Ionic Equilibrium - Medeasy Equilibrium PDF

# Ionic Equilibrium ## Some Imp points 1. **Types of Salts** * **Normal salts** * NaCl, Na₂SO₄, Na₃PO₄ etc. * **Acid salts** * NaHCO₃, NaHSO₄ * **Basic salts** * Zn(OH)Cl, Mg(OH)Cl etc. * **Double salts** * K₂SO₄. Al₂(SO₄)₃. 24H₂O...

# Ionic Equilibrium ## Some Imp points 1. **Types of Salts** * **Normal salts** * NaCl, Na₂SO₄, Na₃PO₄ etc. * **Acid salts** * NaHCO₃, NaHSO₄ * **Basic salts** * Zn(OH)Cl, Mg(OH)Cl etc. * **Double salts** * K₂SO₄. Al₂(SO₄)₃. 24H₂O Potash alum * **Complex salts** * [Ag(NH₃)₂]Cl, [Cu(NH₃)₄]SO₄, etc. * **Mixed salts** * NaKS, NAKRbPO₄ etc. 2. **Isohydric Sol** * Sol" having same [H] Conc" * [H]₁ = [H]₂ * √Ka₁C₁ = √Ka₂C₂ * [Ka₁C₁ = Ka₂C₂] * Relative strength of acids * R.S = [H]₁/√Ka₁C₁ = [H]₂/√Ka₂C₂ * WA + SB Mixing (K(CH₃C0011) = 10⁻⁵) * Case 1: 75mL of 0.1 M CH₃COOH + 25mL of 0.1 M NaOH * NWA = 75 x 0.1 = 7.5 mmol * MSB = 25 x 0.1 = 2.5 mmol * pH = 7 + [5 + log(25/2)] / 2 = 7 + [5 - 1.3] / 2 = 7 + 1.85 = 8.85 * Case 2: 50 mL of 0.1 M CH₃COOH + 50 mL of 0.1 M NaOH * nWA = 5 * C = 5 / 100 = 1 / 20 * NSB = 5 * pH = 5 + log(25/2) = 5 - log2 = 5 - 0.3 = 4.7 * Case 3: 25mL of 0.1 M CH₃COOH + 75 mL of 0.1 M NaOH * nWA = 2.5; NSB = 7.5 * [OH] = 5 / 100 = 1 / 20 * pOH = -log(1 / 20) = log 20 = 1.3 * [pH = 14 - 1.3 = 12.7] ## Solubility and Solubility Product * Solubility * Max. no. of moles of solute dissolved in a solvent to obtain 1 L sol". * Solubility = S (mol/L) = S (gm/L) / MW salt * Solubility in common ion effect * By adding common ion, solubility of Salt decreases. Solubility ↓ * AgCl Ag⁺ Cl⁻ (aq) (aq) * NaCl Na⁺ Cl⁻ * Q. Find solubility of AgCl (Ksp = 10⁻¹⁰ (mol²/L²)) * (a) In H₂O * (b) In 0.1 M AgNO₃ * Ksp = (0.1) x (S)² * S = 10⁻¹⁰ / 0.1 = 10⁻⁹ (mol/L) * (c) In 0.1 M CaCl₂ * Ksp = (S) x (0.2) = 10⁻¹⁰ * S = 10⁻¹⁰ / 0.2 = 5 x 10⁻¹⁰ (mol/L) * lonic product of salt * ABxA + YB * Kjp = [A]ˣ [B]ʸ * Case 1: Kjp = Ksp: Saturated sol" * Case 2: Kjp > Ksp: ppt occurs * Case 3: Kjp < Ksp: Unsaturated sol" * Q. The ppt of CaF₂ (Ksp = 1.7 x 10⁻²⁰) is obtained when following are mixed. * (A) 10⁻³ M 10⁻⁵ M * (B) 10⁻⁵ M 10⁻³ M * (C) 10⁻² M 10⁻³ M * (D) 10⁻¹ M 10⁻⁴ M * CaF₂ Ca²⁺ + 2F⁻ * (A) Kic=10⁻³ x 10⁻⁵ = 10⁻⁸, Kic < Ksp * (B) Kic = 10⁻⁵ x 10⁻³ = 10⁻⁸, Kic < Ksp * (C) Kic=10⁻² x (2 x 10⁻³)² = 10⁻⁸, Kic < Ksp ## Conjugate acid-base pair * Acid H Conjugate Base * SA H W.C.B * WA H S.C.B * HNO₃→ H⁺ NO₃⁻ * HFH⁺ F⁻ * H₂PO₄⁻ H⁺ HPO₄²⁻ * 1. Acid base theory # Buffer Solutions * The solutions which resist the change in pH by adding small amount of strong acid or strong base * CH₃COOH + CH₃COONa (WA+ Salt of WA with SB) * Acidic buffer * NH₄OH + NH₄Cl (WB + Salt of SA with WB) * Basic Buffer * Acidic buffer can be prepared by three ways * 1. CH₃COOH + CH₃COONa * WA (Salt of WA SB) * 2. WA is mixed with SB (NWA > NSB) * CH₃COOH + NaOH = CH₃COONa + H₂O * Calculation of pH of Buffer Solutions * Q. 50 mL of 2M CH₃COOH mixed with 10 mL of 1M CH₃COONa will have an approx pH of (K = 10⁻⁵) * pH = 5 + log(nsalt/nacid) = 5 + log(50 x 2 / 10 x 1) = 5 + log(10) = 5 - log 10 = 5 - 1 = 4 * Q. 0.05M NH₄OH solution is dissolved in 0.001M NH₄Cl solution find PH (Kb = 10⁻⁵) * pH = 5 + log(0.001 / 50) = 5 - log 50 = 5 - 1.7 = 3.3 * Q. When 75% of titration is over then find PH. [CH3COOH + NaOH = CH3COONa + H₂O] (K = 10⁻⁵) * pH = pKa + log(25 / 75) = 5 + log 3 = 5 + 0.48 = 5.48 # Calculation of Kp if pressure is given. * Q. In the Rxn the eq" pressure is 12 atm. If 50% CO₂ reacts find Kp * C(s) + CO₂(g) ⇌ 2CO (g) * Pea = Po - Px * Peq = Po - 2Px * Pea = Po - 2x² * Peq = Po / 2 * D.O.D of CO₂ = Px / Po = 1 / 2 * Po = 12; Peq = 3Po / 2 = 12; Po = 8 * Kp = (P/2)² = 2P = 2 x 8 = 16 (atm) * Q. For the rxn, the observed pressure is 1.12. atm find Kp? * NH4HS (s) ⇌ NH3(g) + H2S(g) * 2P = 1.12 * P = 0.56 * Kp = Px x Px = 0.56 x 0.56 = 0.3136 (atm)² # Relation b/w D.O.D and Vapour density * α = D.O.D = (D - d) / d(n - 1) * D = V.D of Reactant. * d = V.D of Rxn mix # Significance of n * PCl₅ ⇌ PCl₃ + Cl₂ * n = moles of products = 2 * 2NH₃⇌ N₂ + 3H₂ * n = 1 + 3 = 4 # Le-chatelier's principle 1. **Effect of conc"** * R ⇌ P; [R]↑ →; [P]↑ ← * R ⇌ P; [R] ↓ ←; [P] ↓ → 2. **Effect of Pressure** * (1) N₂(g) + 3H₂(g) ⇌ 2NH₃(g) * (2) 2SO₃(g) ⇌ 2SO₂(g) + O₂(g) * (3) H₂(g) + I₂(g) ⇌ 2HI(g) * Δn = 0 No. effect of pressure change 3. **Effect of temp** * (a) Endothermic Rxn * R + Heat ⇌ P * T↑ ↓ ←, ΔH = +ve * (b) Exothermic Rxn * R ⇌ P + Heat * T↑ T↓ → , ΔH = -ve 4. **Effect of Catalyst** * Does not affect the eq" 5. **Effect of addition of inert gas** * Case 1: at const volume * Does not affect the eq" * Case 2: at const pressure * Rxn will shift in that direc" where no. of moles increases. # Reaction Quotient * aA(g) + bB(g) ⇌ yC (g) + sD (g) * Qc = [C]ʸ[D]ˢ / [A]ᵃ[B]ᵇ * Qc = [C]ʸ[D]ˢ / [A]ᵃ[B]ᵇ * not at eq" * Case 1: Qc = Kc; eq" * Case 2: Qc > Kc; B.W.R. * Case 3: Qc < Kc; F.W.R. # Degree of Dissociation * D.O.D = No. of moles dissociated / Initial moles taken * Case 1: n₁ 2 0 * A ⇌ 2B * n₂ 2 - x x * D.O.D of A = X / 2 * Case 2: n₁ 2 0 * 2A ⇌ B * n₂ 2- 2x x * D.O.D of A = 2x / 2 = x * Case 3: n₁ 2 3 0 * A + 2B ⇌ C * n₂ 2- x 3- 2x x * D.O.D of A = X / 2 * D.O.D of B = 2x / 3 * Case 4: n₁ 2 3 0 * 2A + 2B ⇌ 3C * n₂ 2 - 2x 3 - 2x 3x * D.O.D of A = 2x / 2 = x * D.O.D of B = 2x / 3 * Case 5: n₁ 5 8 0 * N₂ + 3H₂ ⇌ 2NH₃ * n₂ 5 - x 8 - 3x 2x * D.O.D of N₂ = X / 5 * D.O.D of H₂ = 3x / 8 * Case 6: n₁ 4 0 0 * 2SO₃⇌ 2SO₂ + O₂ * n₂ 4 - 2x 2x x * D.O.D of SO₂ = 2x / 4 = x / 2 * Case 7: n₁ 6 7 3 0 * 4NH₃ + 5O₂ ⇌ 4NO + 6H₂O * n₂ 6 - 4x 7 - 5x 4x 6x * D.O.D of NH₃ = 4x / 6 = 2x / 3 * D.O.D of O₂ = 5x / 7 * **Apps of D.O.D** * To Find Eq" constants * Case 1: Calculation of Kc * n₁ a b 0 * A(g) + B(g) ⇌ C(g) * n₂ a - x b - x x * Ca(g) Cb(g) Cc(g) * Kc = (x) / (a - x)(b - x) # Lewis Theory * LA * Lone Pair acceptor * H⁺, CO₂, SF₆ PF₅, BCl₃, AlCl₃, FeCl₃ etc. * LB * Lone Pair donor * XO⁻, OH⁻, NH₃, H₂O" etc. * HCO₃⁻ H⁺ CO₃²⁻ * CO₃²⁻ H⁺ HCO₃⁻ * H₂PO₄⁻ H⁺ HPO₄²⁻ * HPO₄²⁻ H⁺ PO₄³⁻ * H₂PO₄⁻ H⁺ HPO₄²⁻ * H₃PO₂ X * H₃PO₄ H⁺ H₂PO₄⁻ * H₂PO₄⁻ H⁺ HPO₄²⁻ * HPO₄²⁻ H⁺ PO₄³⁻ # Acid Base Theory * Base + H Conjugate Acid * WB SCA * SB WCA * NO₃⁻ + H⁺ HNO₃ * HCO₃⁻ + H⁺ H₂CO₃ * WB SCA * SB WCA # PH Calculation * pH = -log[H⁺] * pOH = -log[OH⁻] * pH + pOH = 14 at 25°C * [H⁺] x [OH⁻] = 10⁻¹⁴ at 40°C * T↑ kw↑ [H⁺] x [OH⁻] * Case 1: Strong acid/Base * (1) [H⁺]≥ 10⁻⁶ M * Q. Find pH of 10⁻³ M HCl * [H⁺] = 10⁻³ M * pH = -log 10⁻³ = 3 * pH = 3 * (2) [H⁺] < 10⁻⁶ * (-log (a x 10⁻ᵇ) = b - loga * Q. Find pH of 10⁻⁸ M HCl * [H⁺] = 10⁻⁸ M * pH = -log 10⁻⁸ = 8 * [H⁺]net = [H⁺]from