9.-Ionic-Equilibrium.pdf

Ionic Equilibrium Types of Substances Substances can be classified into following two types, viz., non-eletrolytes and electrolytes. Non-electrolytes They do not conduct electricity in molten state or aqueous solution, for ex. solution of sugar, urea, glucose, glycerine. Elec...

Ionic Equilibrium Types of Substances Substances can be classified into following two types, viz., non-eletrolytes and electrolytes. Non-electrolytes They do not conduct electricity in molten state or aqueous solution, for ex. solution of sugar, urea, glucose, glycerine. Electrolytes Concept Ladder They conduct electricity in molten state or aqueous solution, for ex. NaCl. Electrolytes can In the solid state, be further divided into following types: electrolytes are bad conductors of electricity, Strong electrolytes These get strongly ionized they become good in water, hence they exhibit high conduction, for conductors either in example, strong acids like HCl, H2 SO4 , HNO3 , the molten state or in a strong bases like KOH, NaOH and salt of strong solution. acid or strong base like NaCl, CH3 COONa, NH4 X. Weak electrolytes They show less conduction as they are less strongly ionized in water, for ex., weak acids like HCN, CH3COOH, H3PO4, H2CO3, weak bases like NH4OH and their salts like NH2CN, CH3COONH4. Rack your Brain Arrhenius Theory It is also called as dissociation of electrolytes Does human body also have theory or introduction of ionic theory. The main electrolytes? If yes, then what is postulates of this theory are: their role? y Electrolytes on dissolving in a solvent, ionize into ions. Solvent AB  HO A + B   + − 2 Ions [Ionization] y Ionization process is reversible and the ionization constant (K) is given as Ionic Equilibrium [A + ][B+ ] K= [AB] 1. Ionic Equilibrium 2. y The discharging of ions are always in equivalent amount, it is independent of their their relative speeds during electrolysis. y Ions act like molecules towards depressing the freezing point, elevating the boiling point, y lowering osmotic pressure and vapour Concept Ladder pressure. Evidence in Support of lonic Theory The process of ionisatino y X-ray diffraction method gives information is reversible for weak about presence of ions in solid electrolytes. electrolytes. y Ions obey Ohm’s law (I = E/R) y Ionic reactions, for example, AgNO3 + NaCl  → Ag + Cl − ↓ + Na +NO3− y Color of some solutions is due to the presence of ions, for ex. CuSO4 solution is blue due to Cu2+. Degree of ionization It is denoted by α and is defined as the range Rack your Brain to which ionization of an electrolyte takes place in a solvent. Is there any difference between Number of molecules dissociated ionisation and dissociation? a= Total number of molecules Degree of ionization or α depends on: y Nature of solute and solvent. For weak electrolytes α is less than that for strong electrolytes. y α represents a dielectric constant of solvent it is directly proportional to the ionization of Concept Ladder electrolyte in it. y ⚫ The degree of dissociation for weak The degree of dissociation electrolyte is directly proportional to dilution, of an electrolyte is i.e., α is maximum at infinite dilution. assumed to be unity at 1 infinite dilution, i.e., a = 1 y a∝ Concentration at infinite dilution. Ionic Equilibrium y a ∝ Temperature 3.  