KOH → K⁺ + OH^-
10^-5M    10^-5M 10^-5M
[OH^-] = 10^-5M
pH = 14 - pOH
pH= 14-(-log [OH^-])
= 14 + log [OH^-]
= 14 + log 10^-5
= 14 - 5 = 9

PbI_2(s) ⇌  Pb²⁺_(aq)  + 2I^-_(aq) 
K_sp = (s) (2s)²
3.2 × 10^-8 = 4s³
s = (3.2 × 10^-8/4)^1/3
= ( 8 x  10^-9)^1/3
= 2 x 10^-3 M

BF₃ → electron deficient →  Lewis acid
PF₃ → electron rich →  Lewis base
CO → having lone pair of electron →  Lewis base
F → unshared pair of electron →  Lewis base

H₂O + H₂O ⇌  H₃O⁺ + OH^-
acid 1  base 1  acid 2   base 2
HF + H₂O ⇌  H₃O⁺ + F^-
acid 1  base 1  acid 2   base 2
∴ Conjugate bases are OH^- and F^- respectively.

BaSO₄ ⇌ Ba²⁺ + SO₄^2-
K_sp = (s) (s)
K_sp = (s)²
= (2.42 × 10^-3gL^-1)²
\(=\frac{2.42 \times 10^{-3} gL^{-1}}{233 \text{ g mol}^{-1}}\)
= (0.01038 x 10^-3)²
= (1.038 x  x 10^-5)²
= 1.077 x 10^-10
= 1.08 × 10^-10mol²L^-2

i) HNO₃ \(\stackrel{H_{2} O}{\rightleftharpoons}\) H⁺_(aq) + NO^-_3(aq) [Acid]
ii) Ba(OH)₂ \(\stackrel{H_{2} O}{\rightleftharpoons}\) Ba²⁺_(aq) + 2OH^-_(aq) [base]
iii) H₃PO₄  \(\stackrel{H_{2} O}{\rightleftharpoons}\) 2H⁺_(aq) + HPO₄^2-_(aq) [Acid]
iv) CH₃COOH  \(\stackrel{H_{2} O}{\rightleftharpoons}\) CH₃COO^- _(aq) + H⁺_(aq) [Acid]

pH=-log[H⁺]
For weak acids,
\(= \sqrt{k_a \times C}\)
=\(\sqrt{1.8\times10^{-5}\times0.1}\)
=\(1.34 \times10^{-3}\) M
\(pH=-\log(1.34\times10^{-3})\)
= 3-log1.34
= 3-0.1271
= 2.8729 \(\simeq\) 2.87

Given that \(\alpha=1.20\)%=\(\frac{1.20}{100}\times1.2\times10^{-2}\)
\(K_{a}=\alpha^{2}c\)
\(=(1.2\times10^{-2})^{2}(0.1)=1.44\times10^{-4}\times10^{-1}\)
=\(1.44\times10^{-5}\)

\(\underset{0.001M}{HCl}\overset{H_{2}O}{\rightleftharpoons }\underset{0.001M}{H_{3}O^{+}}+\underset{0.001M}{Cl^{-}}\)
H₃O⁺ from the auto ionisation of H₂O (10^-7M) is negligible when compared to the H₃O⁺ from 10^-3M HCl.
Hence [H₃O⁺] = 0.001 mol dm^-3
pH = -log₁₀ [H₃O⁺]
= -log₁₀^(0.001)
= -log₁₀(10^-3) = 3

a) Primary and secondary nitroalkanes, having α-H, also show an equilibrium mixture of two tautomers namely nitro - and aci - form

b) Difference:

(i) The Chalcogens belong to group (16).
(ii) The group consists of elements: Oxygen, Sulphur, Selenium, Tellurium and Polonium.
(iii) These are ore forming elements as most of the ores are oxides and sulphides.
(iv) Chalcos meaning 'ore formers'.

a) \(BaSO_{4}(s)\overset{H_{2}O}{\rightleftharpoons }Ba^{2+}(aq)+SO^{2+}_{4}(aq)\)
\(K_{sp}=[Ba^{2+}][SO^{2-}_{4}]\) =   (s)   (s)
K_sp = s²
b) \(Ag_{2}CrO_{4}(s)\overset{H_{2}O}{\rightleftarrows }2Ag^{+}(aq)+CrO_{4}^{2-}(aq)\)
\(K_{sp}=[Ag^{+}]^{2}[CrO)^{2-}_{4}]\)
=  (2s)² (s)
K_sp = 4S³

