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Chapter 18

Chapter 18. Acid-Base Equilibria. Acid – Base Reactions. Strong acid + strong base: HNO 3 + Ca(OH) 2 H + (aq) + OH - (aq) ----------> H 2 O(l). Acid-Base Reactions. Weak acid + strong base: HF + KOH HF(aq) + OH - (aq) <----> F - (aq) + H 2 O(l). Acid-Base Reactions.

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Chapter 18

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  1. Chapter 18 Acid-Base Equilibria WOLPA/AP CHEMISTRY/CDO

  2. Acid – Base Reactions • Strong acid + strong base: • HNO3 + Ca(OH)2 • H+(aq) + OH-(aq) ----------> H2O(l) WOLPA/AP CHEMISTRY/CDO

  3. Acid-Base Reactions • Weak acid + strong base: • HF + KOH • HF(aq) + OH-(aq) <----> F-(aq) + H2O(l) WOLPA/AP CHEMISTRY/CDO

  4. Acid-Base Reactions • Strong acid + weak base: • HClO4 + NH3 • H+(aq) + NH3(aq) <----------> NH4+(aq) • H2SO4 + Na2CO3 • H+(aq) + CO32-(aq) <------> HCO3-(aq) WOLPA/AP CHEMISTRY/CDO

  5. The Common Ion Effect The common ion effect is when an ion common to the ionization of an acid is present in the solution in an amount greater than that produced by the acid ionization. The presence of this ion, according to LeChatelier's principle, limits the extent to which the acid will ionize and thus affects the pH of the solution. WOLPA/AP CHEMISTRY/CDO

  6. Example • Determine the pH of a solution of 0.25 M acetic acid. Ka = 1.8 x 10-5 WOLPA/AP CHEMISTRY/CDO

  7. Example • Determine the pH of a solution of 0.25 M acetic acid in a solution of 0.10 M sodium acetate. WOLPA/AP CHEMISTRY/CDO

  8. Example Determine the pH of a solution prepared by mixing 50.0 mL of 0.100 M HOCl with 50.0 mL of 0.100 M NaOCl (Ka = 3.5x108). WOLPA/AP CHEMISTRY/CDO

  9. Buffers • Acid-base buffersconfer resistance to a change in the pH of a solution when hydrogen ions (protons) or hydroxide ions are added or removed. An acid-base buffer typically consists of a weak acid, and its conjugate base. • Prepared by adding both the weak acid HB and the salt of its conjugate base B- to water. WOLPA/AP CHEMISTRY/CDO

  10. General Expressions for Buffer Solutions • Assume equilibrium is established, therefore • Ka = [H3O+] [B-] / [HB] and • [H3O+] = Ka [HB] /[B-] or • Kb = [HB] [OH-] / [B-] and • [OH-] = Kb [B-] /[HB] WOLPA/AP CHEMISTRY/CDO

  11. Example • Determine the pH of a solution of 0.10 M acetic acid in a solution of 0.10 M sodium acetate. WOLPA/AP CHEMISTRY/CDO

  12. Effect of H3O+ or OH- on Buffer System • A buffer system contains one species (HB) that will react with added hydroxide ions and another species (B-) that will react with added hydronium ions. • Both reactions will go virtually to completion hence, the added hydronium or hydroxide ions are consumed and the effect on the overall pH is negligible. WOLPA/AP CHEMISTRY/CDO

  13. Buffer Calculations • Determine the concentrations of HB and B- after the addition of H3O+ or OH- WOLPA/AP CHEMISTRY/CDO

  14. Addition of an Acid • 0.10 M HC2H3O2 with 0.10 M NaC2H3O2 add 50.0 mL 0.10 M HCl WOLPA/AP CHEMISTRY/CDO

  15. Addition of a Base • 0.10 M HC2H3O2 with 0.10 M NaC2H3O2 add 50.0 ml 0.10 M NaOH WOLPA/AP CHEMISTRY/CDO

  16. The Henderson-Hasselbalch Equation • pH = pKa + log [conjugate base]/[acid] • This equation clearly shows that the pH of the solution of a weak acid and its conjugate base is controlled primarily by the strength of the acid. • Can also be written • pOH = pKb + log [conjugate acid]/[base] WOLPA/AP CHEMISTRY/CDO

  17. Henderson-Hasselbalch Equation Allows us to predict pH when HB/B mixed. • When [B] /[HB] = 1 (i.e. [HB]=[B]), pH = pKa WOLPA/AP CHEMISTRY/CDO

  18. Example Calculate pH of solution containing 0.040M Na2HPO4 and 0.080M KH2PO4. pKa2=7.20. WOLPA/AP CHEMISTRY/CDO

  19. Example Determine the ratio of the concentration of the conjugate acid to concentration of the conjugate base for a weak acid in which the pH was 5.45 and pKa was 5.75. WOLPA/AP CHEMISTRY/CDO

  20. Example Determine the pH of a solution consisting of 0.100 M NH3 and 0.150 M NH4Cl. WOLPA/AP CHEMISTRY/CDO

  21. Preparation of a Buffer Solution Selection of weak acid or weak base • (pKa ~ pH) WOLPA/AP CHEMISTRY/CDO

  22. Titrations of Weak Acids and Strong Bases Typical net ionic equation • HC2H3O2 + OH- ------> C2H3O2- + H2O • K = 1/Kb(C2H3O2-) = 1/5.6 x 10-10 = 1.8 x 109 • The reaction goes essentially to completion WOLPA/AP CHEMISTRY/CDO

