Acid Dissociation Constant. Dissociation Constants For a generalized acid dissociation, the equilibrium expression would be This equilibrium constant.

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Presentation transcript:

Acid Dissociation Constant

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

Dissociation Constants The greater the value of K a, the stronger is the acid.

Calculating K a from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25  C is Calculate K a for formic acid at this temperature. We know that [H 3 O + ] [COO - ] [HCOOH] K a =

Calculating K a from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25  C is Calculate K a for formic acid at this temperature. To calculate K a, we need the equilibrium concentrations of all three things. We can find [H 3 O + ], which is the same as [HCOO - ], from the pH.

Calculating K a from the pH pH = -log [H 3 O + ] 2.38 = -log [H 3 O + ] = log [H 3 O + ] = 10 log [H 3 O + ] = [H 3 O + ] 4.2  = [H 3 O + ] = [HCOO - ]

Calculating K a from pH Now we can set up a table… [HCOOH], M[H 3 O + ], M[HCOO - ], M Initially Change   At Equilibrium  = =  10 -3

Calculating K a from pH [4.2  ] [0.10] K a = = 1.8  10 -4

Calculating Percent Ionization Percent Ionization =  100 In this example [H 3 O + ] eq = 4.2  M [HCOOH] initial = 0.10 M [H 3 O + ] eq [HA] initial Percent Ionization =   = 4.2%

Calculating pH from K a Calculate the pH of a 0.30 M solution of acetic acid, HC 2 H 3 O 2, at 25  C. HC 2 H 3 O 2 (aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2 - (aq) K a for acetic acid at 25  C is 1.8 

Calculating pH from K a The equilibrium constant expression is [H 3 O + ] [C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ] K a =

Calculating pH from K a We next set up a table… [C 2 H 3 O 2 ], M[H 3 O + ], M[C 2 H 3 O 2 - ], M Initially Change-x+x At Equilibrium x  0.30 xx We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored.

Calculating pH from K a Now, (x) 2 (0.30) 1.8  = (1.8  ) (0.30) = x  = x  = x

Calculating pH from K a pH = -log [H 3 O + ] pH = -log (2.3  ) pH = 2.64