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Chapter 16 Aqueous Equilibria

Chapter 16 Aqueous Equilibria.

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Chapter 16 Aqueous Equilibria

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  1. Chapter 16 Aqueous Equilibria The intravenous (IV) solution in this bag might save a life. The concentration of each solute in the solution is carefully selected to keep the total solute concentration within an optimal range and to maintain the pH of the blood. Because even slight deviations in blood pH can be fatal, the first treatment administered to an injured person is usually an intravenous solution. Single acid (base)  Salts and mixed salts (anions and cations). Central topic: How pH values can be affected and controlled by ions.

  2. Assignment 16.5,16.14,16.26,16.31,16.38, 16.47,16.52,16.61,16.71,16.76.

  3. Acidity/Basicity of Various Solutions CuSO4 NaCl KBr Characteristic colors of 10 mM water solutions of K2FeO4, KMnO4 and Fe2 (SO4)3. 1 = Permanganate; 2 = Potassium ferrate; 3 = Ferric chloride. Why?

  4. How to modify the pH value of a solution?

  5. Indicators: tell you how acidic or basic your samples are by looking at it.

  6. Salts in Water • Ions as acids and bases • The pH of a salt solution • The pH of mixed solutions

  7. Figure 16.1 Solutions of salts in water give rise to acidic, neutral, and basic solutions, as shown here by the color of the indicator bromothymol blue (see Table 16.3). From left to right are solutions of ammonium chloride, sodium chloride, and sodium acetate. Throughout this chapter, we use the Brønsted definitions of acids and bases. Salts that contain the conjugate acids of weak bases produce acidic aqueous solutions. All anions that are conjugate bases of weak acids act as proton acceptors, giving basic solutions.

  8. Small, highly charged metal ions have strong polarizing effect on water so that they act as a proton donator, an acid. Figure 16.2 A solution of titanium(III) sulfate is so acidic that it can releaseH2S from some sulfides.

  9. H+ H+ H+ Figure 16.3 In water, Al3 cations exist as hydrated ions that can act as Brønsted acids. Although, for clarity, only four water molecules are shown here, metal cations typically have six H2O molecules attached to them. Small, highly charged metal ions exert the greatest pull on the electrons. They look like a proton donator, an acid.

  10. (A cation cannot be a base)

  11. However, • the cations of Group 1 and 2 metals and those of charge +1 from other groups are such weak Lewis acids that they DO NOTact as acids when hydrated. These metal ions are too large or have too low a charge to have an appreciable polarizing effect on water molecules that surround them. As a result, the hydrating water molecules do not readily release their protons.

  12. Questions • Is Na+ an acid in water? • Is Ca2+ an acid in water? No and No

  13. Figure 16.4 The relative strengths of conjugate acids and bases have a reciprocal relation. (a) When a species has a high tendency to donate a proton, the resulting conjugate base has a low tendency to accept one. (b) When a species has a low tendency to donate a proton, the resulting conjugate base has a strong tendency to accept one. In the insets, a solid blue color represents the water molecules. Ka=10-12 Kb=10-13 Ka=0.1 Kb=10-2 Strong Acid (Base)  Very Weak Conjugate Base (Acid) Kb=10-10 ~ 10-4 Ka=10-4 ~ 10-10 – most weak acids  weak conjugate bases – most weak bases  weak conjugate acids Kb=10-4 ~ 10-10 Ka=10-10 ~ 10-4

  14. Only very few anions are acids (Very weak bases Kb 0  neutral) All anions that are conjugate bases of weak acids act as proton acceptors, giving basic solutions.

  15. Exercise • Decide whether aqueous solutions of (a) Ba(NO2)2; (b) CrCl3; (c) NH4NO3 are acidic, basic or neutral. • Basic, for the NO2- is the conjugate base of a weak acid. • (b) Acidic, for Cr3+ is small and highly charged so that it can • polarizewater and release proton(s) from water. • (c) Acidic, for the NH4+ is the conjugate acid of a weak base • (and NO3- is the conjugate base of a strong acidneutral) • (a) Ba(NO2)2; (b) CrCl3; (c) NH4NO3

  16. Classroom Exercise • Decide whether aqueous solutions of (a) Na2CO3; (b) AlCl3; (c) KNO3 are acidic, basic or neutral. • Basic for the CO32- is the conjugate base of a weak acid. • (b) Acidic for Al3+ is small and highly charged so that it can • polarizewater and release proton(s) from water. • (c) Neutral for the K+ is large and charge +1 so that it cannot • polarize water. • (a) Na2CO3; (b) AlCl3; (c) KNO3

  17. Figure 16.5 The initial (left) and equilibrium (right) composition of a solution of a salt composed of the cation HB and the anion A, where HB is a weak acid and A is neutral. (left) The hypothetical initial situation that we imagine before deprotonation. (right) At equilibrium, the acidic cation is partially deprotonated, and the solution is acidic. How to calculate the pH of an acidic solution

