Hydrolysis of Salts and choice of Indicators / Titration Curve

According to this theory, an acid is a "proton donor" and a base is a "proton acceptor."

Acids are often divided into categories such as "strong" and "weak." One measure of the strength of an acid is the acid-dissociation equilibrium constant, Ka, for that acid.

The range of concentrations of the H3O+ and OH- ions in aqueous solutions could be compressed into a more manageable set of data by taking advantage of logarithmic mathematics and calculating the pH or pOH of the solution.

pH = - log [H3O+]

pOH = - log [OH-]

The "p" in pH and pOH is an operator that indicates that the negative of the logarithm should be calculated for any quantity to which it is attached. Thus, pKa is the negative of the logarithm of the acid-dissociation equilibrium constant.

pKa = - log Ka

The only disadvantage of using pKa as a measure of the relative strengths of acids is the fact that large numbers now describe weak acids, and small (negative) numbers describe strong acids.

Just as the magnitude of Ka is a measure of the strength of an acid, the value of Kb reflects the strength of its conjugate base. Consider what happens when we multiply the Ka expression for a generic acid (HA) by the Kb expression for its conjugate base (A-).

If we now replace each term in this equation by the appropriate equilibrium constant, we get the following equation.

KaKb = Kw = 1 x 10-14

Because the product of Ka times Kb is a relatively small number, either the acid or its conjugate base can be "strong." But if one is strong, the other must be weak. Thus, a strong acid must have a weak conjugate base.

A strong base, on the other hand, must have a weak conjugate acid.

MATERIALS: Pipette, beaker, conical flask, pH meter, ammonium chloride, IM sodium hydroxide, 10 c hydrochloric acid, bromophenol blue, phenolphthalein

PROCEDURE:

1. Ammonium chloride

1. Pipette ammonium chloride into a beaker.
2. The pH value is measured using a pH meter.

2. Sodium salts

1. Pipette sodium acetate, sodium formate, sodium carbonate into three separate beakers.
2. The pH values are measured using a pH meter.

3. Acid- base titration with different indicator

1.  Pipette two sets of 10ml of hydrochloric acid into conical flask.
2. Add bromophenol blue and phenolphthalein in the two sets.
3. Fill a burette with sodium hydroxide and the volume is recorded.
4. The filtration was started and the volume is recorded at the end point.
5. The steps were repeated by using acetic against sodium hydroxide.

RESULT AND CALCULATION:

1. Ammonium chloride.

1. Do you expect an ammonium chloride to be neutral?

      No.

2. On what previously obtained EXPERIMENTAL evidence do base your above answer?

NCl (aq) is not expected to be neutral but weak acid. This is because the Ammonium ion is Bronstead-Lowry acidic: it can donate to give a neutral NH3 molecule.

        

           NH4+(aq) + H2O(l)                 NH3(aq) + H3O+ (aq) (Reversible reaction)

If ammonium chloride is added to the solution, the concentration of ammonium ion will increase. Greater concentration of ammonium ion will lead to the more H(+) ions to be released into the solution and causes the pH to drop, the pOH to increase, and the solution became acidic. Therefore a solution of any ammonium salt in water, not just ammonium chloride, is a weak acid. This is because the pH of 1 mol dm is 5.64.  

3. Measure the pH of a 1 mol / dm solution of ammonium chloride.

pH= 5.64

4. Is the solution more acidic or basic than pure water?

Yes, it is more acidic than the pure water.

5. Write an equation for the reaction that causes this.

           NH4+(aq) + H2O(l)                 NH3(aq) + H3O+(aq)

6. Brnsted defined an acid as a proton donor and a base as a proton acceptor. What is water’s function in this reaction?

Water is  and in this reaction it received protons from ammonium ions; so it can act as an acid or as a base.

7. Write an expression for the hydrolysis constant of constant (N).

(N) =  

8. What are the concentrations of the following ions in solution?

1. H3O+  

pH = 5.64

-lg[H3O+] = 5.64

[H3O+] = 2.3 x 10-6 mol/dm3

2. NH4+

[NH4+] = (1.0 – 2.3 x 10-6 ) mol/dm3

        1.0 mol/dm3

3. OH-

pOH = 14 – pH = 14 – 5.64 = 8.4

-lg[OH-] = 8.4

[OH-] = 4.0 x 10-9 mol/dm3

4. NH3

[NH3] = 2.3 x 10-6 mol/dm3

5. Cl-

[Cl-] = 1.0 mol/dm3

9. Now calculate (N).

Ka(NH4+) = [NH3][H3O+] / [NH4+]

