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Section 18-3

Section 18.3 Hydrogen Ions and pH. Explain pH and pOH . Relate pH and pOH to the ion product constant for water. Calculate the pH and pOH of aqueous solutions.

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Section 18-3

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  1. Section 18.3 Hydrogen Ions and pH • Explain pH and pOH. • Relate pH and pOH to the ion product constant for water. • Calculate the pH and pOH of aqueous solutions. pH and pOH are logarithmic scales that express the concentrations of hydrogen ions and hydroxide ions in aqueous solutions. Section 18-3

  2. The pH Scale Acids Have a pH less than 7 Bases have a pH greater than 7

  3. pH and pOH (cont.) • Litmus paper and a pH meter with electrodes can determine the pH of a solution. Section 18-3

  4. Self-Ionization of Water H2O + H2O  H3O+ + OH- At 25, [H3O+] = [OH-] = 1 x 10-7 Kw is a constant at 25 C: Kw = [H3O+][OH-] Kw = (1 x 10-7)(1 x 10-7) = 1 x 10-14

  5. pH and pOH • Concentrations of H+ ions are often small numbers expressed in exponential notation. • pH is the negative logarithm of the hydrogen ion concentration of a solution.pH = –log [H+] Section 18-3

  6. pH and pOH (cont.) • pOH of a solution is the negative logarithm of the hydroxide ion concentration. • pOH = –log [OH–] • The sum of pH and pOH equals 14. Section 18-3

  7. Calculating pH, pOH pH = -log10[H3O+] pOH = -log10[OH-] Relationship between pH and pOH pH + pOH = 14 Finding [H3O+], [OH-] from pH, pOH [H3O+] = 10-pH [OH-] = 10-pOH

  8. Calculating pH from solution concentration…

  9. pH and pOH (cont.) • For all strong monoprotic acids, the concentration of the acid is the concentration of H+ ions. • For all strong bases, the concentration of the OH– ions available is the concentration of OH–. • Weak acids and weak bases only partially ionize and Ka and Kb values must be used. Section 18-3

  10. Dissociation Constants: Strong Acids

  11. Dissociation Constants: Weak Acids

  12. A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step #1:Write the dissociation equation HC2H3O2 C2H3O2- + H+

  13. A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step #2:ICE it! HC2H3O2 C2H3O2- + H+ 0.50 0 0 +x +x - x x x 0.50 - x

  14. A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step #3:Set up the law of mass action HC2H3O2 C2H3O2- + H+ E 0.50 - x x x

  15. A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step #4:Solve for x, which is also [H+] HC2H3O2 C2H3O2- + H+ E 0.50 - x x x [H+] = 3.0 x 10-3 M

  16. A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step #5:Convert [H+] to pH HC2H3O2 C2H3O2- + H+ E 0.50 - x x x

  17. Dissociation of Strong Bases MOH(s)  M+(aq) + OH-(aq) • Strong bases are metallic hydroxides • Group I hydroxides (NaOH, KOH) are very soluble • Group II hydroxides (Ca, Ba, Mg, Sr) are less soluble • pH of strong bases is calculated directly from the concentration of the base in solution

  18. Reaction of Weak Bases with Water The base reacts with water, producing its conjugate acid and hydroxide ion: CH3NH2 + H2O  CH3NH3+ + OH- Kb = 4.38 x 10-4

  19. Kb for Some Common Weak Bases Many students struggle with identifying weak bases and their conjugate acids.What patterns do you see that may help you?

  20. Reaction of Weak Bases with Water The generic reaction for a base reacting with water, producing its conjugate acid and hydroxide ion: B + H2O  BH+ + OH- (All weak bases do this.)

  21. A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? Step #1:Write the equation for the reaction NH3 + H2O  NH4+ + OH-

  22. A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? Step #2:ICE it! NH3 + H2O  NH4+ + OH- 0 0.50 0 +x +x - x x x 0.50 - x

  23. A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? Step #3:Set up the law of mass action NH3 + H2O  NH4+ + OH- E 0.50 - x x x

  24. A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? Step #4:Solve for x, which is also [OH-] NH3 + H2O  NH4+ + OH- E 0.50 - x x x [OH-] = 3.0 x 10-3 M

  25. A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? Step #5:Convert [OH-] to pH NH3 + H2O  NH4+ + OH- E 0.50 - x x x

  26. A B C D Section 18.3 Assessment In dilute aqueous solution, as [H+] increases: A.pH decreases B.pOH increases C.[OH–] decreases D.all of the above Section 18-3

  27. A B C D Section 18.3 Assessment What is the pH of a neutral solution such as pure water? A.0 B.7 C.14 D.1.0 × 10–14 Section 18-3

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