  18. The pH of a Salt Solution Acidic • The initial molarity of the acidic ion is the molarity of the initially completely protonated ion. We assume that the initial molarities of its conjugate base and H3O+ are 0. • Write the increase in molarity of H3O+ as x mol/L and use the reaction stoichiometry to write the corresponding changes for the acidic ion and its conjugate base. Ignore H3O+ from the autoprotolysis of water; this approximation is valid if [H3O+] is substantially (about 10 times) greater than 10-7. Check the validity of this approximation at the end of the calculation. • Write the equilibrium molarities of the species of x. • Express the acidity constant for the ion in terms of x and solve the equations for x; assume that x is small, but check the validity of this approximation at the end of the calculation. If Ka is not available, obtain it from the value of Kb for the conjugate base by using Ka=Kw/Kb. Because x mol/L is the H3O+ molarity, the pH of the solution is –logx. Basic Follow the same procedure as above, except that now proton transfer from water to the ion results in the formation of OH- and the conjugate acid of the ion. We therefore use Kb ad the equilibrium table leads to a value of pOH. An amphiprotic anion can act as both an acid and a base: pH = (pKa+pKb)/2

  19. Example • Estimate the pH of 0.15 M NH4Cl (aq). NH3: Kb = 1.8×10-5 (Table 15.3)

  20. Example • Estimate the pH of 0.10 M methylammonium chloride, CH3NH3Cl (aq). CH3NH3+: Kb=3.6×10-4 (Table 15.3) or Ka = 2.8×10-11 (Table 16.1)

  21. Classroom Example • Estimate the pH of 0.10 M NH4NO3 (aq). NH3: Kb = 1.8×10-5 (Table 15.3)

  22. Example • Estimate the pH of 0.15 M calcium acetate, Ca(CH3CO2)2 (aq). CH3CO2-: Ka = 1.8×10-11 (Table 15.3)

  23. Example • Estimate the pH of 0.1 M potassium heptate, KC6H5CO2 (aq). C6H5COOH: Ka = 6.5×10-5

  24. Classroom Example • Estimate the pH of 0.02 M potassium fluoride, KF(aq). HF: Ka = 3.5×10-4

  25. Mixed Solution: Example • Estimate the pH of a solution that is 0.5 M HNO2 and 0.1 M KNO2 (aq). HNO2 :Ka = 4.3×10-4 (Table 15.3)

  26. Mixed Solution: Example • Estimate the pH of a solution that is 0.3 M CH3NH2 and 0.146 M CH3NH3Cl(aq). CH3NH2 :Kb = 3.6×10-4 (Table 15.3)

  27. Classroom Example • Estimate the pH of a solution that is 0.01 M HClO (aq) and 0.2 mM NaClO(aq). HClO :Ka = 3.0×10-8 (Table 15.3)

  28. Titrations Titration: A titrant adds H3O+ (or OH-) to and analyte so that a neutral solution (stoichiometric point) is obtained (the initial pH of the analyte is thus found). • Strong acid-strong base titrations • Weak acid-strong base and strong acid-weak base titrations • Titrating a polyprotic acid • Indicators

  29. slow fast Figure 16.6 The variation of pH during the titration of a strong base (the analyte) with a strong acid (the titrant). This curve is for 25.00 mL of 0.250 M NaOH(aq) titrated with 0.340 M HCl(aq). The stoichiometric point occurs at pH  7 (point S ). The other points are discussed in the text.

  30. slow fast Figure 16.7 The variation of pH during a typical titration of a strong acid (the analyte) with a strong base (the titrant). The stoichiometric point occurs at pH  7.

  31. Figure 16.8 The composition of the solutions initially (left) and at the stoichiometric point (right) in the titration of a strong acid (the analyte) with a strong base (the titrant). Step 1: calculate the moles of H3O+ (if the analyte is a strong acid) or OH- (if the analyte is a strong base) in the original analyte solution from its molarity and volume. Step 2: calculate the moles of H3O+ (if the titrant is a strong acid) or OH- (if the titrant is a strong base) in the volume of titrant added. Step 3: write the chemical equation for the neutralization reaction and use reaction stoichiometry to find the moles of H3O+ (or OH- if the analyte is strong base) that remain in the analyte solution. Each mole of H3O+ ions reacts with 1 mol OH- ions; therefore subtrsct the number of of moles of H3O+ or OH- ions that have reacted from the initial number of moles of H3O+ or OH- ions. Step 4: Determine the concentration of hydronium or hydroxide. Step 5: Find the pH (pOH).

