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ACIDS AND BASES

ACIDS AND BASES. Dissociation Constants. Write the equilibrium expression ( K a or K b ) from a balanced chemical equation. Use K a or K b to solve problems for pH, percent dissociation and concentration. Additional KEY Terms.

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ACIDS AND BASES

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  1. ACIDS AND BASES Dissociation Constants

  2. Write the equilibrium expression (Kaor Kb) from a balanced chemical equation. • UseKaor Kb to solve problems for pH, percent dissociation and concentration. Additional KEY Terms

  3. HA(aq) H+(aq) + A-(aq) HA(aq) + H2O(l) H3O+(aq) + A-(aq) Strong Acid Weak Acid Ka - acid dissociation constant Larger Ka : stronger acid : more product : more H+

  4. BOH(aq) B+(aq) + OH-(aq) B(aq) + H2O(l) BH+(aq) + OH-(aq) Strong Base Weak Base Kb - base dissociation constant Larger Kb : stronger base : more product : more OH-

  5. CH3COOH(aq) + H2O(l) H3O+(aq) + C2H3O2-(aq) Initially a 0.10 M solution of acetic acid, it reaches equilibrium with a [H3O+] = 1.3 x 10-3 M. What is the acid dissociation constant, Ka? I0.10 0 0 C-1.3 x 10-3+1.3 x 10-3 +1.3 x 10-3 Ka = 1.7 x 10-5 Ignore the units for K. 1.3 x 10-3 E0. 0987 1.3 x 10-3 Type III – all initial and one equilibrium concentration

  6. HA(aq) + H2O(l) H3O+(aq) + A-(aq) Ka = [H3O][A-] [HA] HA is a weak acid with a Ka of 7.3 x 10-8. What are the equilibrium concentrations if the initial [HA] is 0.50 M? [I]0.50 0 0 [C]-x+x +x [E]0.5-x +x +x Type IV– all initial and NO equilibrium

  7. 7.3 x 10-8 = [x][x] 0.50 *Ka is small - assume that x is negligible compared to 0.50 - x (7.3 x 10-8)(0.50) = x2 √ √ 3.65 x 10-8 = x2 [H3O+] = [A-] = 1.9 x 10-4 M 1.9 x 10-4 = x [HA] = 0.50 - x = 0.50 - 1.9 x 10-4 = 0.49981 M *Kais small – OK to ignore it 0.50 M

  8. Calculate the pH of a 0.10 mol/L hydrogen sulfide solution. (Ka=1.0 x 10-7) H2S (aq) + H2O (l) Ka = [H3O+][HS-] [H2S] • H3O+(aq) + HS-(aq) [I] 0.10 0 0 [C] -x +x +x [E] 0.10 - x x x

  9. 1.0 x 10-7 = [x][x] 0.10 *Ka is small - xis negligible - x (1.0 x 10-7)(0.10) = x2 √ √ 1.0 x 10 -8 = x2 [H3O+] = [HS-] = 1.0 x 10-4 M 1.0 x 10 -4 = x pH = - log [H3O+] = - log(1.0 x 10-4) pH = 4.00

  10. Each acid/base has K associated with it • polyprotic acids lose their hydrogenone at a time - each ionization reaction has separate Ka H2SO4(aq) H+(aq) + HSO4-(aq) HSO4- (aq) H+(aq) + SO4-2(aq) Sulfuric acid H2SO4 Ka1 Ka2

  11. Percent Dissociation • Ka / Kb represent the degree of dissociation • (how much product has formed) • Another way to describe dissociation is by percent dissociation

  12. CH2O2H (aq) + H2O(l) H3O+(aq) + CH2O2¯(aq) Calculate the percent dissociation of a solution of formic acid (CH2OOH) if the hydronium ion concentration is . 0.100 M 4.21 x 10-3 M

  13. Calculate the Kbof hydrogen phosphate ion (HPO42-) if 0.25 M solution of hydrogen phosphate dissociates 0.080%. HPO42- + H2O H2PO4- + OH- Use the %diss formula to find [OH-]

  14. HPO42- + H2O H2PO4- + OH- Kb = [H2PO4-][OH-] Kb= [2.0 x 10-4][2.0 x 10-4] [HPO42-] 0.25 [I]0.25 0 0 [C]-x+x +x [E]0.25 +x +x [OH-] = [H2PO4-] = 2.0 x 10-4 M Kb= 1.6 x 10-7

  15. The smaller the Ka or Kb, the weaker the acid / base • percent dissociation describes the amountof • acid/base dissociated

  16. CAN YOU / HAVE YOU? • Writethe equilibrium expression (Kaor Kb) from a balanced chemical equation. • UseKaor Kb to solve problems for pH, percent dissociation and concentration. Additional KEY Terms

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