) and the net calorific value (NCV) of the coal. (Take calorific constants: C = 8080, H = 34500, S = 2240 kcal/kg; latent heat of steam = 587 kcal/kg; 1 part H produces 9 parts H₂O.)

Calorific value: The calorific value of a fuel is the total quantity of heat liberated by the complete combustion of a unit mass (or unit volume) of the fuel. Common units: kcal/kg (solids/liquids), kcal/m³ (gases).

Gross (higher) calorific value (GCV): Heat liberated when unit mass of fuel is completely burnt and the combustion products are cooled to room temperature, so the water vapour formed is condensed and its latent heat is recovered.

Net (lower) calorific value (NCV): Heat liberated when the combustion products are allowed to escape (water remains as vapour); the latent heat of vaporization of water is not recovered. NCV<GCV\text{NCV} < \text{GCV}.

Dulong's formula:

GCV=1100[8080C+34500(HO8)+2240S] kcal/kg\text{GCV} = \frac{1}{100}\left[8080\,C + 34500\left(H - \frac{O}{8}\right) + 2240\,S\right]\ \text{kcal/kg}

where C, H, O, S are percentages by mass. The term O/8O/8 accounts for hydrogen already combined with oxygen as water (unavailable for combustion).

Step 1 — available hydrogen:

HO8=588=51=4H - \frac{O}{8} = 5 - \frac{8}{8} = 5 - 1 = 4

Step 2 — GCV:

GCV=1100[8080(80)+34500(4)+2240(1)]\text{GCV} = \frac{1}{100}\big[8080(80) + 34500(4) + 2240(1)\big] =1100[646400+138000+2240]=786640100=7866.4 kcal/kg= \frac{1}{100}\big[646400 + 138000 + 2240\big] = \frac{786640}{100} = 7866.4\ \text{kcal/kg}

GCV =7866.4 kcal/kg= \mathbf{7866.4\ kcal/kg}

Step 3 — mass of water formed per kg fuel: Total hydrogen = 5% = 0.05 kg; water formed =9×0.05=0.45= 9 \times 0.05 = 0.45 kg.

Step 4 — NCV:

NCV=GCV(mass of water×587)\text{NCV} = \text{GCV} - (\text{mass of water} \times 587) =7866.4(0.45×587)=7866.4264.15=7602.25 kcal/kg= 7866.4 - (0.45 \times 587) = 7866.4 - 264.15 = 7602.25\ \text{kcal/kg}

NCV =7602.25 kcal/kg= \mathbf{7602.25\ kcal/kg}

fuelscombustioncalorific-value
B

Section B: Short Answer Questions

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6 questions
6short5 marks

What is a catalyst? Distinguish between homogeneous and heterogeneous catalysis with one example each. List two important characteristics of a catalyst.

Catalyst: A catalyst is a substance that alters (usually increases) the rate of a chemical reaction without itself undergoing any permanent chemical change; it provides an alternative reaction path of lower activation energy. The phenomenon is called catalysis.

Homogeneous vs heterogeneous catalysis:

FeatureHomogeneous catalysisHeterogeneous catalysis
PhaseCatalyst and reactants in the same phaseCatalyst in a different phase from reactants
ActionThroughout the bulkAt the catalyst surface (adsorption)
ExampleHydrolysis of ester catalysed by aqueous H2SO4\text{H}_2\text{SO}_4 (lead-chamber process: NO catalysing SO₂ oxidation)Hydrogenation of vegetable oil over solid Ni; Haber process N2+3H22NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 over Fe

Two characteristics of a catalyst:

  1. A small quantity of catalyst is usually sufficient, and it is recovered chemically unchanged in mass and composition at the end of the reaction.
  2. A catalyst cannot start a thermodynamically impossible reaction; it only changes the rate and does not alter the position of equilibrium (it speeds forward and reverse reactions equally). Catalysts are also highly specific in action.
catalysiscatalyst-typesindustrial-catalysis
7short5 marks

Name the main raw materials and the four principal compounds (Bogue's compounds) present in Portland cement. Briefly explain the role of gypsum and the chemistry of setting and hardening of cement.

Raw materials of Portland cement: Calcareous materials (limestone, chalk — source of CaO) and argillaceous materials (clay, shale — source of SiO₂, Al₂O₃, Fe₂O₃), plus small amounts of gypsum added after burning.

