Detailed Discussions On Sodium

ChemistryThe s-Block ElementsFor JEE aspirants

Sodium is the most important alkali metal in industry: the metal itself and four of its compounds, sodium carbonate, sodium chloride, sodium hydroxide and sodium hydrogencarbonate, are made on a huge scale from common salt. This page covers how sodium is extracted, the Solvay, Castner-Kellner and salt-purification processes with their equations, the properties and reactions of each compound, and the role of sodium and potassium ions in living cells. JEE Advanced asks for the process steps, the balanced equations and the reasons behind each choice.

On this page1Sodium metal2Why electrolysis3Solvay process4Washing soda5Common salt6Caustic soda7Baking soda8Na and K in the body9Solved examples

Key Formulas - Quick Reference

★ Must learnSodium metal: electrolysis of fused NaCl + $CaC l_{2}$ (Downs cell, about 873 K): cathode $N a^{+} + e^{-} \to Na$ , anode $2C l^{-} \to C l_{2} + 2 e^{-}$ .
★ Must learnSolvay: $N H_{4} HC O_{3} + NaCl \to N H_{4} Cl + NaHC O_{3}$ (NaHCO $_{3}$ precipitates), then $2NaHC O_{3} \to N a_{2} C O_{3} + C O_{2} + H_{2} O$ .
★ Must learnNH $_{3}$ recovery: $2N H_{4} Cl + Ca (OH)_{2} \to 2N H_{3} + CaC l_{2} + 2 H_{2} O$ ; overall $2NaCl + CaC O_{3} \to N a_{2} C O_{3} + CaC l_{2}$ .
$K_{2} C O_{3}$ cannot be made by Solvay: $KHC O_{3}$ is too soluble to precipitate.
Washing soda: $N a_{2} C O_{3} \cdot 10 H_{2} O$ $375 K$ $N a_{2} C O_{3} \cdot H_{2} O$ $> 373 K$ $N a_{2} C O_{3}$ (soda ash); solution alkaline by hydrolysis.
★ Must learnCastner-Kellner (Hg cathode): cathode $N a^{+} + e^{-}$ $Hg$ Na-amalgam; anode $C l^{-} \to \frac{1}{2} C l_{2} + e^{-}$ ; amalgam + water → NaOH + $H_{2}$ + Hg.
Pure NaCl: saturate crude brine with HCl gas (common ion effect); $CaC l_{2}$ , $MgC l_{2}$ stay dissolved.
★ Must learnBaking soda: $N a_{2} C O_{3} + H_{2} O + C O_{2} \to 2NaHC O_{3}$ ; plasma $N a^{+}$ 143, $K^{+}$ 5 mmol L $^{- 1}$ ; red cells $N a^{+}$ 10, $K^{+}$ 105.

1. Sodium Metal: Extraction and Uses

Sodium never occurs free; its main sources are rock salt and sea water (NaCl), Chile saltpetre ( $NaN O_{3}$ ) and borax. The metal is obtained by electrolysis of fused sodium chloride, mixed with calcium chloride, in the Downs cell. Calcium chloride lowers the melting point of the bath from 1074 K to about 873 K, which saves energy and reduces the loss of sodium vapour.

Figure 1: Sodium is made by electrolysing fused NaCl (with

CaC l_{2}

to lower the melting point); the liquid Na floats up and is drawn off. In water,

H_{2}

would be discharged instead of Na (NCERT 10.27b).

2NaCl electrolysis, fused 2Na + C l_{2}

1.1 Why not chemical reduction?

The alkali and alkaline earth metals are themselves the strongest reducing agents. To reduce $N a^{+}$ you would need an even stronger reducing agent, and the only ones (Li, K, Ca) are more expensive and are made by electrolysis anyway. Carbon cannot reduce sodium oxide at practical temperatures (and hot sodium reacts with carbon), while electrolysis of an aqueous solution discharges hydrogen, not sodium, at the cathode. So the metals are prepared by electrolysis of their fused chlorides.

Figure 2: The s-block metals are themselves the strongest reducing agents, so no cheaper chemical can reduce their ions; they are obtained by electrolysis of fused chlorides (NCERT 10.8).

