In this experiment, you will use both indicators and pH meters to study hydrolysis reactions of various salts. A hydrolysis reaction occurs when an ion reacts with water to produce H3O+ (an acid hydrolysis) or OH- (a base hydrolysis) in aqueous solution. The trick here is to be able to figure out which species is hydrolyzing produsing the hydronium or hydroxide ions which are making the solution acidic or basic.
Notice that we are dealing primarily with conjugate acid/base relationships in this experiment; the stronger an acid, the weaker its conjugate base, and the stronger a base, the weaker its conjugate acid. Keep this in mind during the experiment! For example, consider the chloride ion, Cl-: this species is the conjugate base of HCl, which is a strong acid. Therefore, chloride ion is too weak a base to hydrolyze in aqueous solution, and a solution of a chloride salt, say, NaCl, will have a pH of 7. Note that the group I and II cations do not hydrolyze in aq solution. Similarly, a solution of KNO3 would have a pH of 7, since the nitrate ion is the conjugate base of nitric acid (a strong acid) and the potassium ion would not hydrolyze.
OK, what about a salt solution of, say, potassium acetate, KC2H3O2? This salt consists of a potassium ion, which doesn't hydrolyze, and an acetate ion. Acetate ion is the conjugate base of acetic acid, HC2H3O2, which is a weak acid. Therefore, we predict that the acetate ion will hydrolyze in aqueous solution. In any base hydrolysis reaction, the pertinent equilibrium is (for a base B-):
In the above example, for the base hydrolysis of acetate ion, we would write
The equilibrium constant for this reaction is termed a Kb, since it describes a base reaction, and is given by
Take a look at the hydrolysis reaction above and the corresponding expression for Kb: if we make a measurement of the pH of such a solution, then we know the equilibrium concentration of H3O+, and we can calculate the equilibrium [OH-], since [H3O+][OH-]= 1.0x10-14 (yes, there are other ways to do this; feel free to use them!) Now, if we know the hydroxide concentration at equilibrium, then we know the acetic acid concentration, since they are 1:1 in the reaction. Further, if we know the initial concentration of the conjugate base species which is hydrolyzing, then we can calculate Kb. Perhaps an example would suffice...
In the lab, you measure the pH of a 0.1 M solution of potassium acetate, KC2H3O2; you found the pH of this solution to be 8.5 (i'm just making up that number!!). Write the hydrolysis reaction responsible for the basic pH, and calculate the equilibrium constant.
OK. We first ionize the salt....
KC2H3O2(aq) ---> K+(aq) + C2H3O2-(aq)
potassium ion does not hydrolyze (*remember that group I and II metal cations do not hydrolyze); we look at the anion of the salt - acetate ion is the conjugate base of a weak acid (acetic acid), so we expect the acetate ion to be a strong enough base to steal a proton from water. Since the measured pH was basic, we write a base hydrolysis reaction for acetate ion:
C2H3O2-(aq) + H2O(l) <---> HC2H3O2(aq) + OH-(aq)
The equilibrium constant is
Kb = [HC2H3O2][OH-] / [C2H3O2]
Now, the initial concentration of acetate ion was 0.1M; the pH = 8.5. Since pH+pOH = 14, we have 14-8.5 = pOH=5.5. Since pOH=-log[OH-], we have that 5.5 = -log[OH-]; taking an antilog, we find that [OH-]= 3.162 x 10-6. From the hydrolysis reaction, we see that the hydroxide ion and the acetic acid appear in a 1:1 ratio; therefore, at equilibrium, [acetic acid] = [OH-]. We can substitute these values into the expression for Kb:
Kb = (3.162 x 10-6)(3.162 x 10-6) / (.10) = 9.99 x 10-11
You will need to write hydrolysis reactions and calculate Ka or Kb values for any salts which do not have a pH significantly different than that of the water used to dissolve the salt.
Shown below is an example of a well plate with the six acid-base indicators and first four salts as listed on the table on page 98 of the lab manual:
Notice that you're using the indicators to get a rough idea of the pH of a salt solution!
Note that the pH of unboiled water should be slightly acidic; this is due to the hydrolysis of carbonic acid (H2CO3).
For the three transition metal salts (copper (II) nitrate, zinc chloride, and potassium aluminum sulfate), notice that the metal cations act as weak Lewis acids; the +2 metal ions interact with water (an ion-dipole interaction), and one of the water molecules loses a proton according to:
M[H2O]62+(aq) <-----> M[H2O]5OH+(aq) + H+(aq)
the lone +3 ion will hydrolyze according to
M[H2O]63+(aq) <-----> M[H2O]5OH+2(aq) + H+(aq)
use these reactions to represent the hydrolysis of the copper(II), zinc(II) and aluminum ions.
If a salt solution of something other than a transition metal salt shows an acidic pH, then the hydrolysis reaction is given by
HA(aq) + H2O(l) <----> A-(aq) + H3O+(aq)
and the equilibrium constant is a Ka. Think about this reaction when you make measurements on the ammonium chloride.
