Notes:
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Acidic Buffers
- Suppose you have a solution which is 0.20 M in acetic acid (HAc) and 0.10 M in sodium acetate (NaAc). What will be the pH of this solution? This is a typical buffer problem. Notice that the concentrations of both the weak acid and its conjugate base are known. The equilibrium is still that of acetic acid; the only difference is that we have a common ion (acetate ion) present.
- First, set up a table as we have done before for a weak acid problem:
- Notice that the initial concentrations of both HAc and Ac- are known. The acetate ion came from the sodium acetate. The sodium ion is not a part of the equilibrium and is therefore neglected as a spectator ion. In the next step we apply LeChatlier's principle and see that in order to establish equilibrium, some HAc will have to dissociate to produce hydronium ions:
- The concentrations of HAc, H3O+ and Ac- at equilibrium will be the sum of initial concentrations and the change in concentrations:
These algebraic quantities may now be inserted into the mass action equilibrium expression for acetic acid:
The values of x shown in bold above may often neglected if they can be shown to be negligibly small. Compare the concentrations to the Ka value. In this case the concentrations of the weak acid and its conjugate base are known to the 10-2 place and the Ka is to the 10-5 place. Since the difference in magnitude is greater than 100 (actually 1000 times different), both of this x quantities may be neglected. This simplifies the algebraic expression to:
Solving for x gives the hydronium ion concentration:
- To find the pH, take the negative log the hydronium ion concentration:
pH = -log[H3O+] = -log[3.6 x 10-5] = 4.44