| Syllabus / Schedule | 
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| Chem 109 |
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Summer 2002 |
| Lecture Notes::24 July |
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| PREVIOUS |
When a weak acid is titrated with a strong base it is found that the equivalence point always occurs at a pH above neutrality (pH > 7). Similarly, when a weak base is titrated with a strong acid the equivalence point always occurs at a pH below neutrality (pH < 7). Why? Essentially it is a consequence of the the nature of acids as seen in the Brønsted acid description: when the acid is titrated a weak base, the conjugate base is formed. The conjugate base then competes with hydroxide ion for hydrogen ion, leading to a slightly higher concentration of hydroxide ion as the equilibrium is shifted:
This process is sometimes referred to as hydrolysis, which is the term we will also use. (Notice that Zumdahl discusses this phenomena in section 14.8 Acid-Base Properties of Salts pp 683-688.)
Note that the ion of a strong acid, such as HCl, will not affect pH since the chloride ion has no tendency to react with water. Similarly the ion of a strong base such as NaOH will not affect pH since Na has no tendency to react with water either.
So how is the pH affected by the presence of the salts of weak acids? Let's look at our favorite acid, acetic acid reacting with NaOH:
The resulting acetate ion can now react with water:
This reaction can be written as the sum of the association of acetic acid and the dissociation of water:
Notice that the the equilibrium constant for the association of acetic acid is the inverse of the dissociation constant (the reaction is backwards, inverting the equilibrium expression): K = 1/Ka
The overall equilibrium constant is then the product of the equilibrium constants for the two reactions, Kh:
Kh = (Kw)(1/Ka) Kh is in fact Kb for the conjugate base. (Zumdahl use Kb in his discussion.) Kb, conj = Kw/Ka
A similar treatment is seen for weak bases. Using ammonia as an example:
When an ammonium salt is dissolved in water the ammonium ion reacts with the water:
NH4+ + H2O ´ H3O+ + NH3 This time Kh is Ka for the conjugate acid, NH4+, so
Kh = Ka = [H+] [NH3] / [NH4+] = Kw/Kb
Note that the weaker the acid or base, the greater the hydrolysis of its conjugate ion (the more the pH deviates from neutrality)!
Let's look at some examples qualitatively:
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Example: Calculate the pH of a 0.54 M solution of NH4Cl. Kb = 1.8 x 10-5
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Example: A 0.15 M solution of NaBrO (sodium hypobromite) has a pH = 10.93. What is the Ka for HBrO?
pH + pOH = 14.00; pOH = 14.00 - pH = 14.00 - 3.07 = 3.07; [OH-] = 8.5 x 10-4 M
BrO- + H2O Æ + OH- at equilibrium 0.15 - 8.5 x 10-4 = 0.15 M 8.5 x 10-4 M 8.5 x 10-4 M
Just as the Arrhenius model for acids and bases did not include some bases because of its limited definition (ammonia is not an Arrhenius base even though its solutions are very obviously basic because it has no oxygen in its formula to donate as a hydroxide), the Brønsted-Lowry model is also limited by its definition based on protons. A more general model is based on electron pairs, a more fundamental view - remember, chemistry is mostly due to electron exchanges and interaction.
A Lewis acid is defined as an electron pair acceptor, while a Lewis base is defined as an electron pair donor. Notice that Brønsted and Arrhenius acids are also Lewis acids, since a proton is an electron pair acceptor. That is, a hydrogen ion has an empty outer orbital which an electron pair can occupy. Similarly, Brønsted bases are Lewis bases, Ammonia for example has a lone pair of electrons which it "donates" when it reacts with water to pick up a proton to become ammonium ion.
Notice the the Lewis definition also includes other species which are not Brønsted acids as Lewis acids. For example, metal ions are Lewis acids. (Recall that a proton can also be considered the simplest metal ion - it is sometimes associated with the alkali metals in the Periodic Chart.)
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© R A Paselk
Last modified 24 July 2002