Sodium is more useful than potassium as a metal and in its compounds: it is more abundant and cheaper, less reactive and so safer to handle, and its salts (NaCl, $N a_{2} C O_{3}$ , NaOH) are the ones industry needs in bulk.
Uses: Na/Pb alloy for the old anti-knock agents $PbE t_{4}$ and $PbM e_{4}$ ; liquid sodium as a coolant in fast breeder reactors; sodium vapour lamps (589 nm yellow light); as a reducing agent (Na in liquid $N H_{3}$ ).
Sodium peroxide is made by burning sodium in excess air: $2Na + O_{2} \to N a_{2} O_{2}$ (first $N a_{2} O$ at about 450 K in limited air, then $N a_{2} O_{2}$ at 573-673 K in excess).

Key idea

No chemical is a stronger reducing agent than sodium, so electrons at a cathode do the job: fused NaCl in, liquid Na and

C l_{2}

out.

2. Sodium Carbonate (Washing Soda), Na₂CO₃·10H₂O

2.1 The Solvay (ammonia-soda) process

Sodium carbonate is made by the Solvay process. It takes advantage of the low solubility of sodium hydrogencarbonate, which precipitates when sodium chloride reacts with ammonium hydrogencarbonate. The ammonium hydrogencarbonate is itself made by passing $C O_{2}$ into a concentrated solution of NaCl saturated with ammonia, where ammonium carbonate forms first:

2N H_{3} + H_{2} O + C O_{2} (N H_{4})_{2} C O_{3}

(N H_{4})_{2} C O_{3} + H_{2} O + C O_{2} 2N H_{4} HC O_{3}

N H_{4} HC O_{3} + NaCl N H_{4} Cl + NaHC O_{3} ↓

The sodium hydrogencarbonate crystals are filtered off and heated to give sodium carbonate; the $C O_{2}$ released goes back to the carbonating tower:

2NaHC O_{3} heat N a_{2} C O_{3} + C O_{2} + H_{2} O

Ammonia is recovered by treating the ammonium chloride solution with slaked lime from the lime kiln, and calcium chloride is the by-product:

CaC O_{3} heat CaO + C O_{2}

CaO + H_{2} O Ca (OH)_{2}

2N H_{4} Cl + Ca (OH)_{2} 2N H_{3} + CaC l_{2} + 2 H_{2} O

Figure 3: The Solvay process as a loop.

N H_{3}

and half of the

C O_{2}

are recycled; the only raw materials are brine and limestone, and the only by-product is

CaC l_{2}

JEE Advanced

The net reaction is 'impossible' on its own. Adding all the steps gives $2NaCl + CaC O_{3} \to N a_{2} C O_{3} + CaC l_{2}$ . Mixed directly, limestone and salt do nothing: the reverse reaction ( $N a_{2} C O_{3}$ + $CaC l_{2}$ precipitating $CaC O_{3}$ ) is the one that runs. The Solvay process gets round this by using ammonia as a recycled helper and by pulling $NaHC O_{3}$ out of solution as a precipitate at each pass, so no step has to go against its own equilibrium. Two consequences are asked: (i) the only raw materials consumed are NaCl and $CaC O_{3}$ (plus fuel), and (ii) for every 2 mol NaCl, 1 mol $CaC l_{2}$ is produced as waste.

The same logic explains NCERT 10.13: $KHC O_{3}$ (33.7 g per 100 g water) is far more soluble than $NaHC O_{3}$ (9.6 g), so it never precipitates and the potassium equilibrium cannot be pulled forward.

Figure 4: The Solvay process depends on

NaHC O_{3}

being the least soluble salt in the mixture.

KHC O_{3}

is about 3.5 times more soluble, so it does not precipitate and

K_{2} C O_{3}

cannot be made this way (NCERT 10.13).

Exam Trick Solvay in one line: salt + ammonia + $C O_{2}$ → baking soda falls out → heat it. Ammonia and half the

C O_{2}

go round in circles; limestone supplies the rest of the

C O_{2}

and the lime that frees the ammonia;

CaC l_{2}

is the waste.

2.2 Properties and uses

Sodium carbonate is a white crystalline solid that exists as the decahydrate, washing soda. It is readily soluble in water. On heating, the decahydrate loses water of crystallisation to form the monohydrate; above 373 K the monohydrate becomes completely anhydrous and changes to a white powder called soda ash.