(a) CH₃COONa is a salt of weak acid
(CH₃COOH) and a strong base (NaOH).
Hence, the solutions is alkaline due to hydrolysis.
\(CH_{3}COO^{-}_{(aq)}+H_{2}O_{(aq)}\rightleftharpoons CH_{3}COOH_{(aq)}+OH^{-}_{(aq)}\)
(i)\(h=\sqrt{\frac{K_{w}}{K_{a}\times C}}\)
Given that pK_a =4.74
pK_a = -log K_a
ie., K_a = antilog of (-pK_a)
= antilog of (-4.74)
= antilog of (-5 + 0.26)
= 10^-5 \(\times\) 1.8 = 1.8 \(\times\) 10^-5
[antilog of 0.26 = 1.82 \( \simeq\) 1.8]
\(\therefore\) h=\(\sqrt{\frac{1\times10^{-14}}{1.8\times10^{-5}\times0.1}}\)
h = 7.5 x 10^-5
(ii) \(K_{h}=\frac{K_{w}}{K_{a}}=\frac{1\times10^{-14}}{1.8\times10^{-5}}\)
\(=5.56\times10^{-10}\)
iii) \(pH=7+\frac{pK_{a}}{2}+\frac{logC}{2}\)
= \(7+\frac{4.74}{2}+\frac{log0.1}{2}\)
= 7 + 2.37 - 0.5
= 8.87

If we do not consider [H₃O]⁺ from the ionisation of H₂O,
then [H₃O⁺] = [HCl] = 10^-7M
i.e., pH = 7, which is a pH of a neutral solution. We know that HCl solution is acidic whatever may be the concentration of HCl i.e, the pH value should be less than 7. In this case the concentration of the acid is very low (10^-7M) Hence, the H₃O⁺ (10^-7M) formed due to the auto ionisation of water cannot be neglected.
so, in this case we should consider  [H₃O⁺] from ionisation of H₂O
[H₃O⁺] = 10^-7 (from HCl) + 10^-7 (from water)
= 10^-7 (1+1)
= \(2\times10^{-7}\)
pH = -log₁₀[H₃O⁺]
=\(-\log_{10}(2\times10^{-7})=-[\log2+\log_{10}10^{-7}]\)
=\(-\log2-(-7)\log_{10}^{10}\)
= 7-log²
= 7-0.3010 = 0.6990 = 6.70
= 6.70

\(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} \rightleftharpoons 3 \mathrm{Ca}^{2+}_{(aq)}+2 \mathrm{PO}_{4_{(aq)}}^{3-}\\\)
(s)                           (3s)             (2s)
\(K_{sp}=[Ca^{2+}]^{3}[PO_{4}^{3-}]^{2}\)
\(K_{sp}=(3s)^{3}(2s)^{2}\)
\(K_{sp}=27s^{3}.4s^{2}\)
\(K_{sp}=108s^{5}\)
(or)
\(K_{s p} =m^{m} \cdot n^{n} \cdot(s)^{m+n} \)
\(K_{\text {sp }} =3^{3} \cdot 2^{2} \cdot(s)^{3+2} \)
\(K_{s p} =27 \times 4 \times(s)^{5} \)
\(=108(s)^{5}=108 s^{5}\)

Considering Ba(OH)₂ to be a strong base:
\(\text { Normality } =\text { Molarity } \times \text { Acidity }  \)
\(=1.5 \times 10^{-3} \times 2\)
\({\left[\mathrm{OH}^{-}\right] } =3 \times 10^{-3}  \)
\(\mathrm{pOH} =-\log _{10}[\mathrm{OH}^-]  \)
\(=-\log _{10}\left(3 \times 10^{-3}\right)  \)
\(=-\left[\log _{10} 3+3 \log 10\right]  \)
= 3-log 3 
= 3-0.4771 
= 2.5229 
\(\mathrm{pH} =14-\mathrm{pOH} \)
= 14-2.5229 
pH = 11.4771 = 11.48

(i) Sulphuric acid is used in the manufacture of fertilisers, ammonium sulphate and super phosphates and other chemicals such as HCl, HNO₃.
(ii) It is used as a drying agent and also used in the preparation of pigments, explosives etc.

