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ACIDS AND BASES

ACIDS AND BASES. Dissociation Constants. CAN YOU / HAVE YOU? Write equilibrium expression for an acid or base Calculate the acid/base dissociation constant Calculate the percent dissociation. HA (aq) H + (aq) + A - (aq).

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ACIDS AND BASES

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  1. ACIDS AND BASES Dissociation Constants

  2. CAN YOU / HAVE YOU? • Write equilibrium expression for an acid or base • Calculate the acid/base dissociation constant • Calculate the percent dissociation

  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 : strong 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 : strong base : more product : more OH-.

  5. CH3COOH(aq) + H2O(l) H3O+(aq) + C2H3O2¯(aq) Initially a 0.10 M solution of acetic acid, it reached 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 E0. 0987 1.3 x 10-3 1.3 x 10-3

  6. Ka = 1.7 x 10-5 There are no units for the Ka value

  7. 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 (HA, H3O+ and A¯) if the initial concentration of HA is 0.50 mol/L? [I]0.50 0 0 [C]-x+x +x [E]0.5-x +x +x

  8. 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 *Ka is small – right to ignore it compared to 0.50 • 0.50 M

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

  10. 1.0 x 10-7= [x][x] 0.10 *Ka is small - assume that x is negligible compared to 0.50 - 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

  11. Each acid/base has K associated with it. Diprotic/triprotic acids lose their hydrogensone 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

  12. Percent Dissociation Ka / Kb represent the degree of dissociation. Another way to describe the amount of dissociation is by percent dissociation.

  13. 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

  14. Calculate the Kb of hydrogen phosphate ion (HPO42-) if a 0.25 Msolution of hydrogen phosphate is dissociated is 0.080%. HPO42-+ H2O H2PO4-+ OH-

  15. HPO42¯ + H2O H2PO4¯ + OH¯ Kb= [H2PO4-][OH-] Kb= [2.0 x 10-4][2.0 x 10-4] [HPO42-] 0.25 [OH-] = [H2PO4-] = 2.0 x 10-4 M Kb= 1.6 x 10-7

  16. The smaller the Ka or Kb, the weaker the acid / base • The percent dissociation also describes the amount of acid/base dissociated • The percent dissociated is calculated by

  17. CAN YOU / HAVE YOU? • Write equilibrium expression for an acid or base • Calculate the acid/base dissociation constant • Calculate the percent dissociation

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