N a_{2} C O_{3} \cdot 10 H_{2} O 375 K N a_{2} C O_{3} \cdot H_{2} O + 9 H_{2} O

N a_{2} C O_{3} \cdot H_{2} O > 373 K N a_{2} C O_{3} + H_{2} O

The carbonate ion is hydrolysed by water, so the solution is alkaline:

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

Figure 5: Washing soda loses 9 of its 10 waters at 375 K (56.7% of its mass) and the last one above 373 K. In solution

C O_{3}^{2 -}

takes a proton from water, so

N a_{2} C O_{3}

solution is alkaline (NCERT 10.27a).

Water softening, laundering and cleaning.
Manufacture of glass, soap, borax and caustic soda.
Paper, paints and textile industries; an important laboratory reagent in qualitative and quantitative analysis.

Key idea

Solvay works because $NaHC O_{3}$ is the least soluble salt in the mixture. Washing soda holds 10 waters; soda ash holds none; both give an alkaline solution.

Quick Recall: tap to check

Why is the Solvay process not used for

K_{2} C O_{3}

KHC O_{3}

is too soluble to precipitate when ammonium hydrogencarbonate is added to KCl solution.

What is the by-product of the Solvay process?

Calcium chloride, from

2N H_{4} Cl + Ca (OH)_{2} \to 2N H_{3} + CaC l_{2} + 2 H_{2} O

Why is a solution of

N a_{2} C O_{3}

alkaline?

C O_{3}^{2 -}

takes a proton from water:

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

3. Sodium Chloride, NaCl

The most abundant source of sodium chloride is sea water, which contains 2.7 to 2.9% of the salt by mass. In tropical countries like India, common salt is obtained by solar evaporation of sea water (about 50 lakh tonnes a year). The crude salt contains sodium sulphate, calcium sulphate, calcium chloride and magnesium chloride. $CaC l_{2}$ and $MgC l_{2}$ are the troublesome impurities because they are deliquescent: they absorb moisture from the air.

To purify it, the crude salt is dissolved in the minimum amount of water and filtered to remove insoluble impurities. The solution is then saturated with hydrogen chloride gas. Crystals of pure NaCl separate out, while calcium and magnesium chlorides, being more soluble, remain in solution.

Figure 6: Saturating the solution with HCl raises

[C l^{-}]

, so the ionic product of NaCl exceeds its solubility product and pure NaCl crystallises; the more soluble, deliquescent

CaC l_{2}

and

MgC l_{2}

stay behind.

Properties: NaCl melts at 1074 K. Its solubility is 36.0 g in 100 g of water at 273 K and does not increase appreciably with temperature (so it is crystallised by evaporation, not by cooling).
Uses: common or table salt; starting material for $N a_{2} O_{2}$ , NaOH and $N a_{2} C O_{3}$ (and for Na metal and chlorine).

4. Sodium Hydroxide (Caustic Soda), NaOH

Sodium hydroxide is prepared commercially by the electrolysis of brine in the Castner-Kellner cell, using a mercury cathode and a carbon anode. Sodium discharged at the cathode dissolves in mercury to form sodium amalgam, and chlorine is evolved at the anode. The amalgam is then treated with water to give sodium hydroxide and hydrogen.

Figure 7: On mercury,

N a^{+}

is discharged instead of

H^{+}

(high overvoltage of

H_{2}

on Hg) and the sodium is locked away as amalgam, so NaOH forms only in the separate denuder, away from the chlorine.

N a^{+} + e^{-} Hg Na - amalgam

C l^{-} ½ C l_{2} + e^{-}

2Na - amalgam + 2 H_{2} O 2NaOH + 2Hg + H_{2}

Why mercury? On a mercury surface, hydrogen has a very high overvoltage, so $N a^{+}$ is discharged in preference to $H^{+}$ ; the sodium dissolves in the mercury at once and cannot react with water or chlorine inside the cell. The NaOH is made only in the separate chamber, so it does not mix with the chlorine (which would give hypochlorite).

4.1 Properties and reactions

Sodium hydroxide is a white, translucent, deliquescent solid that melts at 591 K. It dissolves readily in water to give a strongly alkaline solution. The solution absorbs $C O_{2}$ from the air to form $N a_{2} C O_{3}$ , which is why NaOH solutions must be kept stoppered.

