Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109

General Chemistry

Spring 2009

Lecture Notes:: 4 May

© R. Paselk 2002
 
     
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Acids and Bases, cont.

Hydrolysis (Acid/Base Behavior of Salts)

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:

A- + H2O Equilibrium double arrow OH- + HA

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 655-660.)

Note that the ion of a strong acid, such as HCl, will not affect pH since the strong anion (e.g chloride ion) has no tendency to react with water. Similarly the ion of a strong base such as NaOH will not affect pH since the strong cation (e.g. Na ion) 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:

HC2H3O2 + OH- Equilibrium double arrow C2H3O2- + H2O

The resulting acetate ion can now react with water:

C2H3O2- + H2O Equilibrium double arrow HC2H3O2 + OH-

This reaction can be written as the sum of the association of acetic acid and the dissociation of water:

C2H3O2- + H+ Equilibrium double arrow HC2H3O2

H2O Equilibrium double arrowH+ + OH-

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:

NH3 + H2O Equilibrium double arrow NH4+ + OH-

When an ammonium salt is dissolved in water the ammonium ion reacts with the water:

NH4+ + H2O Equilibrium double arrow 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. As we look at them let's name the salts; write the formulae and name the acids corresponding to the ions, and write the reactions for the various ions with water as the basis for our predictions.

Predicted pH for salt solutions (1 M)
 Salt   pH
KCN

strong base, weak acid salt

K+ + H2O Equilibrium double arrow NR

CN- + H2O Equilibrium double arrow HCN + OH-

>7
KC2H3O2

 strong base, weak acid salt

K+ + H2O Equilibrium double arrow NR

C2H3O2- + H2O Equilibrium double arrow HC2H3O2 + OH-

>7
 NH4Cl

weak base, strong acid salt

NH4+ + H2O Equilibrium double arrow NH3 + H3O+

Cl- + H2O Equilibrium double arrow NR

<7
  NH4 NO3

weak base, strong acid salt

NH4+ + H2O Equilibrium double arrow NH3 + H3O+

NO3- + H2O Equilibrium double arrow NR

<7
 NH4CN

NH3 stronger than HCN

NH4+ + H2O Equilibrium double arrow NH3 + H3O+

CN- + H2O Equilibrium double arrow HCN + OH-

>7
NH4C2H3O2

pKb NH3 = pKa HC2H3O2 

NH4+ + H2O Equilibrium double arrow NH3 + H3O+

C2H3O2- + H2O Equilibrium double arrow HC2H3O2 + OH-

7

Example: Calculate the pH of a 0.54 M solution of  NH4Cl. Kb = 1.8 x 10-5

   NH4+ + H2  Equilibrium double arrow NH3 + H3O+
at equilibrium 0.54 - x     x   x

Kh = (Kw)(1/Kb) = [H+] [NH3] / [NH4+]

Kh = (1.0 x 10-14) / (1.8 x 10-5) = 5.6 x 10-10

x2/(0.54 - x) = 5.6 x 10-10; assume x << 0.54

x = [H3O+] = 1.7 x 10-5 M

pH = 4.76

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 - 10.93 = 3.07; [OH-] = 8.5 x 10-4 M

   BrO- + H2  Equilibrium double arrow
HBrO
+ 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

Kh = Kw / Ka = [HBrO] [OH-] / [BrO-]

Ka = Kw[BrO-] / [HBrO] [OH-]

Ka = ( 1.0 x 10-14) (0.15) / (8.5 x 10-4)2 = 2.1 x 10-9

Lewis Acids

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.)

Polyprotic acids

There are many acids with more than one dissociable proton. For small acids, with the various dissociating protons on a single group we can consider that the protons dissociate successively, and(for pKa differences of 2 or greater) that the lower pKa protons dissociate completely before subsequent protons dissociate. (overhead, Titration curve for Phosphoric acid, P 3.17) This makes life a bit easier for purposes of calculating which species exist in a given solution. (on-line lecture slides for titration and relative species for phosphoric acid)

H3PO4 Equilibrium double arrow H+ + H2PO4- Equilibrium double arrow 2 H+ + HPO42- Equilibrium double arrow 3 H+ + PO43-

Rxn   Ka  pKa
 H3PO4 Equilibrium double arrow H+ + H2PO4-  7.5 x 10-3  2.2
 H2PO4- Equilibrium double arrow H+ + HPO42-  6.2 x 10-8  7.2
HPO42- Equilibrium double arrow H+ + PO43-  4.8 x 10-13  12.7

We will not cover 15.6 - 15.8 in Zumdahl (Chem 110 material).

 

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