Ostwald Dilution Law y This law states the law of mass action for dilute solutions and weak electrolytes and it gives the relationship between the dissociation of weak electrolytes. Rack your Brain y It is not applicable for highly concentrated or solutions strong electrolytes. Consider a weak What is the value of degree electrolyte AB in V litres of a solution. of dissociaiton for a strong AB   A+ + B− electrolyte? Initial moles 1 0 0 Moles at equilibrium 1−a a a Concentration at (1 − a) / V (a / V) (a / V) Equilibrium [B− ][A + ] K= [AB] (a / V) × (a / V) K= (1 − a) / V Concept Ladder a2 / V2 a2 So, K = = (1 − a) / V (1 − a) / V As 1 >>> a so (1 − a) = 1 Smaller the value of Ka or Kb, weaker will be the a2 =K or= a2 KV or = a KV electrolyte. V a= V Hence, degree of dissociation is directly proportional to the square root of its dilution at constant temperature. y If C is the concentration then AB  A + + B−  C(1 − a) C a Ca Ca 2 Rack your Brain K= (1 − a) K = C a2 or a2 = K / C Can the approximation of (1 - a) ~ a = (K/ C) 1 be applied when a is 10%? that is, a ∝ (1 / C) Ionic Equilibrium Hence, degree of dissociation is inversely proportional to the square root of its concentration at constant temperature. 4.  λV ∧ y a= or V Previous Year’s Questions λ∞ ∨∞ Here λ∞ or Λ∞ is equivalent conductivity at The molar solubility of CaF2 (Ksp = infinity dilution where as λv or Λv is equivalent 5.3 × 10-11) in 0.1 M solution of NaF conductivity at V dilution. will be y λ V = KV × V [NEET-2019] Here Kv = Specific conductivity (1) 5.3 × 10-11 mol L-1 V = Volume of solution in litres (2) 5.3 × 10-8 mol L-1 y λ∞ = λc + λα (3) 5.3 × 10-9 mol L-1 Here λc and λα are the ionic mobilities of cation (4) 5.3 × 10-10 mol L-1 and anion respectively. Common Ion Effect The value of degree of dissociation will be decreased for a weak electrolyte on adding a strong electrolyte containing a common ion. y The concentration of the uncommon ion of the Concept Ladder weak electrolyte decreases due to common ion effect. Degree of dissociation of Examples: a weak electrolyte gets (1) NH4 OH   NH4+ + OH− suppressed by the addiiton Weak of a strong electrolyte NH4Cl   NH4+ + Cl − containing same ion. Common ion Here the value of α for NH4OH will be decreased by NH4Cl. (2) CH3 COOH   CH3COO− + H+ CH3COONa   CH3COO− + Na + Common ion Previous Year’s Questions Here CH3COONa decreases the value of a for CH3COOH because equilibrium shifts in Consider the nitration of benzene backword direction. using mixed conc. H2SO4 and y Due to common ion effect solubility of HNO3. If a large amount of KHSO4 a partially soluble salt decreases, for ex. is added to the mixture, the rate presence of KCl or AgNO3, decreases the of nitration will be Ionic Equilibrium solubility of AgCl in H2O. [NEET-2016] y By addition of NaCl salting out of soap. (1) slower (2) unchanged y Purification of NaCl by passing HCl gas. (3) doubled (4) faster 5.  Isohydirc Solution Isohydric solution in the solution which contains Rack your Brain same conc. of common ions. How common ion effect can Ionic Product Of Water be explained using Le Chatelier It is the product of the molar conc. of H+ or H3O+ Principle? and OH− ions. It is denoted by Kw. H2O + H2O   H3+ O + OH− Kw = [H3O+]. [OH−] or [H+]. [OH−] Here Kw = Ionic product of water y Kw = Ka. Kb y pKw = −log10 Kw y Kw = Ka × Kb y pKw = pKa + pKb y At 25°C, Kw = 1 × 10−14 Concept Ladder pKw = 14 y With increase in temp. the value of Kw Pure water is a weak increases, for example, at 98°C Kw is 1 × 10−12. electrolyte, it undergoes self ionisation which is also pH Scale known as autoprotolysis. y It was introduced by Sorenson, for measurement of acidity or basicity of the given solution. y pH stands for potentiel de-H+ or conc. of H+. It is given by [H+] = 10−pH pH = –log10 [H+] pH of aqueous solution is equal to –ve logarithm of H+(H+3O) concentration in m o l e / litre. Rack your Brain 1 pH = log 10 + [H ] Ionic Equilibrium How dissociation constant of water is different from ionic product of water? 