Notes on The Basic Approach to Chemical Equilibrium | CHE 111, Study notes of Chemistry

Download Notes on The Basic Approach to Chemical Equilibrium | CHE 111 and more Study notes Chemistry in PDF only on Docsity! Chapter 4 The Basic Approach to Chemical Equilibrium Text Notes I. The Chemical Composition of Aqueous Solutions A. Water is the most plentiful solvent on Earth and serves as the major medium for chemical analyses. B. Electrolytes – solutes that form ions when dissolved in water producing a solution with electrical conductivity. -Strong electrolytes ionized completely in water while weak electrolytes only partially ionize, so a strong electrolyte solution will be a better conductor or electricity. -Strong electrolytes consist of acids, bases and salts. C. Acids and Bases 1. Bronstead-Lowry Theory – acids donate protons/bases accept protons -For a species to act as an acid a base (proton acceptor) must be present -and vice versa. -The species produced when an acid gives up a proton is called the conjugate base of the parent acid. Acid1  base1 + proton -When a base accepts a proton a conjugate acid is produced. Base2 + proton  acid2 -If the 2 processes are combined a neutralization reaction occurs: Acid1 + base2  base1 + acid2 NH3 + H2O  NH4+ + OH- Problem 4-4 Acid Conjugate Base a) HOCl OCl- b) H2O OH- c) NH4+ NH3 d) HCO3- CO32- e) H2PO4- HPO42- D. Amphiprotic Solvents – a solvent that can act as either an acid or base depending on the solute it’s in. - methanol, ethanol, anhydrous acetic acid and dihydrogen phosphate ion are all examples of amphiprotic solvents. H2PO4- + H3O+  H3PO4 + H2O H2PO4- + OH-  HPO42- + H2O - zwitterions – an amphiprotic compound that is produced by a simple amino acid’s weak acid and weak base functional groups. - zwitterions carry both a positive charge (amino group) and negative charge (carboxyl group). 1. Amphiprotic solvents can undergo spontaneous self-ionization or autoprotolysis to form 2 different ionic species. H2O + H2O  H3O+ + OH- Both the hydronium ion and hydroxyl ion will have concentrations of about 10-7 M. Problem 4-6 Expressions of Autoprotolysis a) 2H2O  H3O+ + OH- b) 2CH3COOH  CH3COOH2+ + CH3COO- c) 2CH3NH2  CH3NH3+ + CH3NH- d) 2CH3OH  CH3OH2+ + CH3O- E. Strong and Weak Acids and Bases 1. Strong acids dissociate completely in water, weak acids partially dissociate yielding both parent acid and conjugate base, 2. Acids and bases can be anionic, cationic or neutral - perchloric and hydrochloric acids are strong and will completely dissociate in water leaving a conjugate bases and spectator ion. - ammonium ion and acetic acid are weak acids and have a higher affinity for protons. - acetic acid can act as a differentiating solvent in which various acids dissociate to different degrees and thus have different strengths, while water acts as a leveling solvent for strong acids because the strong acid will dissociate completely and have no differences in strength. F. Chemical Equilibrium There is never actually a complete conversion of reactants to product in a chemical reaction, there is only a chemical equilibrium. 1. A chemical equilibrium state is when the ratio of concentration of reactants and products is constant. 2. An equilibrium-constant expression is an algebraic equation that describes the concentration relationships that exist among reactants and products at equilibrium. H3AsO4 + 3I- +2H+  H3AsO3 + I3- + H2O This equilibrium reaction can be monitored as it moves to the right by the orange to red color change of the triiodide ion. Once the color becomes g) S = [Bi2+] = 1/3[I-] Ksp = (S)(3S)3 = 27S4 h) S = [Mn2+]=[NH4+][PO43-] Ksp = (S) (S) (S) = S3 Problem 4-16 # mmol IO3- = 50 ml (.300M) = 1.9x10-3 a) # mmol PdCl62- = 50 ml (.400 mmol/ml) = 20 mmol # mmol excess K+ = 20 mmol – 2(5mmol) = 10 mmol [K+] = 10 mmol K+ / 50 ml = 0.200 M b) # mmol PdCl62- = 50 ml (.200M) = 10 mmol S = [PdCl62-] = ½ [K+] = Ksp = [K+]2 [PdCl62-] = (2S)2(S) = 4S3 = 6x10-6 S = 1.14x10-2 K+ = 2S = 2.2x10-2 M c) # mmol PdCl62- added = 50 ml (.400M) = 20 mmol # mmol excess PdCl62- = 20 mmol – ½(20 mmol) = 10 mmol [PdCl62-] = 10 mmol/ 100 ml + S ≈ 0.100 + S = 0.100 M [K+] = 2S [K+]2 [PdCl62-] = 6x10-6 = 4S2 (0.100 M) S = √1.5x10-5 = 3.9x10-3 M 1. The common ion effect is responsible for the reduction in solubility of an ionic precipitate when a soluble compound combining one of the ions of the precipitate is added to the solution in equilibrium with the precipitate. 2. The common ion effect is a mass action effect predicted from the Le Chatelier principle. 3. The Le Chatelier principle states that the position of equilibrium in a system always shifts in a direction that tends to relieve an applied stress to the system. J. Writing Dissociation Constants for Acids and Bases - Ka is the acid dissociation constant HNO2 + H2O  H3O+ + NO2- Ka = [H3O + ][NO 2- ] [HNO2] - Kb is the base dissociation constant NH3 + H2O  NH4+ + OH- Kb = ([NH4+][OH-]) / [NH3] Water is not necessary in the denominator of either equation because the concentration of water is so large compared to the concentration of the weak acid or base that the dissociation does not change [water] appreciably. Kw = [H3O+][OH-] = KaKb Kw = KaKb K. Calculating the [H30+] in Solutions of Weak Acids 1. When a weak acid, HA, is dissolved in water, 2 equilibria are established that give hydronium ions: HA + H2O  H3O+ +A- Ka = [H3O+][A-] [HA] 2H2O  H3O+ + OH- Kw = [H3O+][OH-] The hydronium produced in the first reaction suppress the dissociation of water so that the contribution of hydronium ions from the second equation are negligible so, [A-] = [H3O+] The sum of the molar concentrations of the weak acid and its conjugate base must equal the analytical concentration of the acid CHA because the solution has no other source of A- ions so, cHA = [A-] + [HA] or cHA = [H3O+] + [HA] rearrange and get, [HA] = cHA – [H3O+] the Ka expression would then be: Ka = [H3O + ] 2 CHA – [H3O+] This can be further simplified to [H3O+] = √KaCHA Problem 4-19 [H3O+] of water at 100°C at 100°C the Ksp of water = 49x10-14 M [H3O+] = √49x10-14 = 7x10-7 M 2. Hydroxide Ion concentrations in solutions of weak acids can be found using the same approach used in finding hydronium ion concentration. NH3 + H2O  NH4+ + OH- Kb = [NH4+ ][OH-] [NH3]