  32. Calculating points on the pH curve for a strong acid-strong base titration (before stoichiometric point) • The initial pOH of the analyte: -log(0.25)=0.602pH=13.40. • The amount of OH- is 0.025 L×0.25 M = 6.25 mmol. Titrant: 0.34 M HCl(aq); analyte: 25 mL, 0.25 M NaOH. After addition of 5.0 mL of titrant, pH=? • The amount of H3O+ supplied by the titrant is: • 0.005 L×0.34 M = 1.70 mmol 3. After the reaction of all the hydronium ions added, the amount of hydroxide ion remaining is 6.25 – 1.70 = 4.55 mmol. 4. Because the total volume of the solution is now 30 mL, the molarity of hydroxide ion is 4.55 mmol/30 mL = 0.152 M 5. pOH=-log(0.152)=0.82 pH = 13.18.

  33. Example: titration (before stoichiometric point) • The initial pOH of the analyte: -log(0.25)=0.602pH=13.40. • The amount of OH- is 0.025 L×0.25 M = 6.25 mmol. Titrant: 0.34 M HCl(aq); analyte: 25 mL, 0.25 M NaOH. After addition of 10.0 mL of titrant, pH=? • The amount of H3O+ supplied by the titrant is: • 0.01 L×0.34 M = 3.40 mmol 3. After the reaction of all the hydronium ions added, the amount of hydroxide ion remaining is 6.25 – 3.40 = 2.85 mmol. 4. Because the total volume of the solution is now 35 mL, the molarity of hydroxide ion is 2.85 mmol/35 mL = 0.0814 M 5. pOH=-log(0.0814)=1.09 pH = 12.91.

  34. Classroom Exercise • The initial pOH of the analyte: -log(0.25)=0.602pH=13.40. • The amount of OH- is 0.025 L×0.25 M = 6.25 mmol. Titrant: 0.34 M HCl(aq); analyte: 25 mL, 0.25 M NaOH. After addition of 12.0 mL of titrant, pH=? • The amount of H3O+ supplied by the titrant is: • 0.012 L×0.34 M = 4.08 mmol 3. After the reaction of all the hydronium ions added, the amount of hydroxide ion remaining is 6.25 – 4.08 = 2.17 mmol. 4. Because the total volume of the solution is now 37 mL, the molarity of hydroxide ion is 2.17 mmol/37 mL = 0.0586 M 5. pOH=-log(0.0586)=1.232 pH = 12.77.

  35. Stoichiometric Point Titrant: 0.34 M HCl(aq); analyte: 25 mL, 0.25 M NaOH. How much titrant is needed to reach stoichiometric point? • The initial pOH of the analyte: -log(0.25)=0.602pH=13.40. • The amount of OH- is 0.025 L×0.25 M = 6.25 mmol. • The amount of H3O+ supplied by the titrant is assumed to be x at the stoichiometric point, then • 6.25 mmol = x×340 mM  x = 18.38 mL For strong acids-strong base titration, the pH is 7.0, but for other cases, the pH may be not 7.0 at the stoichiometric point!

  36. Calculating points on the pH curve for a strong acid-strong base titration (after stoichiometric point) • The initial pOH of the analyte: -log(0.25)=0.602pH=13.40. • The amount of OH- is 0.025 L×0.25 M = 6.25 mmol. Titrant: 0.34 M HCl(aq); analyte: 25 mL, 0.25 M NaOH. After addition of 19.4 mL of titrant, pH=? • The amount of H3O+ supplied by the titrant is: • 0.0194 L×0.34 M = 6.6 mmol 3. After the reaction of all the hydroxide ions, the amount of hydronium ion remaining is 6.60 – 6.25 = 0.35 mmol. 4. Because the total volume of the solution is now 44.4 mL, the molarity of hydronium ion is 0.35 mmol/44.4 mL = 7.9 mM 5. pH=-log(0.0079)=2.10.

  37. Classroom Exercise (pH after stoichiometric point) • The initial pOH of the analyte: -log(0.25)=0.602pH=13.40. • The amount of OH- is 0.025 L×0.25 M = 6.25 mmol. Titrant: 0.34 M HCl(aq); analyte: 25 mL, 0.25 M NaOH. After addition of 25 mL of titrant, pH=? • The amount of H3O+ supplied by the titrant is: • 0.025 L×0.34 M = 8.5 mmol 3. After the reaction of all the hydroxide ions, the amount of hydronium ion remaining is 8.5 – 6.25 = 2.25 mmol. 4. Because the total volume of the solution is now 50 mL, the molarity of hydronium ion is 2.25 mmol/50 mL = 45 mM 5. pH=-log(0.045)=1.35.

