Lakhasly

A- HYDROLYSIS OF SALTS We expect solutions of substances such as HCl and HNO2 to be acidic and solutions of NaOH and NH3 to be basic.2- Using a measuring cylinder, put 25 mL of a buffer solution (of pH = 7) in a 50 mL beaker.9- Put 25 mL of distilled water in a 50 mL beaker and, keeping its exposure time to the air as short as possible, record its pH. If the water is absolutely pure its pH will be 7.0, but it is very difficult to achieve this.Similarly, cations from strong bases, such as Na+ from NaOH or K+ from KOH, do not react with water to affect the pH. Hydrolysis of an anion occurs only when it can form a molecule or ion that is a weak electrolyte in reaction with water.However, we may be somewhat surprised at first to discover that aqueous solutions of some salts such as NaNO2 and KC2H3O2 are basic, whereas others such as NH4Cl and FeCl3 are acidic.For example, when NaOH and HNO2 (nitrous acid) react, the salts NaNO2 is formed: NaOH (aq) + HNO2 (aq) -> NaNO2 (aq) + H2O (l) [1] Nearly all salts are strong electrolytes and exist as ions in aqueous solutions.Recall that salts are the products formed in neutralization reactions of acids and bases.1.2.3.4.

A- HYDROLYSIS OF SALTS
We expect solutions of substances such as HCl and HNO2 to be acidic and
solutions of NaOH and NH3 to be basic. However, we may be somewhat surprised
at first to discover that aqueous solutions of some salts such as NaNO2 and
KC2H3O2 are basic, whereas others such as NH4Cl and FeCl3 are acidic. Recall
that salts are the products formed in neutralization reactions of acids and bases.
For example, when NaOH and HNO2 (nitrous acid) react, the salts NaNO2 is
formed:
NaOH (aq) + HNO2 (aq) → NaNO2 (aq) + H2O (l) [1]
Nearly all salts are strong electrolytes and exist as ions in aqueous solutions.
Many ions react with water to produce acidic or basic solutions. The reactions of
ions with water are frequently called hydrolysis reactions. We will see that anions
such as CN‾ and C2H3O2‾ that are the conjugate bases of the weak acids HCN and
HC2H3O2, respectively, react with water to form OH‾ ions. Cations such as NH4
+
and Fe3+ come from weak bases and react with water to form H+ ions.
Hydrolysis of Anions: Basic Salts
Anions of weak acids react with proton sources. When placed in water these
anions react to some extent with water to accept protons and generate OH– ions
and thus cause the solution pH to be greater than 7. Recall that proton acceptors
are BrØnsted-Lowry bases. Thus, the anions of weak acids are basic in two senses:
they are proton acceptors, and their aqueous solutions have pH’s above 7. The
nitrite ion, for example, reacts with water to increase the concentration of OH–
ions:
NO2‾ (aq) + H2O (l) ⇌ HNO2 (aq) + OH‾ (aq) [2]
This reaction of nitrite ion is similar to that of weak bases such as NH3 with water:
NH3 (aq) + H2O (l) ⇌ NH4

• (aq) + OH‾ (aq) [3]
  Thus, both NH3 and NO2‾ are bases and as such have a basicity or basedissociation constant, Kb, associated with their corresponding equilibria.
  According to the BrØnsted-Lowry theory, the nitrite ion is the conjugate base of
  nitrous acid. Let’s consider the conjugate acid-base pair HNO2 and NO2‾ and their
  behavior in water:
  HNO2 ⇌ H+ + NO2‾ Ka = [H+] [NO2‾] / [HNO2] [4]