6. pH of weak acid  CH3COO− + H+ CH3COOH  Concept Ladder 1 0 0 C(1 − x) Cx Cx The pH of a mixture of two weak acids can be obtained K = Cx2 as : x = Ka / C pH  Ka 1  c1  Ka 2  C2 [H= + ] Cx = C (Ka / C) ≈ (Ka.C) pH = − log 10 Cx pH = − log 10 (Ka.C) Here C = Molar concentration of acid x = deg ree of dissociation Ka = Dissociation cons tant of acid pH of weak base Rack your Brain A stronger acid will have ______ NH4OH   NH4+ + OH− value of pH. 1 0 0 (1 − x) Cx Cx [OH=] Cx − = (Kb. c) pOH = − log 10 Cx pOH = − log 10 (Kb. C) Here Kb = Dissociation cons tan t of the base Concept Ladder pOH pH of strong acid or base [OH− ] = 10−pOH does not depend upon temperature. pH of weak 1 pOH = − log 10 [OH− ]or acid decreases with log 10 [OH− ] increase in temperature, pH + pOH = 14 due to increase in ionization. pH of weak base pH + pOH = pK w increases with increase pKa = − log 10 Ka in temperature, due to 1 1 increase in ionization or Ionic Equilibrium pKa ∝ ∝ [OH−] ion concentration. Ka Acidic strength 7.  y A weak acid consist high value of pKa. y A weak base consist high value of pKb. pH Value and Nature of a Solution y For [H+] > 10−7,value of pH is less than 7 and Previous Year’s Questions the solution is acidic. y For [H+] = 10−7, value of pH is 7 and the solution The pH of 0.01 M NaOH (aq) y is neutral. solution will be – y For [H+] < 10-7, value of pH is more than 7 and [NEET-2019] the solution is basic. (1) 7.01 (2) 2 pH Range of Some Substances (3) 12 (4) 9 Substance pH range Gastric juice 1−3.0 Soft drink 2−4.0 Lemon 2.2−3.4 Vinegar 2.2−2.4 Urine 4.8−8.4 Milk 6.3−6.6 Saliva 6.5−7.5 Previous Year’s Questions Blood 7.3−7.5 Sea water 8.5 pH of a saturated solution of Ca(OH)2 is 9. The solubility Tears 7.4 product (Ksp) of Ca(OH)2 is : [NEET-2019] Limitations of pH scale (1) 0.5 × 10 -10 y pH value of a solution does not instantaneously (2) 0.5 × 10-15 give us an idea of the relative strength of the (3) 0.25 × 10-10 solution. (4) 0.125 × 10-15 y pH is zero for 1N solution of strong acid. Ionic Equilibrium y pH is negative for concentrations 2N, 3N, 10N of strong acids. y At higher concentrations, in place of pH, Hammelt acidity functions are used. 8. Buffer Solution y Buffer solution is the solution in which there is no change in pH takes place even on adding small amount of strong base or a acid. y It is also known as reserve acidity or basicity of the solution, this action of resisting change in value of pH is known as buffer action. Previous Year’s Questions Features y Buffer solution has a definite pH, that is a Equimolar solutions of the definite reserve basicity or acidity. following substances were y There is no change in pH for a long period of prepared separately. Which one time. of these will record the highest y There is no change in pH on dilution. pH value? y There is slight change in pH (unnoticeable) on [AIPMT-2012] adding small quantity of a strong base or acid. (1) BaCl2 (2) AlCl3 (3) LiCl (4) BeCl2 Buffer Capacity Buffer capacity is the ratio of no. of moles of the acid or base added in 1 L of the solution, so as to change its pH by unity. It is denoted by Number of moles of acid / base added to 1 L of solution φ. ϕ = Change in pH Types of Buffer Solutions 1. Acidic buffer: The solution of a weak acid & its salt with a strong base is termed as acidic budffer. For example, y CH3COOH + CH3COONa y Phthalic acid + potassium phthalate y Boric acid + borax H2CO3 + NaHCO3 Previous Year’s Questions y Citric acid + sodium citrate 2. Basic buffer: The solution of a weak base and What is the pH of the resulting it salt with a strong acid is termed as basic solution when equal volumes of buffer. For example, 0.1 M NaOH and 0.01 M HCl are y NH4OH + NH4Cl mixed? y Glycerine + Glycerine hydrogen chloride. Ionic Equilibrium [NEET-2015] 3. Ampholytes: The compounds containing both (1) 12.65 (2) 2.0 acidic and basic groups and therefore, exist (3) 7.0 (4) 1.04 as zwitter ions at a certain pH (isoelectric 9.  point). Hence, amino Acids and proteins also act like a buffer solution. Previous Year’s Questions 4. 