  38. Titration of NaOH by HCl 0.34 M HCl 0.25 M NaOH pH HCl 25 ml 13.40 0 ml 13.18 25 ml 5 ml 12.91 25 ml 10 ml 12.77 25 ml 12 ml 7.00 NaOH 25 ml 18.38 ml 2.10 25 ml 19.4 ml 1.35 25 ml 25 ml

  39. Animation of Acid-Base Titration http://science.csustan.edu/chem/titrate/Acid/titrateZ.swf Follow the instructions and try a few times with different initial concentrations/volumes of acid and base. Click the yellow button on the valve to see the generation of the titration curve. (The titration curve might grow rather slow. Be patient.)

  40. Weak acid-strong base and strong acid-weak base titrations Examples: Chemical Engineering General Industry or Lab Use (e.g. cleaning agents and surface treatments with liquid cleaners, degreasers, strippers, passivators, etchants, solutions and additives for cleaning and surface preparation. ) Biochemistry (intracellular/extracellular equilibrium, neuron activation) Medicine (IV, pharmacology, toxicology) Printed Circuit Board (PCB) Integrated Circuits Semiconductor Manufacture Nanomaterials (nanoparticles) Photography

  41. Weak acid-strong base titrations HCOOH+NaOHNaHCO2+H2O Na+ + HCO2- HCO2- + H2O HCOOH +OH- A strong base dominates a weak acid; the solution at the stoichiometric point is basic.

  42. Strong acid-weak base titrations HClO2+NH4OH  NH4Cl+H2O Cl- + NH4+ NH4+ + H2O NH3 +H3O+ A strong acid dominates a weak base; the solution at the stoichiometric point is acidic.

  43. Figure 16.9 The composition of the solutions in the course of the titration of a weakacid with a strong base. From left to right: at the start of the titration, just before the stoichiometric point, at the stoichiometric point, and well after the stoichiometric point. At the stoichiometric point, the OH ions are mainly those arising from proton transfer from H2O to A. (basic)

  44. Step 1: calculate the moles of weak acid (or weak base) in the original analyte solution from its molarity and volume. Calculating the pH during a titration of a weak acid or a weak base Step 2: calculate the moles of H3O+ (if the analyte is a weak acid) or OH- (if the analyte is a weak base) in the in the volume of titrant added. Step 3: write the chemical equation for the neutralization reaction and use reaction stoichiometry to find: Weak acid-strong base titration: the moles of conjugate base formed by the reaction of the acid with the added base, and the moles of weak acid remaining. Weak base-strong acid titration: the moles of conjugate acid formed by the reaction of the base with the added acid, and the moles of weak base remaining. Step 4: find the molarity of the conjugate acid and base in solution by dividing the moles of each species by the total volume of the solution. Step 5: for a weak acid, use the equilibrium table to find the H3O+ concentration and in each case, assume that the contribution of the autoptolysis of water to the pH is insignificant if the pH is less than 6 or larger than 8. Step 6: find pH from the hydronium concentration. For a weak base, use an equilibrium table to find the OH- concentration and find pOHpH.

  45. Calculating the pH during a titration of a weak acid (at stoichiometric point) 1 2 3 HCOOH+NaOHNaHCO2+H2O (No HCOOH left) 4 Na+ + HCO2- 5 HCO2- + H2O HCOOH +OH- Basic! 6 weak base Kb pOHpH

  46. Calculating the pH during a titration of a weak acid (at stoichiometric point) 1 2 3 HA + MOH MA + H2O (No HA left) 4 M+ + A- MA MB MA- MA- 5 Basic! A- + H2O HA +OH- weak base 6 0 0 MA- Kb pOHpH x x -x x x MA- - x

  47. Example: Weak Acid-Strong Base Titration Moles of HCOOH: 0.0025 L × 0.1 mol/L = 2.5 mmol. Estimate the pH at the stoichiometric point of the titration of 25.00 mL of 0.1 M HCOOH (aq) with 0.15 M NaOH (aq). Because all HCOOH has reacted to form HCO2-, the moles of HCO2- at the stoichiometric point is also 2.5 mmol. From the chemical equation, 1 mol OH- is needed for 1 mol HCOOH: The moles of NaOH required is then: 2.5 mmol/0.15 M = 16.7 mL. The total volume of the solution at the stoichiometric point is 25 mL + 16.7 mL = 41.7 mL. The molarity of sodium formate is 2.5 mmol/41.7 mL = 0.06 M. Because the formate ion is a weak base, the equilibrium to consider is:

  48. Because the formate ion is a weak base, the equilibrium to consider is: From Table 15.3, Ka=1.8 ×10-4 for formic acidKb =5.6 ×10-11. Basic!

  49. Figure 16.10 The pH curve for the titration of a weak acid with a strong base. This curve is for the titration of 25.00 mL of 0.100 M HCOOH(aq) with 0.150 M NaOH(aq). The stoichiometric point (S ) occurs on the basic side of pH  7 because the anion HCO2 is a base.

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