  23. pH Changes • -pH starts off at about 2.4, the pH of 1 M HC2H3O2 • -Region, centered around the half-way point of the titration where pH changes very slowly. In this region there are appreciable amounts of two different species: unreacted HC2H3O2and C2H3O2- ions. Hence, we have a buffer system • -Equivalence point - we have a solution of sodium acetate. This solution is basic because C2H3O2- is a weak base. pH at equivalence is > 7.00 WOLPA/AP CHEMISTRY/CDO

  24. pH Calculations • Write a net ionic equation to determine the extent of the reaction. • Calculate the initial pH from the Ka or Kb. • Calculate pH at midpoint using the buffer relation. • Calculate pH at equivalence point using Ka of conjugate acid or Kb of conjugate base. WOLPA/AP CHEMISTRY/CDO

  25. pH Titration Curves • Titration curve: plot of pH of the solution as a function of the volume of base (acid) added to an acid (base). WOLPA/AP CHEMISTRY/CDO

  26. Sharp rise in curve is equivalence point. • pH at equivalence point is 7.0 for SA but higher for WA. • Equivalence point can be used to determine the concentration of the titrant. WOLPA/AP CHEMISTRY/CDO

  27. Example The equivalence point for 15.00 mL of an acid occurred when 25.00 mL of 0.075 M NaOH was added. What was the molarity of the acid? WOLPA/AP CHEMISTRY/CDO

  28. SA–SB Titrations • Base removes some acid and pH increases. • Let nb = moles of base added na,r = moles of acid remaining na,r = na nb = MaVa MbVb • Moles of hydronium ion same as moles of acid remaining. nH3O+ = na,r; • Valid until very close to equivalence point. WOLPA/AP CHEMISTRY/CDO

  29. Equivalence point(EP): pH = 7.00 • Beyond EP: pH due only to base added (i.e. excess base). Use total volume. WOLPA/AP CHEMISTRY/CDO

  30. Example • Determine pH of 10.0 mL of 0.100M HCl after addition of 5.00, 10.0 and 15.0mL of 0.100M NaOH. WOLPA/AP CHEMISTRY/CDO

  31. Titration of SB with SA • Acid removes some of the base and pH is changed by amount of base removed. Let na = moles of acid added nb,r = moles of base remaining nb,r = MbVb MaVa • Moles of hydroxide ion same as moles of base remaining. • nOH = nb,r; • Valid until EP. • EP: pH = 7.00 • Beyond EP: pH due only to excess acid. Use total volume. WOLPA/AP CHEMISTRY/CDO

  32. Example • Determine pH of 10.0 mL of 0.100M NaOH after addition of 5.00, 10.0 and 15.0mL of 0.100M HCl. WOLPA/AP CHEMISTRY/CDO

  33. WA with SB Titration • As above base removes some of the acid and pH is changed by amount of acid removed. Let nb = moles of base added nHA = moles of acid remaining nHA = MHAVHA MbVb nA = nb = MbVb • Up to equivalence point moles of hydronium ions must be determined from equilibrium expression. • Equivalence point: pH = pH of salt of WA • Beyond Equivalence point: Use amount of excess base to determine pH. WOLPA/AP CHEMISTRY/CDO

  34. Example • Determine pH of 10.0 mL of 0.100M HA after addition of 5.00, 10.0 and 15.0mL of 0.100M NaOH. Ka = 1.75x105. WOLPA/AP CHEMISTRY/CDO

  35. WB–SA Titrations • Acid removes some of the base and decreases the pH. Let na = moles of acid added nb,r = moles of base remaining nb,r = CbVb CaVa nBH+ = na = CaVa • Moles of hydroxide ions must be determined from equilibrium expression. Valid until EP. • EP: pH = pH of salt of weak base. • Beyond EP: pH due only to presence of acid added after endpoint (i.e. excess acid) as seen for strong base. Volume correction needed as above (total volume). WOLPA/AP CHEMISTRY/CDO

  36. Example Determine pH of 10.0 mL of 0.100M B after addition of 5.00, 10.0 and 15.0mL of 0.100M HCl. Kb = 1.75x105. WOLPA/AP CHEMISTRY/CDO

  37. Acid-Base Indicators • An acid-base indicator (HIn) is usually an organic dye that is itself a weak acid governed by an equilibrium constant. • The acid form has one color and the base form has another. In principle, the color of the indicator changes when [H3O+] = Ka of the indicator.(because HIn = In- at this point) WOLPA/AP CHEMISTRY/CDO

  38. Choice of Indicator 1. Strong acid-Weak base • Solution at equivalence point is weakly acidic. Choose indicator which turns color below pH 7. ex methyl red (pH 5) WOLPA/AP CHEMISTRY/CDO

  39. Choice of Indicator 2. Weak acid - Strong base • Solution at equivalence point is weakly basic, Choose indicator which turns color above pH 7. ex. phenolphthalein (pH 9) WOLPA/AP CHEMISTRY/CDO

  40. Choice of Indicator • Strong acid - Strong base • Solution at equivalence point is neutral. However, pH changes so rapidly near the end point that any indicator that changes color between 5 and 9 will work. WOLPA/AP CHEMISTRY/CDO

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