Ionic equilibria 3.pdf

3. Ionic Equilibria Introduction The equilibrium between ions and unionized molecule in solution is called ionic equilibrium. These principle used in dealing with equilibrium constant and concentrations of ions and unionized species. Some common example of ionic...

3. Ionic Equilibria Introduction The equilibrium between ions and unionized molecule in solution is called ionic equilibrium. These principle used in dealing with equilibrium constant and concentrations of ions and unionized species. Some common example of ionic equilibria is i. H+ and OH- ions and unionized water molecules. ii. Ionization of weak acids and weak bases. iii. Reaction between ions of salts and ions of water. iv. Solid salt and its ions in water. Types of electrolyte A)Electrolyte :- The substances which give rise to ions when dissolved in water are electrolyte. On the basis of ionisation in dilute aqueous solutions the electrolytes are classified into strong and weak electrolyte. i. Strong electrolyte:- The electrolytes ionizing completely or almost completely into ions are called strong electrolytes. For exm strong acids (HCl,H2SO4) strong bases (NaOH, KOH) and salts (NaCl). ii. Weak electrolyte:- The electrolytes ionising to small extent in the aqueous solution is called as weak electrolyte. For exm weak acids(H3C-COOH,HCN) weak bases (NH4OH)etc. The weak electrolytes dissociate only partially in dilute aqueous solutions. An equilibrium thus can be established between the ions and nonionized molecules. The ionization reaction there in is represented as double arrow (⇌) between the ions and nonionized molecule.  Degree of dissociation :- it is defined as a fraction of total number of moles of the electrolyte that dissociates into its ions when the equilibrium is attained. It is denoted by symbol (α) 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑑𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑒𝑑 ∴ α= ------(1) 𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 Percent dissociation = α × 100 -------(2) If ‘c’ is the molar concentration of an electrolyte the equilibrium concentration of cation or anion is (α × c) moldm-3.  Acids and Bases :- Acids and bases are familiar chemical compounds. In acids turns blue litmus paper into red reacts with active metals with evolution of H2 gas and having sour in taste. For exm Acetic acid is found in vinegar, citric acid in lemons etc. In bases turns red litmus paper into blue and is bitter in taste. For exm magneshium hydroxide in antacids, ammonia in household cleaning products. There are some acids and bases we come across in everyday life.   Arrhenius theory of acids and bases :- According to this theory acids and bases are defined as follows. Acid:- Acid is a substance which contains hydrogen and gives rise to H+ ions in aqueous solution. For exm Arrhenius described H+ ion water as bare ions, they hydrate in aqueous solutions and thus represented as hydronium ions H3O+. Base :-Base is a substance that contain OH group and produces hydroxide ions (OH-) in aqueous solution. For exm Limitation - This theory is applicable only in aqueous solution but does not applicable in nonaqueous or gaseous media. For exm. It does not account for the basicity of NH3 and Na2CO3 which do not have OH group.  Bronsted – Lowry theory:- In 1923 Bronsted and T.M.Lowry proposed a more general acid base theory which is based on proton transfer. Acids :- Acid is a substance that donates a proton (H+) from another substance. Base :- Base is a substance that accepts a proton (H+)from another substance. For exm Ion the above reaction HCl and NH4+ are proton donars and acts as acids. The NH3+ and Cl- are proton acceptors and acts as bases. Further it follows that the products of the Bronsted Lowry acid base reactions are acids bases. The base produced by accepting the proton from an acid is the conjugate base of that acid. While the acid produced when a base accepts a proton is called the conjugate acid of that base. A pair of an acid and a base differing by a proton is said to be a conjugate acid base pair.   