(i) Helium is used to provide inert atmosphere in electric-arc welding of metals.
(ii) Helium has lowest boiling point hence used in cryogenics.
(iii) It is much less denser than air and hence used for filling air balloons.

Each halogen combines with other halogens to form a series of compounds are called interhalogen compounds.
Example: AB type: BrF
AB₃ type: ICI₃

Na₂CO₃ is a salt of weak acid H₂CO₃ and a strong base NaOH
\(\therefore\) Na₂CO_3(aq)  ⟶  2Na⁺_(aq) + CO₃^2-_(aq)
CO₃^2-_(aq) + H₂O(l) ⇌ HCO₃^-_(aq) + OH^-_(aq)
i) pK_a = 10.26
pK_a =- log K_a
K_a = antilog of (-pK_a)
K_a = antilog (- 10.26)
= antilog \((\bar{11} .74)\)
= 5.5 x 10^-11
\(\therefore h =\sqrt { \frac { { K }_{ w } }{ { K }_{ a } \times C } } \)
\(h =\sqrt { \frac { 1\times { 10 }^{ -14 } }{ 5.5\times { 10 }^{ -11 } \times { 0.05 }} } =\sqrt { 9\times { 10 }^{ -4 } } = 6.03 \times 10^{-2}\)
ii) K_h = K_w/K_a
= 1 x 10^-14/5.5 x 10^-11
= 1.8 x 10^-4
iii) pH = 7 + pK_a/2 + log C/2
= 7 + 10.26/2 = log 0.05/2
= 7 + 5.13 - 0.65
= 11. 48

a) Dissociation of buffer components
\(NH_4OH_{aq} \rightleftharpoons NH^{+}_{4(aq)}+OH^{-}_{(aq)}\)
NH₄Cl  ⟶ NH₄ ⁺ + Cl⁺
Addition of H⁺
The added H⁺ ions are neutralized by NH₄OH and there is no appreciable decrease in pH
NH₄OH_(aq) + H⁺  ⟶ NH_4(aq) ⁺ + H₂O_(l)
Addition of OH^-
NH₄⁺_(aq)OH^-_(aq)  ⟶ NH₄ ⁺ + OH_(aq)
The added OH^- ions react with NH₄⁺ to produce unionized NH₄OH. Since NH₄OH is a weak base, there is no appreciable increase in pH.
b) pH = pK_a + log [Acid]/[Salt]
pk_a = -log₁₀ K_a
pk_a = -log 1.8 x 10^-5
=- [log 1.8-5 log 10]
= 5-log 1.8
pK_a= 5-0.2553
pK_a = 4.7447
pH = 4.7447 + log (0.4/0.4)
pH = 4.7447 + log 1
pH = 4.7447 + 0
pH = 4.74
ii) Addition of 0.01 mol HCl to 500 ml of buffer
Added [H⁺] = 0.01 mol/500mL = 0.01 mol/1/2 L
CH₃COOH_(aq)  ⇌ CH₃COO^-_(aq) + H⁺_{(aq)
\(\underset{0.4-\alpha}{\mathrm{CH}_{3} \mathrm{COONa}_{\text {(aq) }}} \rightarrow \underset{\alpha} {\mathrm{CH}_{3} \mathrm{COO}_{(\text {aq })}^{-}}+\underset{\alpha}{\mathrm{Na}^{+}{ }_{(aq)}}\)
\(\underset{0.02}{\mathrm{CH}_{3} \mathrm{COO^-}+ HCl} \rightarrow \underset{0.02} {\mathrm{CH}_{3} \mathrm{COOH}}+\underset{0.02}{\mathrm{Cl}^{+}}\)}
[CH₃COOH] = 0.4 - a + 0.02
= 0.42 - a \(\simeq\) 0.42
[CH₃COO] = 0.4 - a + 0.02
= 0.38 + a \(\simeq\) 0.38
\(\left[\mathrm{H}^{+}\right]=\mathrm{K}_{\mathrm{a}} \frac{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]}\)
\([H^+] = \frac{(1.8 \times 10^{-5})(0.42)}{(0.38)}\)
[H⁺] = 1.99 x 10^-5
pH  = - log (1.99 x 10^-5)
= 5 - log 1.99
= 5 - 0.30 = 4.70

