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Applications of Aqueous Equilibria

Applications of Aqueous Equilibria. Chemistry 4th Edition McMurry/Fay. The Common-Ion Effect 01. Common Ion: Two dissolved solutes that contain the same ion (cation or anion). The presence of a common ion suppresses the ionization of a weak acid or a weak base.

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Applications of Aqueous Equilibria

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  1. Applications of Aqueous Equilibria Chemistry 4th Edition McMurry/Fay

  2. The Common-Ion Effect 01 • Common Ion: Two dissolved solutes that contain the same ion (cation or anion). • The presence of a common ion suppresses the ionization of a weak acid or a weak base. • Common-Ion Effect: is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. Chapter 16

  3. The Common-Ion Effect 02 • To determine the pH, we apply I.C.E. and then the Henderson–Hasselbalch equation. • When the concentration of HA and salt are high (≥0.1 M) we can neglect the ionization of acid and hydrolysis of salt. Chapter 16

  4. The Common-Ion Effect 03 • Calculate the pH of a 0.20 M CH3COOH solution with no salt added. • Calculate the pH of a solution containing 0.20 M CH3COOH and 0.30 M CH3COONa. • What is the pH of a solution containing 0.30 M HCOOH, before and after adding 0.52 M HCOOK? Chapter 16

  5. Buffer Solutions 01 • A Buffer Solution:is a solution of (1) a weak acid or a weak base and (2) its salt; both components must be present. • A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. • Buffers are very important to biological systems. Chapter 16

  6. Buffer Solutions 02 Chapter 16

  7. Buffer Solutions 03 • Buffer solutions must contain relatively high acid and base component concentrations, the buffer capacity. • Acid and base component concentrations must not react together. • The simplest buffer is prepared from equal concentrations of acid and conjugate base. Chapter 16

  8. Buffer Solutions 04 • Calculate the pH of a buffer system containing 1.0 M CH3COOH and 1.0 M CH3COONa. What is the pH of the system after the addition of 0.10 mole of gaseous HCl to 1.0 L of solution? • Calculate the pH of 0.30 M NH3/0.36 NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? Chapter 16

  9. Buffer Solutions 05 • Buffer Preparation: Use the Henderson–Hasselbalch equation in reverse. • Choose weak acid with pKa close to required pH. • Substitute into Henderson–Hasselbalch equation. • Solve for the ratio of [conjugate base]/[acid]. • This will give the mole ratio of conjugate base to acid. The acid should always be 1.0. Chapter 16

  10. Buffer Solutions 06 • Describe how you would prepare a “phosphate buffer” with a pH of about 7.40. • How would you prepare a liter of “carbonate buffer” at a pH of 10.10? You are provided with carbonic acid (H2CO3), sodium hydrogen carbonate (NaHCO3), and sodium carbonate (Na2CO3). Chapter 16

  11. Acid–Base Titrations 01 • Titration: a procedure for determining the concentration of a solution using another solution of known concentration. • Titrations involving onlystrong acids or bases are straightforward. • Titrations involving weak acids or bases are complicated by hydrolysis of salt formed. Chapter 16

  12. Acid–Base Titrations 02 • Strong Acid–Strong Base: • The equivalence point is the point at which equimolar amounts of acid and base have reacted. Chapter 16

  13. Acid–Base Titrations 03 • The pH of a 25 mL 0.10 M HCl sample can be determined after the addition of: • 0. No addition of 0.10 M NaOH. • 1. 10.0 mL (total) of 0.10 M NaOH. • 2. 25.0 mL (total) of 0.10 M NaOH. • 3. 35.0 mL (total) of 0.10 M NaOH. Chapter 16

  14. Acid–Base Titrations 04 • Weak Acid–Strong Base: • The conjugate base hydrolyzes to form weak acid and OH–. • At equivalence point only the conjugate base is present. • pH at equivalence point will always be >7. Chapter 16

  15. Acid–Base Titrations 05 Chapter 16

  16. Acid–Base Titrations 06 • The pH of a 25 mL 0.10 M CH3COOH sample can be determined after the addition of: • 0. No addition of 0.10 M NaOH. • 1. 10.0 mL (total) of 0.10 M NaOH. • 2. 25.0 mL (total) of 0.10 M NaOH. • 3. 35.0 mL (total) of 0.10 M NaOH. Chapter 16

  17. Acid–Base Titrations 07 • Exactly 100 mL of 0.10 M nitrous acid are titrated with a 0.10 M NaOH solution. Calculate the pH for: • 1. The initial solution. • 2. The point at which 80 mL of base have been added. • 3. The equivalence point. • 4. The point at which 105 mL of base have been added. Chapter 16

  18. Acid–Base Titrations 09 • Strong Acid–Weak Base: • The (conjugate) acid hydrolyzes to form weak base and H3O+. • At equivalence point only the (conjugate) acid is present. • pH at equivalence point will always be <7. Chapter 16

  19. Acid–Base Titrations 10 • Calculate the pH at the equivalence point when 25 mL of 0.10 M NH3 is titrated with a 0.10 M HCl solution. • Calculate the pH at the equivalence point in the titration of 50 mL of 0.10 M methylamine with a 0.20 M HCl solution. Chapter 16