4. A mixture of normal salt and acidic salt of a polybasic acid. For ex, Buffer solutions have constant Na2HPO4 + Na3PO4 acidity and alkalinity because 5. Salt of weak base and weak acid (in H2O).For [AIPMT-2012] example, CH3COONH4 , NH4CN (1) these give unionised acid or Uses of Buffer solutions base or base on reaction with y For finding the pH value of unknown solution added acid or alkali by using an indicator. (2) acids and alkalies in these y NH4Cl + NH4OH and (NH4)2CO3 buffer solutions solutions are shielded from precipitates carbonates of group V elements attack by other ions of periodic table. (3) they have large excess of H+ y CH3COOH and CH3COONa buffer is used to or OH- ions remove PO4–3 in qualitative inorganic analysis (4) they have fixed value of pH after IInd group. y CH3COOH and CH3 COONa buffer is used to precipitate PbCrO4 quantitatively in gravimetric analysis. Rack your Brain y They are also used in dye, printing ink,paper, dairy product. When [Salt] = [Acid], i.e., pH = pKa Solubility And Solubility Product for acidic buffer, and thus buffer Solubility has its maximum capacity. What y It is denoted by ‘s’ and is expressed in mole would be the conditions for a per litre or gram per litre. It is the weight of basic buffer? solute in grams, present in 100 mL of solvent. 1 Solubility (s) ∝ Concentration of common ions or number of common ions Solubility increases due to complex ion formation. For ex. the solubility of AgCl in H2O, in presence Concept Ladder of AgNO3. Here, AgCl is more soluble in NH3, due to formation of complex. A buffer must contain two components to work a AgCl + 2NH3  → Ag(NH3 )2 Cl weak acid that reacts with Ionic Equilibrium Due to formation of complex compound (Nesseler’s added base a weak base reagent) HgCl2 is more soluble in KI. that reacts with added acid. 10.  HgI2 + 2KI  → K2HgI4 In the analysis of IInd group elements, As2S3 , Previous Year’s Questions Sb2S3, SnS, are soluble in (NH4)2S due to complex ion formation. Which will make basic buffer : [NEET-2019] SnS + S−2  → SnS2−2 (1) 100 mL of 0.1 M HCl + 100 mL Sb2S3 + 3S−2  → 2SbS3−3 of 0.1 M NaOH As2S5 + 2S−2  → 2AsS4−3 (2) 50 mL of 0.1 M NaOH + 25 mL of 0.1 M CH3COOH Simultaneous solubility The solubility of a (3) 100 mL of 0.1 M CH3COOH + solution of 2 electrolytes containing common 100 mL of 0.1 M NaOH ions are known as simultaneous solubility. (4) 100 mL of 0.1 M HCl + 200 mL Ex.amples are: of 0.1 M NH4OH 1. CaF2 + SrF2 2. AgBr + AgSCN 3. MgF2 + CaF2 Solubility Product In a saturated solution, the product of the molar concentrations of ions of an electrolyte at a Concept Ladder particular temperature is known as solubility product. It is denoted by Ksp or S. At the saturated stage, the quantity of the solute For a binary electrolyte AB dissolved is always AB   AB   A + + B− constant for the given Solid Un−ionized amount of a particular [A + ][B− ] So,K = solvent at a definite [AB] temperature and this is or K.[AB] = [A + ][B− ] known as the solubility of Ksp = [A + ][B− ] (At constant temperature) solute. General representation A xBy   xA + y + yB− x Ksp = [A + y ]x [B− x ]y Rack your Brain Examples: Ionic Equilibrium What happens to the solubility 1.  Ca +2 + 2F− CaF2  of a sparingly soluble salt on the Ksp = [Ca +2 ][F− ]2 addiiton of a common ion? 11.  2.  2Ag + + CrO4−2 Ag 2CrO4  Previous Year’s Questions Ksp = [Ag + ]2 [CrO4−2 ] Concentration of the Ag+­ions in Relation Between Solubility (s) and Solubility a saturated solution of Ag2C2O4 is Product (Ksp) 2.2 × 10-4 mol-1 solubility product A xBy   xA + y + yB− x of Ag2C2O4 is a 0 0 [NEET-2017] (1) 2.42 × 10 -8 a−s xs ys (2) 2.66 × 10-12 Ksp = (x s)x (y s)y (3) 4.5 × 10-11 or (4) 5.3 × 10-12 Ksp = x y y (s)x + y x If a is given : Ksp xx y y (a s)x + y = For example, Cu2Cl 2   2 Cu+ + 2Cl − 2s 2s So Ksp 2= = 2 (s) 16 s 2 2 2+ 2 4 Fe(OH)3   Fe+3 + 3OH− Concept Ladder s 3s =Ksp 1=3 (s)1+ 3 27 s4 1 3 Solubility product is Al 2 (SO4 )3   2Al +3 + 3SO4−2 2s 3s defined as the product of =Ksp 2= 3 (s) 