PCC/PCD 101 Pharm. Analytical Chemistry I Lecture Week 6 PDF

PCC /PCD 101 Pharm. Analytical Chemistry I Fall 2024/2024 – Lecture week 6  pH Calculations 4] Solutions of Salts A- Salts of strong acid and strong base: → Neutral Salts pH = pOH = 7. NaCl → Na+ + Cl- Completely Ionized B- Salts of Weak acids and bases:...

PCC /PCD 101 Pharm. Analytical Chemistry I Fall 2024/2024 – Lecture week 6  pH Calculations 4] Solutions of Salts A- Salts of strong acid and strong base: → Neutral Salts pH = pOH = 7. NaCl → Na+ + Cl- Completely Ionized B- Salts of Weak acids and bases: Salts of weak acids react with H₂O to give a basic solution, while salts of weak bases react with H₂O to give acidic solutions. A hydrolytic reaction (hydrolysis) proceeds because of the tendency of ions of weak acids or bases to react with the H+ or OH- of water forming slightly ionized acids or bases. Hydrolysis of these salts with H₂O doesn't complete to completion, it reaches an equilibrium point and thus may be represented by an equilibrium constant known as kh (hydrolysis Constant)  B- Salts of Weak acids and bases: 1. Salts of weak acids and strong bases CH3COO- + H2O ↔ CH3COOH + OH- sodium acetate CH3COONa Cs ---- ---- before hydrolysis it is a strong electrolyte and completely ionizes. -x +x +x hydrolysis CH3COONa → CH3COO- + Na+ Cs-x +x +x After hydrolysis Cs acetate Cs Cs Hydrolysis of acetate ion formed Kh = Kb = [AcOH][OH-]/[AcO-] CH3COO- + H2O↔ CH3COOH + OH- BASIC NAtURE Hydrolysis constant is the basicity constant of the salt due to its Solution is basic pH ˃ 7 basic nature. As OH-↑ , H+ ↓ Kw = [H+] [OH-] [OH-]=Kw/ [H+]  B- Salts of Weak acids and bases: 1. Salts of weak acids and strong bases X = [OH-] = √Kw x Cs / Ka Kh = Kb = [AcOH]. Kw/[AcO-] [H+] = (Kw x Cs /Ka)1/2 For HOAc ↔ H+ + AcO- pOH = ½ (-Log Kw – log Cs + Log Ka) Ka = [AcO-] [H+] / [AcOH] pOH = ½ (pKw +pCs –pKa) 1/Ka = [AcOH] / [AcO-] [H+] pH = pKw –pOH Kh = Kw/Ka = [AcOH][OH-]/[AcO-] = pKw – ½ (pKw +pCs –pKa) = Kw/Ka = x.x /Cs-x= x2 / Cs-x = pKw - ½pKw - ½ pCs +1/2 pKa = ½ pKw - ½ pCs +1/2 pKa X is small compared to Cs (neglected) pH =½ (pKw - pCs +pKa) = Kw/Ka = x2 / Cs = [OH-] / Cs %degree of hydrolysis (%h) = [OH-]/Cs x 100  B- Salts of Weak acids and bases: 2. Salts of weak base and strong acid NH4+ + H2O ↔ NH4OH + H+ Ammonium chloride NH4Cl Cs ---- ---- before hydrolysis it is a strong electrolyte and completely ionizes. -x +x +x hydrolysis NH4Cl → NH4+ + Cl- ammonium Cs-x +x +x After hydrolysis Cs Cs Cs Kh = Ka = [NH4OH][H+]/[NH4+] Hydrolysis of ammonium ion formed NH4+ + H2O↔ NH4OH + H+ ACIDIC NAtURE Hydrolysis constant is the acidity constant of the salt due to its acidic nature. Solution is Acidic pH ˂ 7 Kw = [H+] [OH-] [H+]=Kw/ [OH-] As OH-↓ , H+ ↑  B- Salts of Weak acids and bases: Kh = [NH4OH]. Kw/[NH4+] [OH-] X = [H+] = √Kw x Cs / Kb For NH4OH ↔ NH4+ + OH- = (Kw x Cs /Kb)1/2 Kb = [NH4+] [OH-] / [NH4OH] pH = ½ (-Log Kw – log Cs + Log Kb) 1/Kb = [NH4OH] / [NH4+] [OH-] pH = ½ (pKw +pCs –pKb) Kh = Kw/Kb = = [NH4OH] [H+] / [NH4+] %degree of hydrolysis (%h) = [H+]/Cs x 100 = Kw/Kb = x.x /Cs-x = x2 / Cs X is small compared to Cs (neglected) = [H+] 2/ Cs  Problems: Calculate pH of 0.1 M sodium acetate and degree of hydrolysis.  Problems: Calculate pH of 0.01 M ammonium chloride and degree of hydrolysis. (pKa= 9.24)  pH Calculations 5] Solutions of Acid Salts Salts of polyprotic acid (salts containing hydrogen) e.g. NaHCO3 sodium bicarbonate NaH2CO4 sodium hydrogen oxalate KH phthalate potassium hydrogen phthalate NaHPO4 sodium dihydrogen phosphate Solutions of weak acid salts of the HA- can act as acids and as bases (Amphoteric) e.g. HCO3- HCO3- ↔ H+ + CO32- ionization as acid K1: 1st ionization constant HCO3- + H2O ↔ H2CO3 + OH- ionization as base K2: 2nd ionization constant pH = ½ (pK1+pK2) [H+] = √K1.K2 pH Calculations