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EQUILBRIUM OF Acids and Bases

EQUILBRIUM OF Acids and Bases. Chapter 17. Water. AUTOIONIZATION. H 2 O can function as both an ACID and a BASE. Equilibrium constant for water = K w K w = [H 3 O + ] [OH - ] = 1.00 x 10 -14 at 25 o C. Water. K w = [H 3 O + ] [OH - ] = 1.00 x 10 -14 at 25 o C

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EQUILBRIUM OF Acids and Bases

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  1. EQUILBRIUM OF Acids and Bases Chapter 17

  2. Water AUTOIONIZATION H2O can function as both an ACID and a BASE. Equilibrium constant for water = Kw Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC

  3. Water Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC In a neutralsolution [H3O+] = [OH-] so [H3O+] = [OH-] = 1.00 x 10-7 M

  4. Calculating [H3O+] & [OH-]given pOH = 3.00 Kw = [H3O+] [OH-] = 1.00 x 10-14at 25 oC [OH-] = 10-pOH = 0.0010 [H3O+] = Kw / 0.0010 = 1.0 x 10-11 M If [H3O+] < [OH-] This solution is _________

  5. Equilibria Involving Weak Acids and Bases Aspirin is a good example of a weak acid, Ka = 3.2 x 10-4

  6. Weak Acids and Bases AcidConjugate Base acetic, CH3CO2HCH3CO2-, acetate ammonium, NH4+NH3, ammonia bicarbonate, HCO3-CO32-, carbonate A weak acid (or base) is one that ionizes to a VERY small extent (< 5%).

  7. Weak Acids and Bases acetic acid, CH3CO2H (HOAc) HOAc + H2O D H3O+ + OAc- Acid Conj. base (K is designated Ka for ACID) [H3O+] and [OAc-] are SMALL, Ka << 1.

  8. Equilibrium Constants for Weak Acids Weak acid has Ka < 1 Leads to small [H3O+] and a pH of 2 - 7

  9. Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the pH. And the equilibrium concs. of EACH Step 1. ICE table. [HOAc] [H3O+] [OAc-] I C E

  10. Equilibria Involving A Weak Acid [HOAc] [H3O+] [OAc-] I 1.00 0 0 C -x +x +x E 1.00-x x x Note that we neglect [H3O+] from H2O.

  11. Equilibria Involving A Weak Acid Step 2. Write Ka expression This is a quadratic. Use quadratic formula or method of approximations (see Appendix A). HOWEVER

  12. Equilibria Involving A Weak Acid Assume x is very small because Ka is so small. Now we can more easily solve this approximate expression.

  13. Equilibria Involving A Weak Acid Step 3. Solve Kaapproximate expression x = [H3O+] = [OAc-] = [Ka • 1.00]1/2 x = [H3O+] = [OAc-] = 4.2 x 10-3 M pH = -log (4.2 x 10-3) = 2.37

  14. Equilibria Involving A Weak Acid Calculate the pH of a 0.0010 M solution of formic acid, HCO2H. HCO2H + H2O D HCO2- + H3O+ Ka = 1.8 x 10-4 Approximate solution [H3O+] = [Ka • Co]1/2 = 4.2 x 10-4 M, pH = 3.4 Exact Solution [H3O+] = [HCO2-] = 3.4 x 10-4 M [HCO2H] = 0.0010 - 3.4 x 10-4 = 0.0007 M pH = 3.5

  15. Chapter 17 – Acids and Bases What is the [H+] of a 1.0 M acetic acid if Ka = 1.8 x 10-5 for the solution: A. 1.0 M B. 0.3 M C. 0.0042 M D. 1.8 x 10-5 M

  16. Chapter 17 – Acids and Bases A 0.100 M solution of HX has a pH of is 4.5 Determine the Ka for the acid, HX. A. 1.0 x 10-8 B. 3.0 x 10-7 C. 2.5 x 10-8 D. 4.5 x 10-9

  17. Weak Bases

  18. Equilibrium Constants for Weak Bases Weak base has Kb < 1 Leads to small [OH-] and a pH of 12 - 7

  19. Equilibria Involving A Weak Base You have 0.010 M NH3. Calc. the pH. NH3 + H2O D NH4+ + OH- Kb = 1.8 x 10-5 Step 1; ICE table [NH3] [NH4+] [OH-] I C E

  20. Weak Base Step 1. ICE table [NH3] [NH4+] [OH-] I 0.010 0 0 C -x +x +x E 0.010 - x x x

  21. Weak Base Step 2. Solve the equilibrium expression Assume x is small (100•Kb < Co), so x = [OH-] = [NH4+] = 4.2 x 10-4 M [NH3] = 0.010 - 4.2 x 10-4 ≈ 0.010 M The approximation is valid !

  22. Kw = Ka* Kb Acids Conjugate Bases

  23. Relation of Ka, Kb, [H3O+] and pH

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