108 s 2 3 2+ 3 5 concentrations of the ions raised to a power equal Na3Li3 (AlF6 )2   3Na + 3Li + 2AIF6 + +3 −3 to the number of times, =Ksp 3=3 2 (s)3+ 3+ 2 2916 s8 3 3 2 the ions occur in the equation representing the Solubility Product and Precipitation dissociaiotn of the when y If Ksp ≈ ionic product, solution is saturated, & the solution is saturated. for precipitation, more solute to be added. y If ionic product > Ksp, solution is super saturated, & therefore gets easily precipitated. y If ionic product < Ksp, solution is unsaturated & therefore no precipitation takes place. Rack your Brain Ionic Equilibrium What is teh difference between ionic product and solubility product? 12. Salt Hydrolysis y Salt hydrolysis is the ionic interaction between Previous Year’s Questions ions of the salt and water causes acidity or basicity in aqueous solution, when a salt is Find out the solubility of Ni(OH)2 added to water. is 0.1 M NaOH. Given that the ionic y Cationic hyrolysis is interaction of cation and product of Ni(OH)2 is 2 × 10-15 anionic hydrolysis is interaction of anion. [NEET-2020] y Hydrolysis is an endothermic process and is (1) 2 × 10-8 M (2) 1 × 10-13 M reverse of neutralization. (3) 1 × 108 M y If neutralization constant is Kn and hydrolysis (4) 2 × 10-13 M constant is Kh, then Kn = 1/Kh. y A solution of the salt of weak base and strong acid is acidic and for it pH < 7 or [H+] > 10−7, for ex. FeCl3 (salt of a strong acid + weak base). Here, the solution is involves cationic hydrolysis and acidic in nature. y A solution of the salt of weak acid and strong Concept Ladder base is basic and for it, pH > 7 or [H+] < 10−7, for ex. KCN (salt of a weal acid + strong base). The net effect of Here, the solution involves anionic hydrolysis dissolveing a salt (which and is basic in nature. undergoes hydrolysis) is y A solution of the salt of weak base and a weak to break up the water acid is : molecules (hydrolysis) to 1. Basic, if Ka < Kb produce a weak acid or weak base or both and 2. Acidic, if Ka > Kb thus, phenomenon is 3. Neutral, if Ka = Kb is neutral always endothermic. y CH3COONH4 (salt of a weak base + weak acid) Here, the solution involves both anionic and cationic hydrolysis and is neutral in nature. y The solution of salt of strong base & strong acid is neutral or pH = 7 or [H+] = 10−7 Rack your Brain y A salt of a strong acid and base can never be hydrolyzed. During the hydrolysis of NH4Cl, However, ions are hydrated, for ex. K2SO4 (salt Ionic Equilibrium which ion will undergo hydrolysis? of a strong acid + strong base). 13.  Various Expressions for Kh, h and pH for Different Type of Salts Previous Year’s Questions 1. For the salt of weak acid & strong base like KCN, The ionisation constant of K ammonium hydroxide is 1.77 × 10-5 Kh = w Ka at 298 K. Hydrolysis constant of h = (Kh / C) ammonium chloride is [AIPMT] h = (Kw / Ka. C) (1) 5.65 × 10 -10 1 (2) 6.50 × 10-12 pH = [pKw + pKa + log C] 2 (3) 5.65 × 10-13 1 (4) 5.65 × 10-12 pOH = [pKw − pka − log C] 2 2. For the salt of strong acid & weak base like FeCl3, Kw Kh = Ka h = (Kh / C) Concept Ladder h = (Kw / Kb. C) 1 pH = [pKw + pKb + log C] The pH of the salt of W.A. 2 and W.B. is independent of 3. For the salt of weak acid & weak base like its concentration. CH3COONH4, Kw Kh = KaKb h = (Kw / KaKb ) 1 pH = [pKw + pKa − pKb ] 2 Acid and Base Arrhenius Concept of Acid and Base Rack your Brain According to Arrhenius concept, “H+ ion donor Ionic Equilibrium in H2O are acids and OH– ion donor in H2O are How the degree of hydrolysis of bases.’’ salts of weak acids and weak bases changes with dilution? 14. 