  20. Acid–Base Titrations 11 • Polyprotic Acids: Chapter 16

  21. Solubility Equilibria 01 • Solubility Product: is the product of the molar concentrations of constituent ions and provides a measure of a compound’s solubility. MX2(s) æ M2+(aq) + 2 X–(aq) Ksp = [M2+][X–]2 Chapter 16

  22. Al(OH)3 1.8 x 10–33 BaCO3 8.1 x 10–9 BaF2 1.7 x 10–6 BaSO4 1.1 x 10–10 Bi2S3 1.6 x 10–72 CdS 8.0 x 10–28 CaCO3 8.7 x 10–9 CaF2 4.0 x 10–11 Ca(OH)2 8.0 x 10–6 Ca3(PO4)2 1.2 x 10–26 Cr(OH)3 3.0 x 10–29 CoS 4.0 x 10–21 CuBr 4.2 x 10–8 Solubility Equilibria 02 CuI 5.1 x 10–12 Cu(OH)2 2.2 x 10–20 CuS 6.0 x 10–37 Fe(OH)2 1.6 x 10–14 Fe(OH)3 1.1 x 10–36 FeS 6.0 x 10–19 PbCO3 3.3 x 10–14 PbCl2 2.4 x 10–4 PbCrO4 2.0 x 10–14 PbF2 4.1 x 10–8 PbI2 1.4 x 10–8 PbS 3.4 x 10–28 MgCO3 4.0 x 10–5 Mg(OH)2 1.2 x 10–11 MnS 3.0 x 10–14 Hg2Cl2 3.5 x 10–18 HgS 4.0 x 10–54 NiS 1.4 x 10–24 AgBr 7.7 x 10–13 Ag2CO3 8.1 x 10–12 AgCl 1.6 x 10–10 Ag2SO4 1.4 x 10–5 Ag2S 6.0 x 10–51 SrCO3 1.6 x 10–9 SrSO4 3.8 x 10–7 SnS 1.0 x 10–26 Zn(OH)2 1.8 x 10–14 ZnS 3.0 x 10–23 Chapter 16

  23. Solubility Equilibria 03 • The solubility of calcium sulfate (CaSO4) is found experimentally to be 0.67 g/L. Calculate the value of Ksp for calcium sulfate. • The solubility of lead chromate (PbCrO4) is 4.5 x 10–5 g/L. Calculate the solubility product of this compound. • Calculate the solubility of copper(II) hydroxide, Cu(OH)2, in g/L. Chapter 16

  24. Solubility Equilibria 04 • Ion Product (Q): solubility equivalent of the reaction quotient. It is used to determine whether a precipitate will form. Q < Ksp UnsaturatedQ = Ksp SaturatedQ > Ksp Supersaturated; precipitate forms. Chapter 16

  25. Solubility Equilibria 05 • Exactly 200 mL of 0.0040 M BaCl2 are added to exactly 600 mL of 0.0080 M K2SO4. Will a precipitate form? • If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl2, will precipitation occur? Chapter 16

  26. The Common-Ion Effect and Solubility 01 • The solubility product (Ksp) is an equilibrium constant; precipitation will occur when the ion product exceeds the Ksp for a compound. • If AgNO3 is added to saturated AgCl, the increase in [Ag+] will cause AgCl to precipitate. Q = [Ag+]0 [Cl–]0 > Ksp Chapter 16

  27. The Common-Ion Effect and Solubility 02 Chapter 16

  28. The Common-Ion Effect and Solubility 03 Chapter 16

  29. The Common-Ion Effect and Solubility 04 • Calculate the solubility of silver chloride (in g/L) in a 6.5 x 10–3 M silver chloride solution. • Calculate the solubility of AgBr (in g/L) in:(a) pure water(b) 0.0010 M NaBr Chapter 16

  30. Complex Ion Equilibria and Solubility 01 • A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions. • Most metal cations are transition metals because they have more than one oxidation state. • The formation constant (Kf) is the equilibrium constant for the complex ion formation. Chapter 16

  31. Complex Ion Equilibria and Solubility 02 Chapter 16

  32. Complex Ion Equilibria and Solubility 03 Chapter 16

  33. Complex Ion Equilibria and Solubility 04 ION Kf Ag(NH3)2+ 1.5 x 107 Ag(CN)2– 1.0 x 1021 Cu(CN)42– 1.0 x 1025 Cu(NH3)42+ 5.0 x 1013 Cd(CN)42– 7.1 x 1016 CdI42– 2.0 x 106 ION Kf HgCl42– 1.7 x 1016 HgI42– 3.0 x 1030 Hg(CN)42– 2.5 x 1041 Co(NH3)63+ 5.0 x 1031 Zn(NH3)42+ 2.9 x 109 Chapter 16

  34. Complex Ion Equilibria and Solubility 05 • A 0.20 mole quantity of CuSO4 is added to a liter of 1.20 M NH3 solution. What is the concentration of Cu2+ ions at equilibrium? • If 2.50 g of CuSO4 are dissolved in 9.0 x 102 mL of 0.30 M NH3, what are the concentrations of Cu2+, Cu(NH3)42+, and NH3 at equilibrium? Chapter 16

  35. Complex Ion Equilibria and Solubility 06 • Calculate the molar solubility of AgCl in a 1.0 M NH3 solution. • Calculate the molar solubility of AgBr in a 1.0 M NH3 solution. Chapter 16

  36. Complex Ion Equilibria and Solubility 07 Chapter 16

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