Reagent	Reaction	Remark
$C O_{2}$	$2NaOH + C O_{2} \to N a_{2} C O_{3} + H_{2} O$	excess $C O_{2}$ gives $NaHC O_{3}$
$C l_{2}$ , cold, dilute	$C l_{2} + 2NaOH \to NaCl + NaOCl + H_{2} O$	hypochlorite (Cl +1)
$C l_{2}$ , hot, conc.	$3C l_{2} + 6NaOH \to 5NaCl + NaCl O_{3} + 3 H_{2} O$	chlorate (Cl +5)
Al (or Zn)	$2Al + 2NaOH + 6 H_{2} O \to 2Na [Al (OH)_{4}] + 3 H_{2}$	amphoteric metals give $H_{2}$
$N H_{4} Cl$	$N H_{4} Cl + NaOH \to NaCl + N H_{3} + H_{2} O$	test for ammonium salts
$Si O_{2}$	$Si O_{2} + 2NaOH \to N a_{2} Si O_{3} + H_{2} O$	why NaOH attacks glass

Uses: manufacture of soap, paper, artificial silk (rayon) and many chemicals; petroleum refining; purification of bauxite (Bayer process); mercerising cotton in the textile industry; preparation of pure fats and oils; laboratory reagent.

Exam Trick Cold chlorine is kind, hot chlorine is harsh. Cold dilute NaOH takes

C l_{2}

only to +1 (NaOCl, bleach); hot concentrated NaOH takes it to +5 (

NaCl O_{3}

). Both are disproportionation: the rest of the chlorine goes to

- 1

as NaCl.

Key idea

Castner-Kellner uses mercury so that sodium, not hydrogen, is discharged and NaOH forms away from

C l_{2}

5. Sodium Hydrogencarbonate (Baking Soda), NaHCO₃

Sodium hydrogencarbonate is called baking soda because it decomposes on heating to give bubbles of $C O_{2}$ , which leave holes in cakes and pastries and make them light and fluffy. It is made by saturating a solution of sodium carbonate with $C O_{2}$ ; the white crystalline powder of NaHCO $_{3}$ , being less soluble, separates out.

N a_{2} C O_{3} + H_{2} O + C O_{2} 2NaHC O_{3}

It is a mild antiseptic for skin infections, an antacid, and is used in fire extinguishers (acid + NaHCO $_{3}$ gives a stream of $C O_{2}$ ). In baking powder it is mixed with a solid acid (such as tartaric acid) so that $C O_{2}$ is released even without strong heating. Its solution is only weakly alkaline (pH about 8.3), unlike $N a_{2} C O_{3}$ .

Figure 8: The three alkaline sodium compounds convert into each other with

C O_{2}

, water, heat or lime. Lime turns sodium carbonate into caustic soda (not the reverse).

5.1 From common salt to every sodium compound

Figure 9: Common salt is the starting point for every industrial sodium compound (NCERT 10.16).

Figure 10: Two tests sort the four compounds: acid (does

C O_{2}

come off?) and heat (only

NaHC O_{3}

releases

C O_{2}

, which turns lime water milky, on gentle heating).

Quick Recall: tap to check

Why is mercury used as the cathode in the Castner-Kellner cell?

Hydrogen has a high overvoltage on Hg, so

N a^{+}

is discharged and dissolves as amalgam; NaOH forms separately, away from

C l_{2}

Why are NaOH solutions kept stoppered?

They absorb

C O_{2}

from air:

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

How is NaHCO

_{3}

made from

N a_{2} C O_{3}

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

; the less soluble

NaHC O_{3}

crystallises.

6. Biological Importance of Sodium and Potassium

A typical 70 kg human contains about 90 g of Na and 170 g of K, against only 5 g of iron and 0.06 g of copper. Sodium ions are found mainly outside cells, in blood plasma and the interstitial fluid. They take part in transmitting nerve signals, regulate the flow of water across cell membranes, and help transport sugars and amino acids into cells. Potassium ions are the most abundant cations within cell fluids, where they activate many enzymes, take part in the oxidation of glucose to produce ATP and, with sodium, carry nerve signals.

The concentrations differ sharply across the membrane. In blood plasma sodium is about 143 mmol L $^{- 1}$ and potassium only about 5 mmol L $^{- 1}$ ; inside red blood cells they change to about 10 mmol L $^{- 1}$ ( $N a^{+}$ ) and 105 mmol L $^{- 1}$ ( $K^{+}$ ). These gradients show that a discriminating mechanism, the sodium-potassium pump, operates across cell membranes. It consumes more than one-third of the ATP used by a resting animal, about 15 kg of ATP per 24 h in a resting human.