15. Ionic Equilibrium  Acid y HA + H2O  → A − + H3O+ Concept Ladder where H3O+ hydronium ion or hydrated proton. y Due to presence of H-bonding,H2O can accept H2PO2– and HPO32– are only H+ to form hydronium ion (H3O+). For example, bronsted bases. As H3PO2 [H5O+2, H7O+3] and H3PO3 are monobasic y H+ can hold H2O molecule by H-bond, as it and dibasic acid respectively, has high heat of hydration. For ex. H2SO4, they can release only one HNO3, HX. and two H+ ions respectively. y H3O+ has sp3 hybridization and has trigonal pyramidal shape. y The Number of H+ ions donated = Basicity or protosity of the acid. Examples: 1. H3PO4 (tribasic) O || H−O −P −O −H | Rack your Brain O− H 2. H3PO3 (dibasic) Why water is amphoteric in H nature? | H−O −P −O −H || O 3. H3PO2 (mono basic) O || H−P −H | O− H Concept Ladder Base B.OH + H2O → B+ + H3O2− [orH5O3− ] The neutralization of an For example, NaOH, KOH. acid and base is basically a neutralization reaction Strength of Acid and Base between H+ and OH- ions. Ionic Equilibrium 1. Acidic strength ∝ Ka H (aq)  OH (aq)  H2 O(l) Ka = Dissociation constant of the acid 16. 2. Base strength ∝ Kb Kb = Dissociation constant of the base Concept Ladder 3. Relative strength = (K1 / K2 ) The fundamental concept of K1 H3PO4    H2PO4 + H or H3O  − + + Bronsted Lowry Concept is K2 H2PO4−    HPO4 + H  −2 + that when an acid & a base K3 react with each other, the HPO4−    PO4 + H  −3 + acid forms its conjugate H3PO4 > H2PO4– > HPO4–2 is the correct acidic base, & the base forms its strength order of K1 > K2 > K3 conjugate acid by exchange 4. With development of negative charge, the of a proton. removal of H+ becomes more and more difficult so acidic nature decreases. Bronsted Lowery Concept or Proton Concept Acid Acids are proton or H+ donor. HA  → A − + H+ Acid Conjugate Rack your Brain base For example, HA  → H+ + X− Which is neither an acid nor Conjugate base a base according to Arrhenius H2SO4  → H+ + HSO4− concept? Acid Conjugate base H3PO4, HCl, NH3, KOH HNO3  → H+ + NO3− Base Bases are proton or H+ acceptor Base + H+ → (Base H) + Pr oton Conjugate given by acid Acid For example, OH− + H+ → H2O Previous Year’s Questions Base Conjugate Acid Amphoteric or Ampholyte Substances Boric acid is an acid because its Amphoteric substances can donate or accept H+ molecule or proton, hence they can behave both like an [NEET-2016] acid as well as base. (1) Contians replaceable H+ ion Examples: (2) gives up a proton −H+ 1. NH4+ ←+H+ NH3  → NH2− (3) accepts OH- from water Conjugate Ampholyte Conjugate Ionic Equilibrium acid base releasing proton 2. H3O+ ← +H  H2O  −H + → OH− + (4) combines with proton from Conjugate Ampholyte Conjugate base water molecule acid 17.  Some other examples are HSO4– , HCO3–, H2PO4–, HPO42–, H2PO3–, HS– and HC2O4– Concept Ladder Lewis Concept of Acid and Base Lewis acids: The species which can accept lone All arrhenius acids are pair of electrons due to defficiency of electron also Bronsted acids but are termed as Lewis acids. The species which act all arrhenius bases are not as Lewis acids are as follows: Bronsted bases. 1. Cations This is because an arrhenius − C −,NO2+ , X + (Halonium ion) + acid is a substance which | can give a H+ ion whereas a 2. Electron deficient central atoms Bronsted acid is a substance BeX2 , BX3 , FeX3 , AIX3 , SnCI2 , ZnCI2 which can donate a proton ↓ ↓ ↓ ↓ ↓ ↓ 4 6 6 6 4 4 which is also a H+ ion. 3. Central atom having multiple bonds O=C=O, O=S=O, S=C=S, SO3 4. When central atom of a molecule has vacant d-orbital then it can have more than 8 Rack your Brain electrons, i.e., octet state can be expand and molecule can behave as a Lewis acid. For What is conjugate acid and base example, for NH3? (a) PX5 + X − → PX6− Lewisacid Lewis base (b) SiX4 + 2X → SiX6−2 − Lewis acid Lewis base y Some other examples of Lewis acids are: PX3, Previous Year’s Questions SnX4, GeX4 , SF4 , TeCl4, SeF4 y Elements with an electron sextet (for example Conjugate base for Bronsted O, S) acids H2O and HF are : [NEET-2019] Strength of Lewis Acid (1) H3O and H2F , respectively + + 1. Lewis acid strength ∝ Electronegativity (2) OH- and H2F+, respectively difference (3) H3O+ and F-, respectively Example: (4) OH- and F-, respectively AlF3 AlCl 3 AlBr3 AII3  Acidic nature increases  → Ionic Equilibrium 2. In case of boron halide (BX3) BI3 > BBr3 > BCI3 > BF3 No back less back Maximum back bonding bonding bonding 18. 