Lewis theory :-In 1923 G.N. Lewis introduced acid base concept. This concept is much greater than that of Bronsted and it is explained not in terms of ionic reactions but in terms of bond formation. According to this theory acids and bases are defined as follows. Acid:- any species that accepts a share in an electron pair is called lewis acid. Base :-Any species that donates a share in an electron pair is called lewis base. For example Amphoteric nature of water :- Water has the ability to act as an acid as well as a base. Such behaviour is known as amphoteric nature of water. H2O act as an acid towards NH3 and acts as a base toward HCl. Therefore H2O is amphoteric. Ionisation of acids and bases :- On the basis of their extent of dissociation or ionisation and number of H+ ions and OH-ions acids and bases are classified as strong acids, strong bases, weak acids and weak bases. i. Strong acids:- If acids completely dissociates and gives more number of H+ ion then they are called as strong acids. For exm:- HCl(aq) H+(aq) + Cl-(aq) HCl, H2SO4, HNO3,HBr,HClO4 etc are examples of strong acids. ii. Weak acids:- Those acids which dissociate to small extent and gives less number of H+ions are called as weak acids. For exm H3C-COOH(aq) ⇌ H3C-COO-(aq) + H+(aq) HCOOH, HF, H2S are example of weak acids. iii. Strong base:- Those bases which dissociate to large extent and gives more number of OH- ions are called as strong bases. For exm: NaOH(aq) → Na+(aq) + OH-(aq) KOH, LiOH, CsOH are example of strong bases. iv. Weak base:- Those bases which dissociate to small extent and gives less number of OH - ions are called as weak bases.  For exm: NH4OH(aq) ⇌ NH4+(aq) + OH-(aq) Fe(OH)3, Cu(OH)2 are example of weak bases. Dissociation constant of weak acids and weak bases:- The dissociation constant of weak acid or a weak base is defined as the equilibrium constant for dissociation equilibrium of weak acid or weak base respectively. Or “ The equilibrium constant which is obtained by law of mass action to any equilibrium reaction”. For exm the dissociation of a weak acid HA in water is expressed as HA(aq) ⇌ H+(aq) + A-(aq) Applying law of mass action [𝐻+] [𝐴−] Ka = [𝐻𝐴] --------(1) Similarly the dissociation of the weak base BOH in water is represented as dissociated would be α, where α is degree of dissociation of acid. The fraction of an acid that remains undissociated would be (1-α). `Initial mole HA ⇌ H+ + A- 1 0 0 Number of moles 1-α α α at equilibrium 1−𝛼 𝛼 𝛼 Concentration at 𝑣 𝑣 𝑣 equilibrium in mol/dm3 1−𝛼 Thus, at equilibrium [HA] = mol/dm3, 𝑣 𝛼 [H+] = [A-] = mol/dm3 𝑣 Substituting these value in equation (1) 𝛼 𝛼 ( )( ) 𝛼2 𝑣 𝑣 Ka = 1−𝛼 = 1−𝛼 [ ] × 𝑣2 𝑣 𝑣 𝛼2 Ka = (1−𝛼)×𝑣 ------(2) If C is the intial concentration of an acid in mol/dm3 and v is the volume in dm3/mol then C= 1/v. Replacing 1/v in equation (2) by C we get 𝛼2 Ka = -----(3) (1−𝛼) For the weak acid HA, α is very small, or (1-α) ≅1 with this equation (2) and (3) reduce BOH(aq) ⇌ B+(aq) + OH-(aq) Applying law of mass action [𝐵+][𝑂𝐻−] Kb = -------(2) [𝐵𝑂𝐻] Where Ka and Kb are dissociation constant of weak acid, weak base respectively.  Ostwald’s dilution law :- Arrhenius concept of acids and bases was expressed quantitatively by F.W.Ostwald in the form of the dilution law in 1888. Statement :-“Degree of dissociation of weak electrolyte (acid or base) is inversely proportional to square root of its concentration or directly proportional to square root of volume of the solution containing 1 mol of the weak electrolyte”. a) Weak acids :- Consider an equilibrium of weak acids HA that exist in solution partly as the undissociated species HA and partly H+ and A- ion. Then HA ⇌ H+(aq) + A-(aq) The acid dissociation constant is given by (Ka) [𝐻 + ] [𝐻 − ] Ka= ------(1) [𝐻𝐴] Suppose 1 mol of acid HA is initially present in volume V dm 3 of the solution. At equilibrium the fraction Ka = α2/v and Ka = α2c --------(4) α = √𝐾𝑎/𝑐 or α = √𝐾𝑎 × 𝑣 ------(5) thus, the equation that the