NO2
– (aq) + H2O (l) ⇌ HNO2 (aq) + OH‾ (aq)
Kb = [HNO2] [OH‾] / [NO2‾] [5]
Multiplication of these dissociation constants yields:
Ka × Kb = [H+] [OH‾] = Kw = 1.0 × 10‾14 [6]
Where Kw is the ion-product constant of water.
We note that the stronger the acid is, the larger its Ka, and the weaker its
conjugate base, the smaller its Kb. Likewise, the weaker the acid (the smaller the
Ka), the stronger the conjugate base (the larger the Kb).
Anions derived from strong acids, such as Cl‾ from HCl, do not react with water
to affect the pH. Nor do Br‾, I‾, NO3‾, SO4
2
‾, and ClO4‾ affect the pH, for the same
reason. They are spectator ions in the acid-base sense and can be described as
neutral ions. Similarly, cations from strong bases, such as Na+ from NaOH or K+
from KOH, do not react with water to affect the pH. Hydrolysis of an anion occurs
only when it can form a molecule or ion that is a weak electrolyte in reaction with
water. Strong acids and bases do not exist as molecules in dilute water solutions.
26
Hydrolysis of Cations: Acidic Salts
Cations that are derived from weak bases react with water to increase the
hydrogen-ion concentration; they form acidic solutions. The ammonium ion is
derived from the weak base NH3 and reacts with water as follows:
NH4

• (aq) ⇌ NH3 (aq) + H+ (aq) [7]
  This reaction is completely analogous to the dissociation of any other weak acid,
  such as HC2H3O2 or HNO2. The acid dissociation constant of NH4


• in [7] is related
  to the Kb of NH3, which is the conjugate base of NH4
  +:
  NH3 (aq) + H2O (l) ⇌ NH4


• (aq) + OH‾ (aq) [3]′
  Cations of the alkali metals (Group 1A) and the larger alkaline earth ions, Ca2+,
  Sr2+, and Ba2+, do not react with water, because they come from strong bases. Thus
  these ions have no influence on the pH of aqueous solutions. Consequently, they
  are described as being neutral in the acid-base sense. The cations of most other
  metals do hydrolyze to produce acidic solutions. Metal cations are coordinated
  with water molecules, and it is the hydrated ion that serves as the proton donor.
  The following equation illustrate this behavior for the hexaaqua iron (III) ion:
  Fe(H2O)6
  3+ (aq) + H2O (l) ⇌ Fe(H2O)5(OH)2+ (aq) + H3O+ (aq) [8]
  We frequently omit the coordinated water molecules from such equations. For
  example, Equation [8] may be written as
  Fe3+ (aq) + H2O (l) ⇌ Fe(OH)2+ (aq) + H+ (aq) [9]
  Additional hydrolysis reactions can occur to form Fe(OH)2


• and even to the
  precipitation of Fe(OH)3. The equilibria for such cations are often complex, and
  not all species have been identified. However, equations such as [8] and [9] serve
  to illustrate the acidic character of dipositive and tripositive ions and account for
  most of the H+ in these solutions.
  Summary of Hydrolysis Behavior of Salts
  Whether a solution of a salt will be acidic, neutral or basic can be predicted on the
  basis of the strengths of the acid and base from which the salts was formed.

1. Salt of a strong acid and a strong base: Examples: NaCl, KBr, and
   Ba(NO3)2. Neither the cation nor anion hydrolyzes, and the solution has a
   pH of 7.


2. Salt of a strong acid and a weak base: Examples: NH4Br, ZnCl2, and
   Al(NO3)3. The cation hydrolyzes, forming H+ ions, and the solution has a
   pH less than 7.


3. Salt of a weak acid and a strong base: Examples: NaNO2, KC2H3O2, and
   Ca(OCl)2. The anion hydrolyzes, forming OH‾ ions, and the solution has a
   pH greater than 7.