3. In BF3, electron pairs is shifted from fluorine back to boron atom. Due to small size and Concept Ladder more electron density (2p5) of F-atom, there is more e–—e– repulsion. Whereas in case of The term nucleophile and B, the atom size is large and 2p orbital is electrophile are more or almost vacant. less inter changable with 4. 4. The strength of some Lewis acids is given lewis bases and lewis acids below in their decreasing order as: respectivley. BX3 , AlX3 , FeX3 , GaX3 , SbX5 , InX3 , SnX4 , AsX5, ZnX2 , HgX2 Lewis bases The species which can donate lone pair of electrons are termed as Lewis bases. y Any anion or molecule central atom with lone pair of electrons and with octet state, as it can donate its lone pair hence it is a Lewis base. For example, OH–, X– Rack your Brain Write name of some convinental amines which behaves as lewis bases? Some of the multiple bonded molecules which form complexes with transition metals are as CO, NO, C2 H4. Strength of Lewis Base 1 Lewis base ∝ Electronegativity difference Previous Year’s Questions NH3 > NI3 > NBr3 > NCl3 > NF3 Least electronegativity difference Which of the folloiwng oxides Factors Affecting Acidic Strength is not expected to react with Effect of electronegativity difference sodium hydroxide? y Acidic strength is directly proportional to [NEET-2009] Ionic Equilibrium Electronegativity difference (1) B2O3 (2) CaO 1. HF > H2O > NH3 > CH4 (3) SiO2 (4) BeO As electronegativity of F > O > N > C. HNO3 > H2CO3 > H3BO3 As electronegativity of N > C > B. 19.  2. HClO > HBrO > HIO HClO3 > HBrO3 > HIO3 The above order is due the electronegativity Concept Ladder order of Cl > Br > I. 3. Acidic strength is directly proportional to the All Lewis bases are also size of central atom or ease of removal of H+ Bronsted bases but all HF < HCl < HBr < HI Bronsted acids may not be H2O < H2S < H2Se Lewis acids. This is because NH3 < PH3 < AsH3 < SbH3 a substance that is capable 4. Strength of oxyacid’s ∝ Oxidation number of of giving an electron pair central atom. +1 +3 +5 +7 has the tendency to accept HO X HCN > HS- Therefore, the decreasing order of basic character S2- > SN- > NH3 > CHCOO- > F- Q.15 Arrange the following 0.1M solutions in order of increasing pH: H2CO3 , KBr, HI,NH3 , KCN, NaOH, NH4Br A.15 Stronger the acid, less is the pH, stronger the base, high is the pH. Increasing order of pH: HI < H2CO3 < NH4Br < KBr < NH3 < KCN < NaOH 33.   Q.16 Calculate the pH at which an acid indicator with Ka = 1.0 × 10-5 changes colour when the indicator is 1.00 × 10-3M. A.16 The point at which acid and conjugate base forms of acid indicator are present in equal concentration is known as the midpoint of the colour change range of an indicator, hence H3O+  In−  Kind = = H3O+=  1.0 × 10−5 [ ] HIn and pH = 5.00 Q.17 At what pH will a 1.0 × 10-3 M solution of an indicator with Kb = 1.0 × 10-10 changes colour? A.17 The indicator changes colour when the conjugates are equal to concentration. [HIn] OH − Kind =  =  OH −  1.0 × 10−10 = [HIn]    thus pOH 10.00 = = and pH 4.00 Q.18 Calculate the pH at which an indicator with pKb = 4 changes colour. A.18 In− + H2O   HIn + OH − At the colour change, In−  = [HIn] OH −  HIn  [  ] OH−  Kb = =1.0 × 10 −4 =  In−  and pOH = 4.0 ∴ pH = 10.00 34.  