Four principal (Bogue's) compounds:

CompoundFormulaAbbreviationNote
Tricalcium silicate3CaOSiO23\text{CaO}\cdot\text{SiO}_2C₃S (alite)Early strength, fast setting
Dicalcium silicate2CaOSiO22\text{CaO}\cdot\text{SiO}_2C₂S (belite)Slow, gives ultimate strength
Tricalcium aluminate3CaOAl2O33\text{CaO}\cdot\text{Al}_2\text{O}_3C₃AFlash set; least durable
Tetracalcium aluminoferrite4CaOAl2O3Fe2O34\text{CaO}\cdot\text{Al}_2\text{O}_3\cdot\text{Fe}_2\text{O}_3C₄AF (felite)Colour, low heat

Role of gypsum (CaSO42H2O\text{CaSO}_4\cdot2\text{H}_2\text{O}): Added (2–3%) to retard the flash setting caused by C₃A. It reacts with C₃A to form calcium sulphoaluminate (ettringite), giving workable setting time.

Setting and hardening (hydration): When water is added the compounds hydrate exothermically:

2(3CaOSiO2)+6H2O3CaO2SiO23H2O+3Ca(OH)22(3\text{CaO}\cdot\text{SiO}_2) + 6\text{H}_2\text{O} \rightarrow 3\text{CaO}\cdot2\text{SiO}_2\cdot3\text{H}_2\text{O} + 3\text{Ca(OH)}_2

The C-S-H gel (tobermorite gel) is the main binding phase. Setting is the initial stiffening into a solid mass; hardening is the subsequent gradual gain of strength as hydration continues and the gel crystallizes over days and weeks.

cement-chemistryportland-cementsetting-hardening
8short5 marks

What is a paint? State the essential constituents of a paint and the function of each. How does a paint differ from a varnish?

Paint: A paint is a mechanical dispersion (suspension) of one or more finely ground pigments in a liquid medium (vehicle), applied as a thin film to a surface for protection and decoration. On drying it forms an adherent, opaque, coloured coating.

Essential constituents and their functions:

ConstituentFunction
PigmentProvides colour, opacity (hiding power) and body; protects film from UV; e.g. TiO₂ (white), red lead.
Vehicle / drying oil (e.g. linseed oil)Binder/film-forming medium; holds pigment, gives adhesion, durability and water resistance.
Thinner / solvent (e.g. turpentine)Reduces viscosity for easy application; evaporates after application.
Drier (e.g. naphthenates of Co, Pb, Mn)Catalyses oxidation and polymerization of the oil, accelerating drying.
Filler / extender (e.g. BaSO₄, CaCO₃)Increases volume, reduces cost, improves durability and reduces cracking.
PlasticizerGives elasticity, prevents cracking of the film.

Paint vs varnish:

PaintVarnish
Contains pigment → opaque, coloured filmNo pigment → transparent/translucent film
Mainly for protection + decorationMainly to give a glossy, protective finish showing grain beneath
Made of pigment + vehicle + thinner + drierSolution of resin/drying oil in a solvent (no pigment)
paints-varnishesconstituentscoatings
9short5 marks

Define an explosive. Classify explosives on the basis of their performance with one example of each class. State two important characteristics that a good explosive should possess.

Explosive: An explosive is a substance (or mixture) which, on application of heat, shock, friction or detonation, undergoes a very rapid self-propagating exothermic chemical reaction, producing a large volume of hot gases and a sudden, enormous increase in pressure (an explosion).

Classification on the basis of performance:

  1. Primary (initiating) explosives: Extremely sensitive to heat/shock/friction; detonate readily and are used to set off other explosives (detonators). Example: mercury fulminate Hg(ONC)2\text{Hg(ONC)}_2, lead azide Pb(N3)2\text{Pb(N}_3)_2.
  2. Low (deflagrating) explosives: Burn rapidly (deflagrate) rather than detonate; produce a propelling/pushing action. Used as propellants. Example: gunpowder (black powder, KNO3\text{KNO}_3 + charcoal + sulphur), nitrocellulose.
  3. High (detonating) explosives: Detonate with a very high velocity shock wave producing a shattering (brisance) effect; relatively insensitive (need a detonator). Example: TNT (trinitrotoluene), RDX, dynamite (nitroglycerine based).

Two characteristics of a good explosive:

  1. It should be stable and safe to store/handle under normal conditions but explode reliably when initiated (controlled sensitivity).
  2. It should produce a large volume of gas and high heat (high power/brisance) per unit mass rapidly, and ideally be cheap and chemically stable over time.
explosivesclassificationsafety
10short5 marks

Define cell constant, specific conductance and equivalent conductance of an electrolytic solution. The resistance of a 0.01 N0.01\ \text{N} KCl solution in a conductivity cell is 250 Ω250\ \Omega. If the cell constant is 1.25 cm11.25\ \text{cm}^{-1}, calculate the specific conductance and the equivalent conductance of the solution.