4. Salt of a weak acid and a weak base: Examples: NH4F, NH4C2H3O2, and
   Zn(NO2)2. Both ions hydrolyze. The pH of solution is determined by the relative
   extent to which each ion hydrolyzes.
   In this experiment, we will test the pH of water and of several aqueous salt
   solutions to determine whether these solutions are acidic, basic, or neutral. In each
   case, the salt solution will be 0.1 M. Knowing the concentration of the salt solution
   and the measured pH of each solution allows us to calculate Ka or Kb for the ion
   that hydrolyzes.
   PROCEDURE
   1- Boil approximately 500 mL of distilled water for about 10 min to expel
   dissolved carbon dioxide. Allow the water to cool to room temperature.
   2- Determine the pH of unboiled and boiled distilled water.
   3- Determine the pH of the following solutions that are 0.1 M: NaCl,
   NaC2H3O2, NH4Cl, ZnCl2, KAl(SO4)2, and Na2CO3. Use about 25 mL of
   each of these solutions. Rinse the beaker with boiled distilled water when
   you go from one solution to the next.
   4- Add three drops of each of the following indicators: methyl red,
   bromothymol blue, and phenolphthalein to each solution (one indicator per
   solution). Record the colors.
   From the pH values that you determined, calculate the hydrogen- and
   hydroxide- ion concentrations for each solution (pH = – log10 [H+], and [H+] ×
   [OH‾] = 1×10‾14). Complete the tables on the report sheets and calculate the Ka or
   Kb as appropriate.
   B- THE ACTION OF A BUFFER SOLUTION
   DISCUSSION
   The pH of solution is often accomplished by the use of buffers. A buffer
   solution has the important property of resisting large changes in pH upon the
   addition of small amounts of strong acids or bases. A buffer solution must have
   two components– one that will react with H+, and the other that will react with
   OH‾. The two components of a buffer solution are usually a weak acid and its
   conjugate base, such as HC2H3O2–C2H3O2‾ or NH4
   +–NH3. Thus buffers are often
   prepared by mixing a week acid or a week base with a salt of that acid or base. For
   example, the HC2H3O2–C2H3O2‾ buffer can be prepared by adding NaC2H3O2 to a
   solution of HC2H3O2; the NH4
   +–NH3 can be prepared by adding NH4Cl to a
   solution of NH3. By appropriate choice of components and their concentrations,
   buffer solutions of virtually any pH can be made.
   To examine how a buffer works, consider, for example, the HC2H3O2–C2H3O2‾
   buffer. If OH‾ ions are added, they react with the acid component of the buffer:
   OH‾ (aq) + HC2H3O2 (aq) → C2H3O2‾ (aq) + H2O (l) [10]
   If H+ are added, they react with the base component of the buffer:

H+ (aq) + C2H3O2‾ (aq) → HC2H3O2 (aq) [11]
Buffers resist changes in pH most effectively when the concentrations of the
conjugate acid-base pair, HC2H3O2 and C2H3O2‾ in the above example, are about
the same.
Here you will demonstrate the action of a buffer solution by comparing the pH
changes after the addition of small measured amounts of HCl and NaOH to a
buffer solution and to pure water.
PROCEDURE
1- Fill a burette with 0.1 M HCl and another with 0.1 M NaOH.
2- Using a measuring cylinder, put 25 mL of a buffer solution (of pH = 7)
in a 50 mL beaker.
3- Rinse the pH meter electrode with distilled water from a wash bottle,
and put it into the beaker. Make sure that the glass bulb is completely
immersed and record the pH.
4- Place the beaker under the burette containing NaOH and, making sure
the alkali does not fall directly on the electrode, add 1 drop of 0.1 M
NaOH. Stir gently to ensure thorough mixing and record the pH.
5- Add more NaOH to make the total volume added 1.0 mL and record the
pH as before.
6- Add more NaOH to make the total volume added 5.0 mL and record the
pH.
7- Rinse the electrode in distilled water and stand it in a flask of distilled
water.
8- Take another 25 mL portion of the buffer and record its pH. Record the
pH on the addition of 1 drop, 1.0 mL, and 5.0 mL of 0.1 M HCl in the
same way you did for NaOH. Again, rinse the electrode carefully and
stand it in distilled water.
9- Put 25 mL of distilled water in a 50 mL beaker and, keeping its
exposure time to the air as short as possible, record its pH. If the water
is absolutely pure its pH will be 7.0, but it is very difficult to achieve
this. If the pH is less than 6.0, wash the beaker and the electrode more
carefully and try again.
10- When you have a pH between 6.0 and 7.0 for the ‘pure’ water, record
the pH changes on the addition of 0.1 M NaOH and 0.1 M HCl
(separately) just as you did for the buffer solution. Take special care to
wash the electrode when you change from using alkali to acid.