Tài liệu Chapter 8B: Systematic Treatment of Equilibriu docx

Chapter 8B: Systematic Treatment of Equilibrium I. Introduction ¾Concentrations of individual species ¾Determining which species are present ¾Reactions ¾Equations ¾Steps in Figure 9-1 Example 9-1 ¾ BaSO 4 (s) <=> Ba 2+ + SO 4 2- ¾ 2H 2 O <=> H 3 O + + OH - ¾ SO 4 2 - + H 3 O + <=> HSO 4 - + H 2 O ¾HCl + H 2 O <=> H 3 O + + Cl - I. Introduction ¾Concentrations of individual species ¾Determining which species are present ¾Reactions ¾Equations ¾Assumption ¾Steps in Figure 9-1 A Systematic Method II. A Systematic Method for Solving Multiple Equilibria Problems ¾Equilibrium-constant expressions ¾Mass (concentration)-balance equations ¾A single charge-balance equation Mass (concentration)-balance equations ¾Equilibrium concentrations of various species in a solution to one another and to the analytical concentration ¾Two Steps: ¾Reactions ¾Equations Example 9-1: Write mass-balance expressions for a 0.0100 M solution of HCl that is in equilibrium with an excess of solid BaSO 4 (s). ¾ BaSO 4 (s) <=> Ba 2+ + SO 4 2- ¾ HCl + H 2 O <=> H 3 O + +Cl - ¾ 2H 2 O <=> H 3 O + + OH - ¾ SO 4 2 - + H 3 O + <=> HSO 4 - + H 2 O [Ba 2+ ] = [SO 4 2 - ] + [HSO 4 - ] [H 3 O + ] = [Cl - ]+[OH - ]-[HSO 4 - ] = C HCl + [OH - ] - [HSO 4 - ] =0.0100 +[OH - ] - [HSO 4 - ] [Cl - ] = C HCl Example 9-2: Write mass-balance expressions for the system formed when a 0.010 M NH 3 solution is saturated with AgBr. ¾ AgBr (s) <=> Ag + + Br - [1] ¾ Ag + + 2NH 3 <=> [Ag(NH 3 ) 2 ] + [2] ¾ NH 3 +H 2 O <=> NH 4 + + OH - [3] ¾ 2H 2 O <=> H 3 O + + OH - [4] [Br - ] = [Ag + ] + [[Ag(NH 3 ) 2 ] + ] C NH3 = [NH 3 ] + [NH 4 + ] + 2[[Ag(NH 3 ) 2 ] + ] = 0.010 M [ OH - ] = [NH 4 + ] + [H 3 O + ] A single charge-balance equation ¾Solutions are neutral ¾ The Sum of molar concentration of positive charge = The sum of molar concentration of negative charge ¾Concentration of charge by an ion = molar concentration of that ion x its charge ¾Steps: ¾List all ions in the solution and their charges and concentrations ¾ The Sum of molar concentration of positive charge = The sum of molar concentration of negative charge [...]... Pertinent Equilibria The definition of unknown Equilibrium Constant Expressions Mass balance equation Charge balance equation Check No of unknown species concentrations and No of independent equations Approximations: Check of assumptions Calculate unknown concentrations III Calculation of Solubility by the Systematic Method Metal Hydroxides (Examples 9-5) The effect of pH on solubility (1) pH is fixed... (Coordinate compound): a compound consisting either of complex ions and other ions of opposite charge or of neutral complex species Formation of Complex Ion: Example 9-8 The solubility product of CuI is 1.0x10-12 The formation constant K2 for the reaction of CuI with I- to give CuI2 – is 7.9x10-4 Calculate the molar solubility of CuI in a 1.0x10-4 M solution of KI Step 1 Pertinet Equilibria CuI (s) ... increased due to the reaction of S-2 with H2O Any Questions? Review of Textbook for Chem115/116: Chapters 17 & 23: Complex (coordinate compound) Coordinate covalent bonds: a bond formed when both electrons of the bond are donated by one atom [H3N: Ag :NH3]+ Ag+ + 2(:NH3) Electron configuration of Ag [Kr]4d105s15P0 Ag+ [Kr]4d105s0 5P0 Sp hybrid orbitals: accommodate 2 pairs of electrons Linear Complex... is variable or unknown (Examples 9-7) Formation of Complex Ion (Examples 9-8) Separation of ions (Examples 9-9) Metal Hydroxides: Example 9-5: Calculate the molar solubility of Fe(OH)3 in water 1 Pertinent Equilibria: Fe(OH)3 (s) < = = = > Fe+ 3 + 3 OH2 H2 O (l) < = = = = > H3 O+ + OH2 The definition of unknown: Molar solubility Fe(OH)3 = [Fe+ 3 ] 3 Equilibrium Constant Expressions [Eq 1] Ksp = [Fe+... (1x10−7 ) 3 K sp In conclusion : The effect of the self-ionization of water is significant for relatively insoluble metal hydroxides This is an example of the common ion effect !!! Any Questions? The effect of pH on solubility: (a) pH is fixed (known) Example 9-6: Calculate the molar solubility of PbCO3 in a solution buffered to a pH of 7.00 1 Pertinent equilibria: PbCO3 Pb+2 + CO3-2 Ksp H2CO3 +... Questions? The effect of pH on solubility: (b) pH is unknown Example 9-7: Calculate the solubility of CuS in w ater 1 Pertinent equilibria: CuS Cu+2 + S-2 S-2 + H2O HS- + OHHS- + H2O H2S + OHH2O H3O+ + OH2 Definition of unknown: Solubility = [Cu+2] 3 Mass Balance Equation: [Cu+2] = [S-2] + [HS-] + [H2S] 4 Charge Balance Equation: 2[Cu+2] + [H3O+] = 2[S-2] + [HS-] + [OH-] 5 Equilibrium Expressions:... (s) + I- CuI2KI K+ + IStep 2 Define of Unknown Step 4 Mass-balance Expression [I-] = CKI + [Cu+] - [CuI2-] Step 5 Charge-balance Equation Solubility of CuI (s) = [Cu+] + [CuI2-] [Cu+] + [K+] = [I-] + [CuI2-] [Cu+] + 1.0x10-4 = [I-] + [CuI2-] Step 3 Equilibrium-Constant Expression Ksp = [Cu+][I-] = 1.0x10-12 [CuI -2 ] K2 = = 7.9 x10−4 [I - ] Step 6 Number of Independent Equations and unknown Three... 7.9x10-8 M Solubility of CuI (s) = [Cu+] + [CuI2-] = 1.0x10-8 + 7.9x10-8 = 8.9x10-8 M Step 9 Check Assumptions [Cu+] . Chapter 8B: Systematic Treatment of Equilibrium I. Introduction ¾Concentrations of individual species ¾Determining. are neutral ¾ The Sum of molar concentration of positive charge = The sum of molar concentration of negative charge ¾Concentration of charge by an ion = molar