Cell constant (GG^*): The ratio of the distance between the electrodes (ll) to the cross-sectional area of the electrodes (AA): G=l/AG^* = l/A. Unit: cm1\text{cm}^{-1}. It converts measured conductance into specific conductance.

Specific conductance (κ\kappa): The conductance of a solution contained between two electrodes of 1 cm² area held 1 cm apart, i.e. the conductance of 1 cm³ (unit volume) of the solution. Unit: S cm1\text{S cm}^{-1} (ohm⁻¹ cm⁻¹).

κ=1R×cell constant\kappa = \frac{1}{R} \times \text{cell constant}

Equivalent conductance (Λeq\Lambda_{eq}): The conductance of all the ions produced by 1 gram-equivalent of an electrolyte dissolved in a given volume of solution. Unit: S cm2eq1\text{S cm}^2\,\text{eq}^{-1}.

Λeq=κ×1000N\Lambda_{eq} = \frac{\kappa \times 1000}{N}

where NN = normality.

Step 1 — specific conductance:

κ=1R×cell constant=1250×1.25=0.005 S cm1\kappa = \frac{1}{R} \times \text{cell constant} = \frac{1}{250} \times 1.25 = 0.005\ \text{S cm}^{-1}

Specific conductance =5.0×103 Scm1= \mathbf{5.0 \times 10^{-3}\ S\,cm^{-1}}

Step 2 — equivalent conductance:

Λeq=κ×1000N=0.005×10000.01=50.01=500 S cm2eq1\Lambda_{eq} = \frac{\kappa \times 1000}{N} = \frac{0.005 \times 1000}{0.01} = \frac{5}{0.01} = 500\ \text{S cm}^2\,\text{eq}^{-1}

Equivalent conductance =500 Scm2eq1= \mathbf{500\ S\,cm^2\,eq^{-1}}

electrochemistryconductancespecific-conductivity
11short5 marks

Explain the ion-exchange (demineralization) process for the softening of hard water with the relevant reactions, and state how the exhausted resins are regenerated. State two advantages and one limitation of the ion-exchange process.

Ion-exchange (demineralization) process: Hard water is passed successively through two columns containing synthetic ion-exchange resins:

  • a cation exchanger (RH2\text{RH}_2 — resin with mobile H+\text{H}^+),
  • an anion exchanger (R’(OH)2\text{R'(OH)}_2 — resin with mobile OH\text{OH}^-).

Cation exchange (removes Ca²⁺, Mg²⁺):

RH2+Ca2+RCa+2H+\text{RH}_2 + \text{Ca}^{2+} \rightarrow \text{RCa} + 2\text{H}^+ RH2+Mg2+RMg+2H+\text{RH}_2 + \text{Mg}^{2+} \rightarrow \text{RMg} + 2\text{H}^+

Anion exchange (removes Cl⁻, SO₄²⁻, HCO₃⁻):

R’(OH)2+SO42R’SO4+2OH\text{R'(OH)}_2 + \text{SO}_4^{2-} \rightarrow \text{R'SO}_4 + 2\text{OH}^- R’(OH)2+2ClR’Cl2+2OH\text{R'(OH)}_2 + 2\text{Cl}^- \rightarrow \text{R'Cl}_2 + 2\text{OH}^-

The liberated H+\text{H}^+ and OH\text{OH}^- combine to give water (H++OHH2O\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}), so the outflow is almost ion-free (demineralized) water.

Regeneration of exhausted resins:

  • Exhausted cation resin is regenerated by washing with dilute acid (HCl or H₂SO₄):
RCa+2H+RH2+Ca2+\text{RCa} + 2\text{H}^+ \rightarrow \text{RH}_2 + \text{Ca}^{2+}
  • Exhausted anion resin is regenerated by washing with dilute alkali (NaOH):
R’SO4+2OHR’(OH)2+SO42\text{R'SO}_4 + 2\text{OH}^- \rightarrow \text{R'(OH)}_2 + \text{SO}_4^{2-}

The columns are rinsed and reused.

Advantages (two):

  1. Produces very low residual hardness (almost zero, ~2 ppm) and essentially ion-free (demineralized) water suitable for high-pressure boilers.
  2. The resins can be regenerated and reused many times, and the process can treat both acidic and alkaline water.

Limitation (one): The equipment and resins are expensive, and turbid/coloured water or water containing Fe/Mn must be pre-treated, otherwise the resin bed gets fouled and loses efficiency.

water-treatmention-exchangesoftening

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