Hoffmann's bromide reaction:
When Amides are treated with bromine in the presence of aqueous or ethanolic solution of KOH, primary amines with one carbon atom less than the parent amides are obtained.
\(\underset { \quad \quad \quad amide\\ R=Alkyl(or)Aryl }{ R-\overset { \underset { || }{ O } }{ C } -{ NH }_{ 2 } } \overset { { Br }_{ 2 }/KOH }{ \longrightarrow } \underset { Primary\quad amine }{ R-{ NH }_{ 2 }+{ K }_{ 2 }{ CO }_{ 3 } } +KBr+{ H }_{ 2 }O\)
(ii) Hoffmann's ammonolysis:
When Alkyl halides (or) benzylhalides are heated with alcoholic ammonia in a sealed tube, mixtures of 1°, 2° and 3° amines and quaternary ammonium salts are obtained.
\( { CH }_{ 3 }-Br\overset { \ddot { N } { H }_{ 3 } }{ \underset { \Delta }{ \longrightarrow } } \underset { { 1 }^{ 0 }-amine }{ { CH }_{ 3 }-\ddot { N } { H }_{ 2 } } \overset { { CH }_{ 3 }-Br }{ \longrightarrow } \underset { { 3 }^{ o }-amine }{ \left( { CH }_{ 3 } \right) _{ 2 }\ddot { N } H } \overset { { CH }_{ 3 }Br }{ \longrightarrow } \underset { 3^{ o }-amine }{ \left( { { CH }_{ 3 } } \right) _{ 3 }\ddot { N } } \overset { CH_{ 3 }Br }{ \longrightarrow } \underset { Quartenary \ ammonium\ bromide }{ \left( { CH }_{ 3 } \right) _{ 4 }\overset { + }{ N } { Br }^{ - } } \) 
This is a nucleophilic substitution, the halide ion of alkyl halide is substituted by the -NH₂ group. The product primary amine so formed can also has a tendency to act as a nucleophile and hence if excess alkyl halide is taken, further nucleophilic substitution takes place leading to the formation of quarternary ammonium salt. However, if the process is carried out with excess ammonia, primary amine is obtained as the major product. The order of reactivity of alkylhalides with amines
RI > RBr > RCl
(iii) Gabriel phthalimide synthesis:
Gabriel synthesis is used for the preparation of Aliphatic primary amines. Phthalimide on treatment with ethanolic KOH forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis gives primary amine. Aniline cannot be prepared by this method because the arylhalides do not undergo nucleophilic substitution with the anion formed by phthalimide.

(iv) Schotten - Baumann reaction :
Aniline reacts with benzoylchloride (C₆H₅COCI) in the presence of NaOH to give N - phenyl benzamide. This reaction is known as Schotten - Baumann reaction. The acylation and benzoylation are nucleophilic substitutions.
\(\underset { Aniline }{ { C }_{ 6 }{ H }_{ 5 }-{ NH }_{ 2 } } +\underset { Benzoyl\quad chloride }{ { C }_{ 6 }{ H }_{ 5 }-\overset { \underset { || }{ O } }{ C } -Cl } \overset { Pyridine }{ \longrightarrow } \underset { N-phenyl\quad benzamide }{ { C }_{ 6 }{ H }_{ 5 }-NH-\overset { \underset { || }{ O } }{ C } -{ C }_{ 6 }{ H }_{ 5 } } +HCl\)
(v) Carbylamine reaction :
Aliphatic (or) aromatic primary amines react with: chloroform and alcoholic KOH to give isocyanides (carbylamines), which has an unpleasant smell. This reaction is known as carbylamines test. This test used to identify the primary amines.
\(\underset { Ethylamine }{ { C }_{ 2 }{ H }_{ 5 }-{ NH }_{ 2 } } +\underset { Chloroform }{ { CHCl }_{ 3 }+3KOH } \longrightarrow \underset { Ethylisocyanide }{ { C }_{ 2 }{ H }_{ 5 }-NC } +3KCl+3{ H }_{ 2 }O\)
(vi) Mustard oil reaction:
(a) When primary amines are treated with carbon disulphide (CS₂), N - alkyldithio carbonic acid is formed which on subsequent treatment with HgCI₂, gives an alkyl isothiocyanate.