        = (2.3 x 10-6 mol/dm3)( 2.3 x 10-6 mol/dm3) / (1.0 mol/dm3)

=5.3 x 10-12  mol/dm3

10. What is the degree of hydrolysis of NH4+

Ka = cα2, c = concentration of NH4+,    α = dissociation degree

α = 2.3 x 10-6 mol/dm3

Ka = cα2

=(1.0 mol/dm3) x(2.3 x 10-6)2

= 5.3 x 10-12

11. Realizing that you have a solution of ammonium chloride at the end point of the titration of the strong acid, hydrochloric acid and the weak base, ammonia, what indicator would you use to detect this end point in a titration of approximately molar solutions?

A suitable indicator would be methyl orange (pH range 3.1 - 4.4) or methyl red (pH               range 4.4 - 6.0).

2. Sodium salts.

1. Measure the pH of 1M sodium acetate, 1M sodium formate, 0.5M sodium carbonate  solutions, and complete the following table.

            Answer together with question no.6

2. What are the concentrations of the following ions in the solution?

1. [HOAc] =
2. [OAc] =

3. What is the  hydrolysis of acetate ions?

4. a)  What indicator would you use for titration of acetic acid with a strong base? Why.

A suitable indicator would be phenolphthalein (pH range 8.3 - 10.0) or thymol blue (pH 8.0 - 9.6) because salt of a weak acid & a strong base, will have a pH > 7.

5. Write the equation for the reaction of formate ion with water.

HCOO-(aq) + H2O(l)           HCOOH(aq) + OH-(aq)

6. Considering that the second protonation is small, fine the concentrations.     

1. [H3O+]
2. [HCO3-
3. [OH-]
4. [CO32-]

Sodium acetate salt

pH = 8.77

-lg[H3O+] = 8.77

[H3O+] = 1.7 x 10-9 mol/dm3

Kw = [H3O+][OH-] = 1.0 x 10-14

Therefore, [OH-] = 1.0 x 10-14 / 1.7 x 10-9 mol/dm3

                      = 5.6 x l0-6

Degree of hydrolysis of acetate ion = [CH3COO-]hydrolysis/[ CH3COO-]initial

= 5.6 x 10-6/1.0

= 5.6 x 10-6 mol/dm3

Sodium  formate :

 Equation for the reaction of formate ion with water:

HCOO-(aq) + H2O(l)           HCOOH(aq) + OH-(aq)

pH = 8.18

-lg[H3O+] = 8.18

[H3O+] = 6.6 x 10-9 mol/dm3

Kw = [H3O+][OH-] = 1.0 x 10-14

Therefore, [OH-] = 1.0 x 10-14/ 6.6 x 10-9 mol/dm3

= 1.5 x 10-6 mol/dm3

 Sodium carbonate :

 pH = 11.55

 Considering the second protonation was small in sodium carbonate solution,

 CO32-(aq) + H2O(l)               HCO3-(aq) + OH-(aq)

-lg[H3O+] = 11.55

[H3O+] = 2.8 x 10-12 mol/dm3

Kw = [H3O+][OH-] = 1.0 x 10-14

[OH-] = 1.0 x 10-14/ 2.8 x 10-12 mol/dm3

        = 3.6 x 10-3 mol/dm3

[HCO3-] = 3.6 x 10-3 mol/dm3

 [CO32-] = (0.5 – 3.6 x 10-3) mol/dm3

        = 5.01 x 10-1mol/dm3

7. Comment on the relative basic of acetate, formate, and carbonate ions.

CO32- > CH3COO- > HCOO-

8. Does this agree with the fact that formic acid is stronger acid than acetic acid and acetic acid is stronger acid than the bicarbonate ion?

Yes.

9. Consequently, what relationship between the strength of an acid and the strength of a conjugate base?

There exists an interdependent relationship between the strengths of an acid and conjugate base. The stronger an acid, weaker is its conjugate base, and stronger a base, weaker is its conjugate acid.

10. Derive this in a mathematical form for a general weak acid

3. Acid – base titration with different indicators.

Are the results with different indicators the same:

1. For HCl?
2. For HOAc?

Explain for the above findings.

Calculate the concentrations of HOAc and HCl solutions.