Figure 11:

N a^{+}

dominates outside cells,

K^{+}

inside. Keeping this gradient needs the

N a^{+} / K^{+}

pump, which uses over one-third of the ATP of a resting animal.

Sodium ion,

N a^{+}

Mainly outside cells (plasma 143 mmol L

^{- 1}

). Nerve signals, water balance, transport of sugars and amino acids into cells.

Potassium ion,

K^{+}

Mainly inside cells (105 mmol L

^{- 1}

). Activates enzymes, glucose oxidation to ATP, nerve signals with

N a^{+}

Figure 12: Sodium on one page: one metal, four industrial compounds, three processes (Downs, Solvay, Castner-Kellner) and one pump.

7. Solved Examples

Solved Example 1

In the Solvay process, which substance is recycled within the plant?
(A) NaCl
(B)

N H_{3}

(C)

CaC l_{2}

(D)

N a_{2} C O_{3}

Solution:

Answer: (B). Ammonia is regenerated from $N H_{4} Cl$ with slaked lime ( $2N H_{4} Cl + Ca (OH)_{2} \to 2N H_{3} + CaC l_{2} + 2 H_{2} O$ ) and sent back to the absorber. NaCl is consumed, $CaC l_{2}$ is the by-product and $N a_{2} C O_{3}$ is the product.

Solved Example 2

What percentage of its mass does washing soda,

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

, lose when heated to the monohydrate? (Na = 23.0, C = 12.0, O = 16.0, H = 1.0)

Solution:

$M = 2 (23) + 12 + 48 + 10 (18) = 286$ g mol $^{- 1}$ . Loss = 9 $H_{2} O$ = 162 g.

% loss $= \frac{162}{286} \times 100 =$ 56.6% (56.7% with exact masses).

Solved Example 3

Calculate the pH of 0.10 M

N a_{2} C O_{3}

solution. For

H_{2} C O_{3}

K_{a 2} = 4.7 \times 1 0^{- 11}

at 298 K.

Solution:

$K_{b} = \frac{K _{w}}{K _{a 2}} = \frac{1.0 \times 1 0 ^{- 14}}{4.7 \times 1 0 ^{- 11}} = 2.13 \times 1 0^{- 4}$ .

$[O H^{-}] = K_{b} C = 2.13 \times 1 0^{- 5} = 4.6 \times 1 0^{- 3}$ M, so pOH = 2.34 and pH = 11.66.

The second hydrolysis step (to $H_{2} C O_{3}$ ) is negligible.

Solved Example 4

Which statements about the Castner-Kellner cell are correct? (One or more options.)
(A) Mercury acts as the cathode.
(B) Hydrogen is liberated at the cathode inside the electrolysis tank.
(C) Chlorine is liberated at the anode.
(D) Sodium hydroxide is formed when the amalgam reacts with water.

Solution:

Answer: (A), (C), (D). Because of the high overvoltage of $H_{2}$ on mercury, $N a^{+}$ (not $H^{+}$ ) is discharged at the cathode; $H_{2}$ appears only in the denuder, when the amalgam meets water. So (B) is wrong.

Solved Example 5

Chlorine is passed into (i) cold dilute NaOH and (ii) hot concentrated NaOH. Give the chlorine-containing products and the oxidation state of Cl in each.

Solution:

(i) $C l_{2} + 2NaOH \to NaCl + NaOCl + H_{2} O$ : NaCl (Cl $- 1$ ) and NaOCl (Cl +1).

(ii) $3C l_{2} + 6NaOH \to 5NaCl + NaCl O_{3} + 3 H_{2} O$ : NaCl (Cl $- 1$ ) and $NaCl O_{3}$ (Cl +5). Both are disproportionation reactions.

Solved Example 6

Identify A, B and C: brine saturated with

N H_{3}

and

C O_{2}

gives a white precipitate A; A on heating gives B,

C O_{2}

and water; B with slaked lime gives C and a white precipitate.

Solution:

A = $NaHC O_{3}$ (Solvay precipitate), B = $N a_{2} C O_{3}$ ( $2NaHC O_{3} \to N a_{2} C O_{3} + C O_{2} + H_{2} O$ ), C = NaOH ( $N a_{2} C O_{3} + Ca (OH)_{2} \to 2NaOH + CaC O_{3}$ ; the precipitate is $CaC O_{3}$ ).

