Hydrolysis means "cleavage by means of water". There are many different types of hydrolysis reaction, but in this problem, hydrolysis refers to a reaction between water and the anion of the weak acid. For example, the acetate ion hydrolyzes according to:
You would expect a solution of sodium acetate to be somewhat basic because of the hydroxide produced by this reaction. But how basic will the solution be?
The pH of the solution depends on the how strong a base the hydrolyzing anion is. The stronger the acid that forms from the hydrolysis, the weaker the hydrolyzing anion is as a base. It does make sense: The stronger an acid is, the more easily it gives up hydrogen ions, so the anion formed in the dissociation must not hang on to hydrogen ions very tightly.
This reciprocal relationship between strength of the acid and the conjugate anion base can be expressed quantitatively using equilibrium constants. The equilibrium constant for the basic hydrolysis Kb can be calculated from the the equilibrium constant for the acid dissociation Ka as Kb = Kw/Ka, where Kw is the equilibrium constant for dissociation of water into H+ and OH-(aq) (equal to 1.0x10-14.
Try predicting which of the following solutions will have the highest and which will have the lowest pH:
Hydrolysis equilibria must be accounted for if you want to accurately estimate the pH of any solution that contains appreciable quantities of the anion of a weak acid. There are really two places on the titration curve where hydrolysis equilibria are important: near the equivalence point, and in the buffer region.
pH in the buffer region is usually computed by solving a system of equilibrium constant, mass balance, and charge balance equations (or more simply and approximately using the Henderson-Hasselbach equation). The buffer region is a story for another time, but here is how you can use hydrolysis equilibria to compute the pH at the equivalence point of the titration.
Let's consider a specific example. Suppose we're titrating 25.0 mL of 0.1000 M HC2H3O2(aq) with 0.1000 M NaOH(aq). The equivalence point will occur at 25.0 mL added NaOH(aq) solution. Picture this as a two step process. The first step involves stoichiometric reaction between the NaOH and the acetic acid; the second step involves hydrolysis of the acetate formed by the first step.
|moles before reaction||0.00250||0.00250||0|
|change in moles||-0.00250||-0.00250||+0.00250|
|moles after reaction||0||0||0.00250|
|molarity after reaction||0||0||0.0500|
|C2H3O2-(aq)||+ H2O(l) =||HC2H3O2(aq)||+||OH-(aq)|
|molarity before equilibrium||0.050||0||~0|
|change in molarity||- x||+ x||+x|
|molarity at equilibrium||0.050 - x||x||x|
The equilibrium constant expression is
and solving this for x with Kb = Kw/Ka = 1.0x10-14/1.8x10-5 = 5.6x10-10 gives x = [OH-(aq)] = 5.27x10-6. Then [H3O+(aq)] = Kw/[OH-(aq)] = 1.0x10-14/5.27x10-6 = 1.9x10-9, and pH ~ -log [H3O+(aq)] = 8.72.
Author: Fred Senese firstname.lastname@example.org
Copyright © 1997-2010 by Fred Senese
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Last Revised 08/17/15.URL: http://antoine.frostburg.edu/chem/senese/101/acidbase/faq/print-hydrolysis-in-pH-calculations.shtml