acid + [H⁺]from H₂O * = 10⁻⁸ + 10⁻⁷ = 11 x 10⁻⁸ * pH = -log(11 x 10⁻⁸) = 8 - log 11 * pH = 6.9 * Case 2: pH of mix of two SA & SB * [H]/[OH] = [N₁V₁ ± N₂V₂] / V₁ + V₂ * = [M₁V₁nf₁ + M₂V₂nf₂] / V₁ + V₂ * Q. Find PH * 50 mL of HCl + 50 mL of H₂SO₄ * [H] = [M₁V₁nf₁ + M₂V₂nf₂] / V₁ + V₂ * = (1 x 50 x 1 + 1 x 50 x 2) / 50 + 50 * = 10/100 = 20/100 = 2 x 10⁻² * pH = -log(2 x 10⁻²) = 1 - log2 = 1 - 0.3 * PH = 0.7 * Q. Find pH * 75 mL of M Ba(OH)₂ + 25 mL of M HCl * Case 3: pH of weak monobasic acid * Weak monoacidic base * H ⇌ A + H⁺, Kₐ * K = [H +]² / [HA] * α² = [H⁺] / [HA] * [H⁺] * K = [H⁺]^2 / [H⁺] * [H⁺] = (K * C) ^ (1/2) * Case 4: pH of weak poly basic acid * Weak poly acidic base * H₃PO₄ H⁺ + H₂PO₄⁻ * H₂PO₄⁻ H⁺ + HPO₄²⁻ * HPO₄²⁻ H⁺ + PO₄³⁻ * Ka₁ > Ka₂ > Ka₃ * α < 1; [H⁺] = Ka₂C = [H₃PO₄] * [H₂PO₄⁻] = Ka₂ * Case 5: PH of mix of two weak acid or two weak bases * [H⁺] = √Ka₁C₁ + Ka₂C₂ * or * [OH⁻] = √Kb₁C₁ + Kb₂C₂ * C1 & C2 conc" after mixing * Q. Find Pt of Soln obtained by mixing equal volumes of 0.02 M HOCI (Ka₁ = 2 x 10⁻⁸) and 0.2 M CH₃COOH (Ka₂ = 2 x 1.0⁻⁵) * C₁= 0.02 / 2= 0.01; C₂ = 0.2 / 2 = 0.1 * [H] = √2x10⁻⁸x10⁻² + 2x10⁻⁵x10⁻¹ * = 2x10⁻¹⁰ + 2x1.0⁻⁶ * = 2 x 10⁻³ * pH = -log(2 x 10⁻³) = 3 - log 2 = 3 - 0.3 * P = 2.7 * Case 6: PH of mix of WA and SA or WB and SB * WA * C₁ * Ka₁ * SA * C₂ * [H⁺] = ( -C₂+ c₂+ 4Ka₂C₂ ) / 2 * Q. Find Pt of mix of 10⁻¹ M HCl & 10⁻² M CH₃COOH (K = 2 × 10⁻⁵) * -10 ± √(10⁻¹)² + 4 x 2 x 10⁻⁵ x 10⁻² * [H†] = * = 2 * -10 ± √ 10⁻² + 8 x 10⁻⁷ ) / 2 * -10 + √ 81 x 10⁻⁸) / 2 * -10 + 9 x 10⁻⁴) / 2 * 8 x 10⁻⁴ / 2 = 4 x 10⁻⁴ * pH = -log(4 x 10⁻⁴) = 4 - log 4 = 4 - 0.6 = 3.4 * Case 7: Ph of Amphiprotic species * HCO₃⁻ / H₂CO₃ * [PH H⁺ / H₂CO₃ * H₂PO₄⁻ / HPO₄²⁻ * [PH H⁺ / H₂PO₄⁻ * H₂PO₄⁻ / PO₄³⁻ * Q. Find Pt of * a) 0.01 M NaH₂PO₄ * b) 0.1 M Na₂HPO₄ * for H₃PO₄ K₁ = 10⁻³ * Ka₂ = 10⁻⁸ * Ka₃ = 10⁻¹³ * a) pH = 3 + 8 / 2 = 11 / 2 = 5.5 * b) pH = 8 + 13 / 2 = 21 / 2 = 10.5 * Salt hydrolysis * WA + SB (CHCOONa) * CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻ (KH) * (1) Anionic hydrolysis * (2) KH= K_w/ K_a * (3) h = (K_h)^(1/2) / C * (4) PH = 7 + [PK_h + log C] / 2 * PH > 7 Nature of sol" after hydrolysis is basic * SA + WB (NH₄Cl)