Solved Example 7

Assuming 100% conversion, what mass of

N a_{2} C O_{3}

can be made from 1.00 tonne of NaCl by the Solvay process? (Na = 23.0, Cl = 35.5, C = 12.0, O = 16.0)

Solution:

Overall: $2NaCl + CaC O_{3} \to N a_{2} C O_{3} + CaC l_{2}$ . 117 g NaCl gives 106 g $N a_{2} C O_{3}$ .

Mass $= 1000 \times \frac{106}{117} =$ 906 kg.

Practice Questions

Explain why alkali and alkaline earth metals cannot be obtained by chemical reduction methods. (NCERT 10.8)Answer: They are themselves the strongest reducing agents, so no cheaper reducing agent can reduce their ions; they are made by electrolysis of fused chlorides.
Discuss the various reactions that occur in the Solvay process. (NCERT 10.12)Answer: $2N H_{3} + H_{2} O + C O_{2} \to (N H_{4})_{2} C O_{3}$ ; $(N H_{4})_{2} C O_{3} + H_{2} O + C O_{2} \to 2N H_{4} HC O_{3}$ ; $N H_{4} HC O_{3} + NaCl \to N H_{4} Cl + NaHC O_{3}$ ; $2NaHC O_{3} \to N a_{2} C O_{3} + C O_{2} + H_{2} O$ ; $2N H_{4} Cl + Ca (OH)_{2} \to 2N H_{3} + CaC l_{2} + 2 H_{2} O$ .
Potassium carbonate cannot be prepared by the Solvay process. Why? (NCERT 10.13)Answer: $KHC O_{3}$ is too soluble to be precipitated by ammonium hydrogencarbonate from KCl solution.
Starting with sodium chloride, how would you prepare (i) sodium metal (ii) sodium hydroxide (iii) sodium peroxide (iv) sodium carbonate? (NCERT 10.16)Answer: (i) Electrolysis of fused NaCl with CaCl2 (Downs cell); (ii) electrolysis of brine in the Castner-Kellner cell; (iii) burn the sodium from (i) in excess air; (iv) Solvay process.
Describe two important uses of (i) caustic soda (ii) sodium carbonate. (NCERT 10.18)Answer: (i) Soap and paper manufacture; petroleum refining, bauxite purification, mercerising cotton. (ii) Water softening and laundering; manufacture of glass, soap and borax.
State why (a) a solution of $N a_{2} C O_{3}$ is alkaline, (b) alkali metals are prepared by electrolysis of their fused chlorides, (c) sodium is found to be more useful than potassium. (NCERT 10.27)Answer: (a) Hydrolysis: $C O_{3}^{2 -} + H_{2} O ⇌ HC O_{3}^{-} + O H^{-}$ . (b) They are the strongest reducing agents; aqueous electrolysis gives $H_{2}$ instead. (c) Na is more abundant, cheaper and less reactive, and its compounds are needed in bulk.
What volume of $C O_{2}$ at STP (22.7 L mol $^{- 1}$ ) is released when 8.4 g of $NaHC O_{3}$ is heated completely?Answer: 0.100 mol NaHCO $_{3}$ gives 0.050 mol $C O_{2}$ = 1.14 L.

Common Mistakes to Avoid

Watch out

Writing the ammonia recovery step as $2N H_{4} Cl + Ca (OH)_{2} \to 2N H_{3} + CaC l_{2} + H_{2} O$ ; it needs 2 $H_{2} O$ to balance.
Saying $N a_{2} C O_{3}$ precipitates in the carbonating tower. The precipitate is $NaHC O_{3}$ ; $N a_{2} C O_{3}$ is made by heating it.
Explaining the failure for $K_{2} C O_{3}$ by the solubility of $K_{2} C O_{3}$ itself. It is $KHC O_{3}$ that is too soluble.
Thinking hydrogen is evolved at the mercury cathode in the Castner-Kellner cell. Sodium is discharged; $H_{2}$ comes from the amalgam reacting with water.
Electrolysing aqueous NaCl to make sodium metal: it gives $H_{2}$ at the cathode. The chloride must be fused.
Mixing up the products of chlorine with NaOH: cold dilute gives NaOCl (+1), hot concentrated gives $NaCl O_{3}$ (+5).
Writing that lime is used to make sodium carbonate from caustic soda. Lime makes caustic soda from sodium carbonate: $N a_{2} C O_{3} + Ca (OH)_{2} \to 2NaOH + CaC O_{3}$ .
Placing $K^{+}$ outside and $N a^{+}$ inside cells. $N a^{+}$ is outside (plasma), $K^{+}$ inside.

