Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109 - General Chemistry - Spring 2013

Lecture Notes 40: 1 May

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Hydrolysis (Acid/Base Behavior of Salts), cont.

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. (slide; Titration curve for Phosphoric acid) 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

Let's find the phosphate species concentrations for a 0.10 M solution with [H+] = 3.16 x 10-7 M. From the pKa values we can see that the first titration is complete, and the third has not started - only the middle titration contributes to the hydrogen ion concentration, [H+] = 3.16 x 10-7 M.

We can find the concentrations of the two phosphate species involved in this equilibria using the Henderson-Hasselbalch equation:

pH = pKa + log [HPO42-] / [H2PO4-]

log [HPO42-] / [H2PO4-] = pH - pKa= 6.5 - 7.2 = -0.7

[HPO42-] / [H2PO4-] = 0.20

but [HPO42-] + [H2PO4-] = 0.10 M,

[H2PO4-] = 0.10 - [HPO42-]

substituting for [H2PO4-] and rearranging

[HPO42-] = (0.2)(0.10 - [HPO42-])

1.2 [HPO42-] = 0.020

[HPO42-] = 0.017 M; [H2PO4-] = 0.083 M

Even though the other two equilibria do not contribute significantly to the pH, they are still important for finding the concentrations of the other two phosphate species. And such species are frequently important in other aspects of the system of interest, such as potential for precipitating metals as phosphates etc.

Let's find phosphate ion first. From the table above:

HPO42- Equilibrium double arrow H+ + PO43-

and

Ka = 4.8 x 10-13 = [H+] [PO43-] / [HPO42-]

But we know [H+] = 3.16 x 10-7 M from the pH value given, and we know [HPO42-] = 0.017 M from above. Substituting

4.8 x 10-13 = (3.16 x 10-7) [PO43-] / (0.017), &

[PO43-] = 2.6 x 10-8 M

Similarly we can find the phosphoric acid concentration. From the table above:

H3PO4 Equilibrium double arrow H+ + H2PO4-

and

Ka = 7.5 x 10-3 = [H+] [H2PO4-] / [H3PO4]

Again know [H+] = 3.16 x 10-7 M, and [H2PO4-] = 0.083 M from above. Substituting:

7.5 x 10-3 = (3.16 x 10-7)(0.083) / [H3PO4],

[H3PO4] = 3.5 x 10-6 M

 

Elements Video

 


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© R A Paselk

Last modified 10 May 2013