Equation for titration of hydrochloric acid and sodium hydroxide:

 NaOH(aq) + HCl(aq)               NaCl(aq) + H2O(l)

Volume of HCl, V1 = 10.0cm3

Concentration of NaOH, M2 = 0.1 mol/dm3

While Bromophenol blue was used,

Volume of NaOH, V2 = 42.0 cm3

1 mol of NaOH reacted with 1 mol of HCl

Therefore, concentration of HCl, M1 = [(M2V2)]/V1

= (0.1 x 42.0)/10.0 = 0.42 mol/dm3

While phenolphthalein was used,

Volume of HCl, V1 = 10.0cm3

Concentration of NaOH, M2 = 0.1 mol/dm3

While Bromophenol blue was used,

Volume of NaOH, V2 = 41.8cm3

1 mol of NaOH reacted with 1 mol of HCl

Therefore, concentration of HCl, M1 = [(M2V2)]/V1

= (0.1 x 41.8)/10.0 = 0.418 mol/dm3

While Bromophenol blue was used,

Equation for titration of sodium hydroxide and acetic acid with :

NaOH(aq) + HCOOH(aq)               HCOONa(aq) + H2O(l)

Volume of HOAc, V1 = 10.0cm3

Concentration of NaOH, M2 = 0.1 mol/dm3

Volume of NaOH, V2 = 41.8 cm3

1 mol of NaOH reacted with 1 mol of HOAc

Therefore, concentration of HCl, M1 = [(M2V2)]/V1

= (0.1 x41.8)/10.0 = 0.418mol/dm3

While phenolphthalein was used,

Volume of HOAc, V1 = 10.0 cm3

Concentration of NaOH, M2 = 0.1 mol/dm3

Volume of NaOH, V2 = 39.9 cm3

1 mol of NaOH reacted with 1 mol of HOAc

Therefore, concentration of HCl, M1 = [(M2V2)]/V1= (0.1 x 39.9)/10.0 = 0.39 mol/dm3

DISCUSSION:

        

In most titrations experiment it is necessary to add an indicator which produces a sudden colour change at the .  Colorless phenolphthalein (3,3-bis(4-hydroxyphenyl)-2-benzofuran-1-one) at low-. A typical indicator for - s is phenolphthalein, HC20H13O4. Phenolphthalein, whose structure is shown below, is a colour less  (Ka = 3 × 10–10 mol dm–3). Its , C20H13O4– has a strong pinkish-red colour. In order to simplify, the phenolphthalein  can be written as HIn (protonated indicator) and its pink conjugate base as In–. According to Le Chatelier’s principle, the equilibrium will be shifted to the left if H3O+ is added. Thus in a strongly acidic solution we expect nearly all the pink In– to be consumed, and only colorless HIn will remain. On the other hand, if the solution is made strongly basic, the equilibrium will shift to the right because OH– s will react with HIn molecules, converting them to In–. Thus the phenolphthalein solution will become pink. Clearly there must be some  situation where half the phenolphthalein is in the acid form and half in the colored conjugate-base form. Any acid-base indicator that changes colour between pH 4 and pH 10 is suitable to detect the end-point for a strong acid - strong base titration. Both bromophenol and phenolphthalein could be used. Just one drop of the added base will bring about a change in colour of the indicator. The pH curve for the strong acid - weak base titration explains that phenolphthalein is not a suitable indicator but bromophenol is fine. For the titration of weak acid - strong base, phenolphthalein, but not bromophenol, is a suitable indicator. For a weak acid - weak base titration, based on the pH curve there will be no rapid change in pH corresponding to the addition of just one or two drops of the base. For this reason it is not usually possible to detect the end-point using an acid-base indicator.

The degree of hydrolysis is defined as the fraction (or percentage) of the total salt which is hydrolysed at equilibrium. For example, if 90% of a salt solution is hydrolysed, its degree of hydrolysis is 0.90 or as 90%. It is generally represented by ‘h’.

h = Number of moles of the hydrolysed / Total number of moles of the salt taken

 

 

 (iv) Salts of strong acids and strong bases do not undergo hydrolysis (they undergo only ionization) hence the resulting aqueous solution is neutral.

CONCLUSION:

The choice of both colors and pKa is made, it is possible to mix several indicators and obtain a universal indicator which changes color continuously over a very wide pH range. With such a mixture it is possible to find the approximate pH of any solution within this range. So-called pH paper is impregnated with one or several indicators. When a strip of this paper is immersed in a solution, its pH can be judged from the resulting color.

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