Frequently Asked Questions

What is the Solvay process?

The Solvay or ammonia soda process makes sodium carbonate from common salt and limestone. Brine saturated with ammonia is treated with carbon dioxide, and the sparingly soluble sodium hydrogencarbonate precipitates. Heating it gives sodium carbonate. Ammonia is recovered with slaked lime from the limestone, leaving calcium chloride as the by-product.

Why can potassium carbonate not be made by the Solvay process?

The Solvay process works only because sodium hydrogencarbonate is sparingly soluble and precipitates from the reaction mixture. Potassium hydrogencarbonate is about three and a half times more soluble, so it stays in solution when ammonium hydrogencarbonate is added to potassium chloride, and the reaction cannot be driven forward.

Why is a mercury cathode used in the Castner-Kellner process?

Hydrogen has a very high overvoltage on mercury, so sodium ions are discharged instead of hydrogen ions. The sodium dissolves in the mercury as an amalgam, which is carried to a separate chamber and treated with water to give sodium hydroxide and hydrogen. This keeps the sodium hydroxide away from the chlorine formed at the anode.

How is common salt purified?

Crude salt from sea water is dissolved in the minimum amount of water and filtered. Hydrogen chloride gas is passed into the saturated solution. The extra chloride ions push the equilibrium towards solid sodium chloride, the common ion effect, so pure crystals separate, while the more soluble calcium and magnesium chlorides stay dissolved.

What happens when washing soda is heated?

Washing soda, sodium carbonate decahydrate, loses nine of its ten water molecules at about 375 K and becomes the monohydrate. Above 373 K the monohydrate loses its last water and becomes anhydrous sodium carbonate, a white powder called soda ash. Unlike sodium hydrogencarbonate, it does not give off carbon dioxide.

Why is sodium obtained by electrolysis of fused sodium chloride and not from its solution?

Sodium is such a strong reducing agent that no ordinary chemical can reduce its compounds. Electrolysis of an aqueous solution would discharge hydrogen at the cathode, not sodium. So molten sodium chloride, mixed with calcium chloride to lower its melting point to about 873 K, is electrolysed in the Downs cell.

How is sodium chemistry examined in JEE Advanced?

JEE Advanced usually tests sodium through process questions: the reactions and recycled substances of the Solvay process, why potassium carbonate cannot be made this way, electrode reactions in the Castner-Kellner cell, reactions of sodium hydroxide with chlorine or aluminium, and identifying compounds in a sequence of reactions.

Which sodium equations are most important for JEE Advanced?

For JEE Advanced learn the five Solvay equations, including ammonia recovery with two molecules of water, the decomposition of sodium hydrogencarbonate, the Castner-Kellner electrode reactions, and sodium hydroxide with cold and hot chlorine. Also learn the dehydration of washing soda and the hydrolysis of carbonate that makes its solution alkaline.

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1. Sodium Metal: Extraction and Uses

1.1 Why not chemical reduction?

2. Sodium Carbonate (Washing Soda), Na2CO3·10H2O

2.1 The Solvay (ammonia-soda) process

2.2 Properties and uses

3. Sodium Chloride, NaCl

4. Sodium Hydroxide (Caustic Soda), NaOH

4.1 Properties and reactions

5. Sodium Hydrogencarbonate (Baking Soda), NaHCO3

5.1 From common salt to every sodium compound

6. Biological Importance of Sodium and Potassium

7. Solved Examples

Common Mistakes to Avoid

Frequently Asked Questions

What is the Solvay process?

Why can potassium carbonate not be made by the Solvay process?

Why is a mercury cathode used in the Castner-Kellner process?

How is common salt purified?

What happens when washing soda is heated?

Why is sodium obtained by electrolysis of fused sodium chloride and not from its solution?

How is sodium chemistry examined in JEE Advanced?

Which sodium equations are most important for JEE Advanced?

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2. Sodium Carbonate (Washing Soda), Na₂CO₃·10H₂O

5. Sodium Hydrogencarbonate (Baking Soda), NaHCO₃