Ostwald’s dilution law of weak acid. b) Weak base :- Consider 1 mol of weak base BOH dissolved in v dm 3 of solution. The base dissociates partially as BOH(aq) ⇌ B+(aq) + OH-(aq) The base dissociation constant is [𝐵+] [𝑂𝐻−] Kb = -------(1) [𝐵𝑂𝐻] Let the fraction dissociated at equilibrium is α and that remains undissociated is (1-α). BOH ⇌ B+(aq) + OH-(aq) Initial mole 1 0 0 Number of moles at 1-α α α equilibrium concentration at equilibrium 1-α/v α/v α/v 3 in mol/dm At equilibrium [BOH] = 1-α/v mol/dm3 [B+] = [OH-] = α/v mol/dm3 Substituting these value in eq (1) gives 𝛼 𝛼 ( )( ) 𝛼2 𝑣 𝑣 Kb = 𝛼 = 1−𝛼 --------(2) (1− ) 𝑣 𝑣 If C is initial concentration of an acid in mol dm -3 and v is the volume in dm3/mol then c= 1/v. replacing 1/v in eq(2) by c we get Kb = α2c/(1-α) ------(3) For weak base BOH, 𝛼 is very small or (1-α) ≅ 1. With this eq (2) and (3) reduce Kb = α2/v and Kb = α2c -------(4) α = √𝐾𝑏/𝑐 or α =√𝐾𝑏 × 𝑣 --------(5) Thus, the equation (5) shows that the ostwald’s dilution law. Autoionization of water :- Pure water ionizes to a very small extent. The ionization equilibrium of water is represented as H2O(l) + H2O(l) ⇌ H3O+(aq) + OH-(aq) But general dissociation of water can be represented as H2O ⇌ H+ + OH- The equilibrium constant (k) for the ionization of water is given by [𝐻3𝑂+][𝑂𝐻−] [𝐻+][𝑂𝐻−] K=[ or K = 𝐻2𝑂]𝑙 [𝐻2𝑂]𝑙 [𝐻2𝑂] But concentration of water [H2O] molecules are practically constant because a majority of H2O molecules are undissociated. Then [H2O]2 = K1. Substituting this in eq (2) we get K × K1 = [H3O+] [OH-] ------(3) Kw = [H3O+] [OH-] Where Kw= KK1 ia called ionic product of water. “ The product of molar concentration of H+ ions and OH- ions in their aqueous solution or pure water at a given temperature”. In pure water H3O+ ion concentration always equals the concentration of OH- ion. Thus at 298k this concentration is found to be 1.0 × 10-7 mol/l Kw = (1.0 × 10-7) (1.0 × 10-7) Kw = 1.0 × 10-14 --------(4) The value of Kw at different temperature as follows 273 K 0.11 × 10-14 mol2/dm6 283 K 0.31 × 10-14 mol2/dm6 298 K 1 × 10-14 mol2/dm6 398 K 7.5 × 10-14 mol2/dm6  pH Scale :- The pH concept is very convenient to express it on the logarithmic scale. This is known as pH scale. It was introduced by P.L. Sorenson in 1909. In the symbol of pH and pOH the ‘p’ indicated potenz or power i.e. strength of hydrogen concentration or hydroxyl ion concentration. pH:- It is defined as the negative logarithm to the base 10, of the concentration of H+ ions in solution in mol dm-3. Expressed mathematically as pH= -log10[H+] -------(1) similarly pOH:- It is defined as, the negative logarithm to the base 10 of the molar concentration of OH- ions in solution. Thus, pOH = - log 10 [OH-] --------(2) Relationship between pH and pOH:- The ionic product of water is Kw = [H3O+][OH-] Now, Kw = 1×10-14 at 298 k and thus [H3O+] [OH-] = 1.0 × 10-14 Taking logarithm of both sides, we write Log10[H3O+] + log10[OH-] = -14 -log10[H3O+] + [-log10(OH-)] = 14 From equation (1) and (2) pH + pOH = 14 -------(3) Acidity, basicity and neutrality of aqueous solutions a) Neutral solution :- For pure water or any aqueous neutral solution at 298k [H3O+] = [OH-] = 1.0 × 10-7m  Hence pH= -log10[H+] = -log10[1×10-7] = 7 b) Acidic solution : In acidic solution there is a excess of H3O+ or H+ ions or [H3O+] > [OH+] hence, [H3O+]> 1× 10-7 and pH 7. Hydrolysis of salts :- Hydrolysis of salts is defined as the reaction in which cation or anions or both ions of salt react with ions of water to produce acidity or basicity (alkalinity) or even neutrality is called hydrolysis of salt. When one or more of the salt ions react with water, and the equality of concentration of H3O+ and OH- ions is disturbed. The solution become acidic or basic depending on the type of salt. When concentration of H3O+ and OH- is equal. The solution become neutral. Pure water is neutral because [H3O+] = [OH-]. There are four types of salts. A. Salts derived from strong acids and strong bases:- (salts of strong acids and strong base) For exm :- NaCl, Na2SO4, NaNO3, KCl, KNO3 Consider a salt of NaCl which is salt of strong acid (HCl) and strong base (NaOH). On hydrolysis, the hydrolysis reaction can be written as NaCl + H2O ⇌ HCl +NaOH Above reaction can be written in terms of ionic form Na+(aq) + Cl-(aq) +H2O ⇌ H+(aq) + Cl-(aq) + Na+(aq) +OH-(aq) Cancelling the common ion H2O ⇌ H+ + OH- Thus the reactant and the products are the same. This implies that neither the cation nor anion of the salt react with water or there is no hydrolysis. Equality H3O+ or H+ = OH- produced by ionization of water is not disturbed and solution is neutral. It may be concluded that salt of strong acid and strong base does not undergo hydrolysis and pH is equal to 7. B. Salt of strong acids and weak bases:- When salt of strong acid as like (H2SO4) and weak base Cu(OH)2 undergoes hydrolysis, it dissociated in the solution and produce acidity, therefore the nature of such a salt solution remains acidic. For exm NH4Cl, CuSO4, NH4NO3, CuCl2 CuSO4 + H2O ⟶ Cu(OH)2(aq) + H2SO4(aq) This equation can be written in terms of ions CuSO4(aq) ⟶ Cu2+(aq) + SO2-(aq) SO42- ions of salt have no tendency to react with water because the possible product H2SO4 is strong electrolyte. The reaction of Cu2+ ions with OH- ions form unionized Cu(OH)2. The hydrolytic equilibrium for CuSO4 is then written as Cu2+(aq) + 4H2O(l) ⇌ Cu(OH)2(aq) + 2H3O+(aq) Due to presence of excess of H3O+ ions, the resulting solution of CuSO4 becomes acidic and turns blue litmus red.  Formation of sparingly soluble Cu(OH)2 by hydrolysis makes the aqueous solution of CuSO4 turbid. If H2SO4, that is H3O+ ions are added, a turbidity of Cu(OH)2 dissolves to give a clear solution. C. salts of weak acids and strong bases:- when salt of strong base (NaOH) and weak acid (H3O-COOH) undergoes hydrolysis, it dissociated in the solution and produce basicity, therefore the nature of such a salt solution remains basic. Example H3C-COONa, H3C-COOK, Na2CO3, NaC2O4 etc. consider a salt of H3C-COONa hydrolysis in water, it dissociates completely. H3C-COONa(aq) ⟶ H3C-COO-(aq) + Na+(aq) Water dissociates slightly as, H2O(l) + H2O(l) ⇌ H3O+(aq) + OH-(aq) Solution of H3C-COONa contains Na+, H3O+, H3C-COO-, OH-. The Na+3 ions of salt have no tendency to react with OH- ions of water and H3C-COO- ion of SaH react with H3O+ ions from water produces unionized H3C-COOH. H3C-COO(aq)- + H2O(l) ⇌ H3C-COOH(aq) + OH(aq)- Thus, the hydrolytic equilibrium for H3C-COONa is H3C-COONa(aq) + H2O(l) ⇌ H3C-COOH(aq) + Na(aq)+ + OH(aq)- As a result of excess OH- ions produced the solution becomes basic. The solution of H3C-COONa is therefore basic. D. Salts of weak acids and weak bases :- When a salt of weak acid (like H3C-COOH) and weak base (NH4OH) undergoes hydrolysis, it dissociates in the solution and produce basicity or acidity or neutrality depends upon dissociation constant values of acid and base. Examples H3C-COONH4, HCOONH4, (NH4)2CO3, NH4CN etc. Consider when salt of BA dissolved in water, it dissociates completely as BA(aq) ⟶ B+(aq) + A-(aq) The hydrolysis reaction involves the interaction of both the ions of the salt with water. B+(aq) + A-(aq) + H2O ⇌ BOH + HA (weak base) (weak acid) I. If, Ka > Kb, the solution will be acidic II. If, Ka < Kb, the solution will be basic III. If, Ka = Kb, the solution will be neutral. Salt of weak acid and weak base for which Ka>Kb. NH4F is a salt of weak acid HF (Ka=7.2 × 10-4) and weak base NH4OH (1.8 × 10-5). Here, Ka is greater than Kb. The salt hydrolysis as NH4+(aq) + F-(aq) + H2O ⇌ NH4OH(aq) + HF(aq) (weak base) (weak acid) The acid HF is slightly stronger than base NH4OH. The two ions react with water as NH4+(aq) + 2H2O(l) ⇌ NH4OH(aq) + H3O+(aq) ------(1) F-(aq) + H2O(l) ⇌ HF(aq) + OH-(aq) -------(2) The NH4+ ions hydrolysis to a slightly greater extent than the F- ions. That means the reaction produces more H3O+ ions than the OH- ion produced in reaction. Thus, the solution of NH4F is only slightly acidic and turns blue litmus red. Salt of weak acid and weak base for which Ka