(b) When aniline is treated with carbon disulphide, or heated together, S-diphenylthio urea is formed, which on boiling with strong HCI, phenyl isothiocyanate (phenyl mustard oil), is formed.

These reactions are known as Hofmann - Mustard oil reaction. This test is used to identify the primary amines.
(vii) Coupling reactions (or) p-hydroxyazobenzene, p-aminoazobenzene, 2-phenyl azo-4-methyl phenol:
Benzene diazonium chloride reacts with electron rich aromatic compounds like phenol, aniline to form brightly coloured azo compounds. Coupling generally occurs at the para position. If para position is occupied then coupling occurs at the ortho position. Coupling tendency is enhanced if an electron donating group is present at the para - position to -N₂CI- group. This is an electrophilic substitution.

Aryl fluorides and iodides cannot be prepared by direct halogenation and the cyano group cannot be introduced by nucleophilic substitution of chlorine in chlorobenzene. For introducing such a halide group. cyano group -OH, NO₂, etc.. benzenediazonium chloride is a very good intermediate Diazo compounds obtained from the coupling reactions of diazonium salts are coloured and are used as dyes.
(viii) Diazotisation :
Aniline reacts with nitrous acid at low temperature (273 - 278 K) to give benzene. Diazonium chloride which is stable for a short time and slowly decomposes even at low temperatures.This reaction is known as diazotization

(ix) Gomberg reaction :
Benzene diazonium chloride reacts with benzene in the presence of sodium hydroxide to give biphenyl. This reaction in known as the Gomberg reaction

Nitrobenzene into 1,3,5 - trinitro benzene

Nitrobenzene                                 1,3,5 - trinitro benzene
ii. Nitrobenzene into o - and p - nitro phenols

iii. Nitrobenzene into m - nitro aniline
  + 6NH₃ + 2H₂O + 3S
iv. Nitrobenzene into azoxybenzene
+3H₂O
v. Nitrobenzene into hydrazobenzene

vi. Nitrobenzene into N - phenyl hydrazobenzene

vii. Nitrobenzene into aniline

i) 2 - nitro propane is boiled with HCI:

It gives acetone.
ii) Nitrobenzene on electrolytic reduction in strongly acidic medium:

It gives aniline, phenyl hydroxylamine and p - amino phenol
iii) Oxidation of text – butylamine with KMnO₄:

It gives 2 - methyl - 2- nitro propane is obtained.
iv) Oxidation of acetoneoxime with trifluoroperoxy acetic acid.

It gives 2 - nitro propane.

(i) Ostwald's dilution law relates the dissociation constant of the weak acid (K_a) with its degree of dissociation (α) and the concentration (c).
where  \(\alpha=\frac{\text { Number of moles dissociated }}{\text { Total number of moles }}\)
(ii) The dissociation of acetic acid can be represented as
\(\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{H}^{+}+\mathrm{CH}_{3} \mathrm{COO}^{-}\)
The dissociation constant of acetic acid is,
\({ K }_{ a }=\frac { \left[ { H }^{ + } \right] \left[ { CH }_{ 3 }COO^{ - } \right] }{ \left[ { CH }_{ 3 }COOH \right] } \)  ........(1) 

Substituting the equilibrium concentration in equation (1)
\({ K }_{ a }=\cfrac { \left( \alpha C \right) \left( \alpha C \right) }{ \left( 1-\alpha \right) C } \) 
\({ K }_{ a }=\cfrac { { \alpha }^{ 2 }C }{ 1-\alpha } \) .......(2) 
(iii) We know that weak acid dissociates only to a very small extent compared to one, a is so small and hence in the denominator (1 - α) ⋍1. The above expression (2) now becomes,
k_a =a²C \(\Rightarrow { \alpha }^{ 2 }=\cfrac { { k }_{ a } }{ C } \) ; \(\alpha =\sqrt { \cfrac { { K }_{ a } }{ C } } \) 
(iv) When dilution increases, the degree of dissociation of weak electrolyte also increases. This is called Ostwald's dilution law
Also \(;\left[\mathrm{H}^{+}\right]=\alpha \mathrm{C}\) and \(\left[\mathrm{H}^{+}\right]=\left(\sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{C}}}\right) \mathrm{C}=\sqrt{\frac{\mathrm{K}_{\mathrm{a}} \mathrm{C}^{2}}{\mathrm{C}}}=\sqrt{\mathrm{K}_{\mathrm{a}} \cdot \mathrm{C}}\)
Similarly for a weak base
\(\begin{aligned} & \mathrm{K}_{\mathrm{b}}=\alpha^2 \mathrm{C} ; \quad \therefore \alpha=\sqrt{\frac{\mathrm{k}_{\mathrm{b}}}{\mathrm{C}}}, \\ \end{aligned}\)
\(\begin{aligned} & {\left[\mathrm{OH}^{-}\right] \alpha \mathrm{C}=\sqrt{\frac{\mathrm{K}_{\mathrm{b}}}{\mathrm{C}}} \times \mathrm{C}=\sqrt{\frac{\mathrm{K}_{\mathrm{b}} \mathrm{C}^2}{\mathrm{C}}}=\sqrt{\mathrm{K}_{\mathrm{b}} \mathrm{C}}} \end{aligned}\)