Disproportionation of HCO3- 2 HCO3- ↔ H2CO3 + CO3- Problem!! Calculate pH of 0.1 M NaHCO3 K1 = 3.5 x 10-7 and K2 = 6 x 10-11 pH = ½ (pK1 +pK2) = ½ (6.45 +10.22) = 8.335 pK1= -Log K1 = -log 3.5 x 10-7 = 6.45 pK2 = -Log K2= -Log 6 x 10-11 = 10.22  pH Calculations 6] Buffer solutions: Solution that resist change in pH when a small amount of an acid or base is added or when solution is diluted. A) Acidic Buffer: Mixture of weak acid and its conjugate base (Salt) e.g, HOAC / OAC- Acetic acid/Acetate pH = pKa + Log [A-] / [HA] Handerson Hassebalch Equation pH = pKa + Log Cs/Ca Cs: Salt Ca: weak acid pH = pKa + Log [Conjugate Base] / [Acid] pH = pKa + Log [Proton Acceptor] / [Proton Donor] Mechanism of Buffer Action: If added strong acid: protons will combine with an equal amount of the A- to form HA (weak Acid). H+ + A- ↔ HA A- ↓ HA↑ Ratio [A-]/[HA] change will be small and pH does not sharply change. Thus, increasing the A- or HA composition of buffer will lead to increase the efficiency of the buffer as it will resist pH change efficiently. Mechanism of Buffer Action: If added strong base: will combine with an equal amount of the HA to form A-(salt). OH- + HA ↔ A - + H2 O A- ↑ HA ↓ Ratio [A-]/[HA] change will be small and pH does not sharply change. Thus, increasing the A- or HA composition of buffer will lead to increase the efficiency of the buffer as it will resist pH change efficiently.  pH Calculations 6] Buffer solutions: Solution that resist change in pH when a small amount of an acid or base is added or when solution is diluted. B) Basic Buffer: Mixture of weak base and its conjugate acid e.g, NH4OH / NH4+ pOH = pKb + Log [B+] / [BOH] Handerson Hassel balch equation pH = pKa + Log Cs/Ca Cs: Salt Ca: weak acid pH = pKa + Log [Conjugate Acid] / [Base] pH = pKa + Log [Proton Donor] / [Proton Acceptor] pH = pKw –(pKb + Log [B+] / [BOH])  pH Calculations Mechanism of Buffer Action: If added strong acid: protons will react with an equal amount of the BOH to form B+. H+ + BOH ↔ B+ BOH ↓ B+↑ Ratio [B+]/[BOH ] change will be small and pH does not sharply change. Thus, increasing the A- or HA composition of buffer will lead to increase the efficiency of the buffer as it will resist pH change efficiently.  pH Calculations Mechanism of Buffer Action: If added strong base: will combine with an equal amount of B+ to form BOH. OH- + B+ ↔ BOH BOH ↑ B+↓ Ratio [B+]/[BOH ] change will be small and pH does not sharply change. Thus, increasing the A- or HA composition of buffer will lead to increase the efficiency of the buffer as it will resist pH change efficiently.  Buffer Capacity Number of moles of a strong base that cause one liter of buffer to undergo one unit change in pH. The buffer capacity is governed by the ratio of salt to acid or base, where buffer capacity increases as concentration of buffer components increases and reaches a maximum capacity when the ratio is UNITY. pH = pKa Maxmium buffer capacity is when Ca=Cs For acidic buffers: Ca = Cs so, pH = pKa For basic buffers: Cb= Cs so pH = pKw – pKb Buffer Range: Acidic Buffers: pH = pKa ± 1 Basic Buffers: pH = pKw – (pKb ± 1)us, increasing the A- or HA composition of buffer will lead to increase the efficiency of the buffer as it will resist pH change efficiently.  Factors affecting pH of buffer solutions: 1)DILUTION: 2)Temperature pH of both acidic and basic buffers depend on Acidic Buffer slightly changes by the ratio of molar concentration of the 2 components of the buffer only with no any temperature. effect on dilution. Basic buffers makes a strong change in pH Thus, Dilution should have no effect on pH with temp. due to Kw which appears in the equation and changes with temp. BUT!!!!! Extensive dilution may cause a small deviation due to: 1) It alters the activity coeffecients. 2) Water itself can act as weak acid or base.