Q.19 Determine the solubility of AgCl. A.19 For AgCl, Ksp = 1.8 × 10-10, hence AgCl ( s ) + aq   Ag + ( aq)+ Cl − ( aq) S S Ksp = Ag  Cl  + − Ksp = S2M2 ( 1.8 × 10−10 ) 1/2 i. S/ M = 1.34 × 10−5. = Ag= +  Cl − = ; S 1.34 × 10−5 mol L−1 ii. S of AgCl in gL−1 = 1.34 × 10−5 × 143.35g mol −1 = 1.92 × 10−3 gL−1 Q.20 Let the solubilities of AgBr in water and in 0.01M CaBr , 0.01 M KBr, and 0.05M 2 AgNO3 be S1, S2, S3 and S4 respectively. Give the decreasing order of the solutbilitys. A.20 i. → Ag + ( aq) + Br − ( aq) AgBr + H2O ....... ( s1 ) ii. AgBr in 0.01 M CaBr2....... ( s2 ) Br −  added = 0.01 × 2 = 0.02 M iii. AgBr in 0.01 M KBr....... ( s3 ) Br −  added = 0.01 M iv. AgBr in 0.05 M AgNO3....... ( s4 )  Ag +  added = 0.05 M Since both Br- ions and Ag+ ions act as common ions, so larger the concentration of Br- or Ag+­ion added, more is the suppression of ionissation of Agbr and hence less will the solubility of AgBr. Therefore, the decreasing solubility order: S1 > S3 > S2 > S4 35.   Q.21 A solution contains 0.01 M each of CaCl2 and SrCl2. A 0.005M solution of SO42- is slowly added the given solution. (a) If H2SO4 is containuously added, determien when will other salt be precipitated? A.21 Now, If SO42- ions are continuously added, at some instant, its concentration will become equal to that minimum required for precipitating out Ca2+ ions. Hence CaSO4 will start precipitating if, [SO42-] = 1.3 × 10-2­ M. Q.22 The solubility of silver benozate (PhCOOAg) in H O and in solution of pH = 4, 2 5, and 6 are S1, S2, S3 and S4, respectively. Given the decreasing order of their solubilities. A.22 Solubility of a salt of W A increases with the increase in [H+] or increases at lower pH. Thus, decreasing order of solubility is S2 > S3 > S4 > S1 Q.23 Explain why CoS is more soluble than predicted by the K. b A.23 Not all the sulphide which dissolves remains as S − , most of it hydrolyses. 2- S 2− + H2O   HS + OH − Q.24 Explain why CuS is more soluble than predicated by the K. sp A.24 The S 2- hydrolyses extensively. The amount which dissolves and the amount which exists as S2- in solution are very different. Q.25 The solubility of CuS in pure water at 25°C is 3.3 × 10 g L-1. Calculate Ksp of -4 CuS. The accurate value of Ksp of Cus was found to be 8.5 × 10-36 at 25°C A.25 CuS  → Cu2+ + S2− 3.3 × 10−4 g L−1 S = = 3.5 × 10−6 M 95.6 g mol −1 ( 3.5 × 10−6 ) = 2 Apparent Ksp = 1.2 × 10−11. 36.  Q.26 Nicotinic acid (K = 1.4 × 10-5) is represented by the formula HNiC. Calculate its a percent dissociation in a solution which contains 0.10 moles of nicotinic acid per 2.0 L of solution. A.26 Given, HNiC   H+ + NiC− 1 0 0 1−a a a Also, C= 0.1 / 2 = 5 × 10−2 mol L−1 , Ka = 1.4 × 10−5 Ca 2 ∴ Ka = = Ca 2 (1 − a) 1.4 × 10−5 = a Ka /= C = 1.67 × 10−2 or 1.67% 5 × 10 −2 Q.27 For the indicator 'HIn' the ratio (Ind )/(HIn) is 7.0 at pH of 4.3. What is K - eq for the indicator? A.27 pH = 4.3, H= +  5.0 × 10−5 M HIn   H+ + In− H+  In−  Keq =     [HIn] (5.0 × 10−5 ) ( 7.0) = = 3.5 × 10−4 M Q.28 Determine OH−  of a 0.050 M solution of ammonia to which sufficient NH Cl 4 has been added to make the total NH4  equal to 0.1 M. + A.28 NH3 + H2O   NH4+ + OH− NH4+  OH−  (0.1) x K= = = 1.8 × 10−5 b [NH3 ] (0.050 ) thus x OH= = −  9.0 × 10−6 37.   Chapter Summary 1. Based on Ingold concept, an electrophile is an acid whereas a nucleophile is a base. For ex. AlCl3 is an acid and NH3 is a base. 2. In the reaction I2 + I−  → I3− , the Lewis base is Iˉ. 3. [Fe(H2O)6 ]3+ + H2O  → [Fe(H2O)5 OH]2+ + H3O+ Acid Base Conjugate Conjugate base acid 4. [Zn(H2O)5 OH]2+ + H3O+  → [Fe(H2O)6 O]3+ + H2O+ Base Acid Conjugate Conjugate acid base 5. Salt hydrolysis is the reaction of a anion or cation with H2O accompanied by cleavage of O–H bond. 6. A change of even 0.2 unit in pH of blood causes death also. 7. H2CO3 + NaHCO3 is the buffer system present in blood. 8. In case of strong acid with 10–8M concentration pH is 6.95 and in case of strong base with same concn pH is 7.0414. (H+) from H2O = 1 × 10–7 (H+) from acid = 1 × 10–8 Net (H+) = 1.1 × 10–7 pH = –log [1.1 × 10–7] = 6.95 Similarly, for base [O– H] Net = 1.1 × 10–8 pOH = 6.95 pH = 14–6.95 = 7.05 38.