(i) The dissociation of a weak acid (CH₃COOH) is suppressed in the presence of a salt (CH₃COONa) containing an ion common to the weak electrolyte. It is called the common ion effect.
(ii) Consider the dissociation of a weak acid, acetic acid whose ionisation is incomplete.
CH₃COOH_(aq) ⇌ H⁺_(aq) + CH₃COO^-_(aq)
(iii) If the salt sodium acetate with common ion CH₃COO^- is added to the above equilibrium, it dissociates completely increasing.
CH₃COONa_(aq)⟶Na⁺_(aq) + CH₃COO^-_(aq)
(iv) Hence, the overall concentration of CH₃COO^- is increased, and the acid dissociation equilibrium is disturbed.
(v) So, in order to maintain the equilibrium, the excess CH₃COO^- ions combines with H⁺ ions to produce much more unionized CH₃COOH i.e, the equilibrium will shift towards the left. In other words, the dissociation of CH₃COOH is suppressed.

(i) An acid is defined as a substance that has a tendency to donate a proton to another substance and base is a substance that has a tendency to accept a proton form other substance.
(ii) In other words, an acid is a proton donor and a base is a proton acceptor.
(iii) When hydrogen chloride is dissolved in water, it donates a proton to the later. Thus, HCI behaves as an acid and H₂O is base. The proton transfer from the acid to base can be represented as
HCI + H₂O ⇌ H₃O⁺ + Cl^-
(iv) When ammonia is dissolved in water, it accepts a proton from water. In this case, ammonia (NH₃) acts as a base and H₂O is acid. The reaction is represented as
H₂O + NH₃ ⇌ NH₄⁺ + OH^-
(v) Let us consider the reverse reaction following equilibrium.
\(\underset { proton\ donar\\ \quad \quad \ (acid) }{ HCl } +\underset { Proton\ acceptor\\ \quad \quad \quad \quad (base) }{ { H }_{ 2 }O } \leftrightharpoons \underset { Proton\ donar\\ \quad \quad \quad \quad \ (acid) }{ { H }_{ 2 }{ O }^{ + } } +\underset { Proton\ acceptor\\ \quad \quad \quad \quad \ (base) }{ { Cl }^{ - } } \) 
H₃O⁺ donates a proton to Cl^- to form HCI i.e., the products also behave as acid and base.
(vi) In general, Lowry - Bronsted (acid - base) reaction is represented as 
Acid₁ + Base₂ ⇌ Acid₂ + Base₁
(vii) The species that remains after the donation of a proton is a base (Base₁) and is called the conjugate base of the Bronsted acid (Acid₁). In other words, chemical species that differ only by a proton are called conjugate acid - base pairs.

Chlorine reacts with cold dilute alkali to give chloride and hypochlorite, while with hot concentrated alkali chlorides and chlorates are formed.
\(\mathrm{Cl}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{HCl}+\underset{\text { Hypochlorous acid }}{\mathrm{HOCl}} \)
\(\mathrm{HCl}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O} \)
\(\mathrm{HOCl}+\mathrm{NaOH} \rightarrow \mathrm{NaOCl}+\mathrm{H}_{2} \mathrm{O} \)
                   (Sodium hypo chlorite)
Overall Reaction
3Cl₂ + 6NaOH \(\rightarrow\) NaClO₃ + 5NaCl + 3H₂O
                             (Sodium Chlorate)