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Acid Dissociation Constant

Acid Dissociation Constant. K c =. [H 3 O + ] [A - ] [HA]. HA ( aq ) + H 2 O ( l ). A - ( aq ) + H 3 O + ( aq ). Dissociation Constants. For a generalized acid dissociation, the equilibrium expression would be

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Acid Dissociation Constant

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  1. Acid Dissociation Constant

  2. Kc = [H3O+] [A-] [HA] HA (aq) + H2O (l) A-(aq) + H3O+(aq) Dissociation Constants • For a generalized acid dissociation, the equilibrium expression would be • This equilibrium constant is called the acid-dissociation constant, Ka.

  3. Dissociation Constants The greater the value of Ka, the stronger is the acid.

  4. [H3O+] [COO-] [HCOOH] Ka = Calculating Ka from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25C is 2.38. Calculate Ka for formic acid at this temperature. We know that

  5. Calculating Ka from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25C is 2.38. Calculate Ka for formic acid at this temperature. To calculate Ka, we need the equilibrium concentrations of all three things. We can find [H3O+], which is the same as [HCOO-], from the pH.

  6. Calculating Ka from the pH pH = -log [H3O+] 2.38 = -log [H3O+] -2.38 = log [H3O+] 10-2.38 = 10log [H3O+] = [H3O+] 4.2  10-3 = [H3O+] = [HCOO-]

  7. Calculating Ka from pH Now we can set up a table…

  8. [4.2  10-3] [4.2  10-3] [0.10] Ka = Calculating Ka from pH = 1.8  10-4

  9. [H3O+]eq [HA]initial 4.2  10-3 0.10 Calculating Percent Ionization • Percent Ionization =  100 • In this example [H3O+]eq = 4.2  10-3 M [HCOOH]initial = 0.10 M Percent Ionization =  100 = 4.2%

  10. Calculating pH from Ka Calculate the pH of a 0.30 M solution of acetic acid, HC2H3O2, at 25C. HC2H3O2(aq) + H2O (l) H3O+(aq) + C2H3O2-(aq) Ka for acetic acid at 25C is 1.8  10-5.

  11. [H3O+] [C2H3O2-] [HC2H3O2] Ka = Calculating pH from Ka The equilibrium constant expression is

  12. Calculating pH from Ka We next set up a table… We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored.

  13. 1.8  10-5 = (x)2 (0.30) Calculating pH from Ka Now, (1.8  10-5) (0.30) = x2 5.4  10-6 = x2 2.3  10-3 = x

  14. Calculating pH from Ka pH = -log [H3O+] pH = -log (2.3  10-3) pH = 2.64

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