Exp 3 Repaired)

Universiti Tunku Abdul Rahman (Kampar Campus) Faculty of Science, Engineering, and Technology Bachelor of Science (Hons) Biotechnology Year 1 Semester 2 Laboratory 1B (UESB 1212) (II) The Properties of Matter Lecturer: Ms. Chew Yin Hoon Student’s Name: Cheah Hong Leong Student’s ID: 08AIB03788 Partner’s Name: Chong Shi Fern Partner’s ID: 08AIB02580 Experiment No. 3 Title: Hydrolysis of Salts and Choice of Indicators or Titration Curve Date: 20 February 2009

Title: Hydrolysis of Salts and Choice of Indicators or Titration Curve

Objectives: –

Determine the pH of ammonium chloride, sodium acetate, sodium formate, and sodium carbonate.

–

Derive other information from the pH values obtained.

–

Carry out titrations between sodium hydroxide with hydrochloric acid and ethanoic acid to determine the suitable indicator for each type of titration.

Data: Table 1: The pH Values for Different Salts Solution Salts Solution: Ammonium chloride Sodium acetate Sodium formate Sodium carbonate

Concentration, M (mol/dm3) 1.0 1.0 1.0 0.5

*pH Value 5.27 8.95 8.26 11.48

*Obtained through electronic pH meter.

Table 2: Titration of Sodium Hydroxide with Hydrochloric Acid and Ethanoic Acid with Different Indicators

*Final reading *Initial reading **Volume of NaOH, V

Bromophenol Blue HCl HOAc

Phenolphthalein HCl HOAc

15.2 24.2 9.0

5.6 15.2 9.6

(cm3) *Burette reading **Volume = Final reading – Initial reading HCl: Hydrochloric acid solution, 0.1 mol/dm3 HOAc: Ethanoic acid solution, 0.1 mol/dm3 NaOH: Sodium hydroxide solution, 0.1 mol/dm3 Observations:

24.2 27.2 3.0

27.2 38.7 11.5

For both the titration of sodium hydroxide with hydrochloric acid and ethanoic acid, when Bromophenol blue was used as indicator, the solution turned from yellow colour to blue colour when end point was reached; when phenolphthalein was used as indicator, the solution turned from colourless to pink colour when end point was reached.

Analysis and Calculation: Part (A): Ammonium chloride After measuring the pH of ammonium chloride solution, 1 mol/dm3 by electronic pH meter, the pH value obtained was 5.27. Equation for hydrolysis of ammonium chloride: NH4+(aq) + H2O(l)

NH3(aq) + H3O+(aq)

Therefore, Ka(NH4+) = [NH3][H3O+]/[NH4+] pH = 5.27 -lg[H3O+] = 5.27 [H3O+] = 5.37 x 10-6 mol/dm3 NH4+(aq) + H2O(l) [Initial](mol/dm3) 1.0 [Equilibrium](mol/dm3) 1 - 5.37 x 10-6

[NH3] = 5.37 x 10-6 mol/dm3 [NH4+] = (1.0 – 5.37 x 10-6 ) mol/dm3 ≈ 1.0 mol/dm3 [Cl-] = 1.0 mol/dm3

NH3(aq) 0 5.37 x 10-6

+

H3O+(aq) 0 5.37 x 10-6

pOH = 14 – pH = 14 – 5.27 = 8.73 -lg[OH-] = 8.73 [OH-] = 1.86 x 10-9 mol/dm3 Ka(NH4+) = [NH3][H3O+]/[NH4+] = 2.88 x 10-11 mol/dm3 Ka = cα2, c = concentration of NH4+, α = dissociation degree α = 5.37 x 10-6

Part (B): Sodium Salts Sodium acetate saltpH = 8.95 -lg[H3O+] = 8.95 [H3O+] = 1.12 x 10-9 mol/dm3 Kw = [H3O+][OH-] = 1.00 x 10-14 Therefore, [OH-] = 1.00 x 10-14/1.12 x 10-9 mol/dm3 = 8.93 x 10-6 mol/dm3 Therefore, CH3COO- + H2O [Initial](mol/dm3) 1.0 [Equilibrium](mol/dm3) 1 - 8.93 x 10-6

CH3COOH 0 8.93 x 10-6

Degree of hydrolysis of acetate ion = [CH3COO-]hydrolysis/[ CH3COO-]initial = 8.93 x 10-6/1.0 = 8.93 x 10-6

OH0 8.9310-6

Sodium formate saltEquation for the reaction of formate ion with water: HCOO-(aq) + H2O(l)

HCOOH(aq) + OH-(aq)

pH = 8.26 -lg[H3O+] = 8.26 [H3O+] = 5.50 x 10-9 mol/dm3 Kw = [H3O+][OH-] = 1.00 x 10-14 Therefore, [OH-] = 1.00 x 10-14/ 5.50 x 10-9 mol/dm3 = 1.82 x 10-6 mol/dm3

Sodium carbonate saltpH = 11.48 Considering the second protonation was small in sodium carbonate solution, CO32-(aq) + H2O(l)

HCO3-(aq) + OH-(aq)

-lg[H3O+] = 11.48 [H3O+] = 3.31 x 10-12 mol/dm3 Kw = [H3O+][OH-] = 1.00 x 10-14 [OH-] = 1.00 x 10-14/ 3.31 x 10-12 mol/dm3 = 3.02 x 10-3 mol/dm3 Therefore, CO32- + H2O [Initial](mol/dm ) 0.5 3 [Equilibrium](mol/dm ) 0.5 - 3.02 x 10-3 3

[HCO3-] = 3.02 x 10-3 mol/dm3 [CO32-] = (0.5 – 3.02 x 10-3) mol/dm3 = 0.497 mol/dm3

Part (C): Acid-base Titration with Different Indicator

HCO30 3.02 x 10-3

OH0 3.02 x 10-3

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 = 9.0 cm3 1 mol of NaOH reacted with 1 mol of HCl Therefore, concentration of HCl, M1 = [(M2V2)]/V1 = (0.1 x 9.0)/10.0 = 0.090 mol/dm3 While phenolphthalein was used, Volume of NaOH, V2 = 9.6 cm3 Therefore, concentration of HCl, M1 = [(M2V2)]/V1 = (0.1 x 9.6)/10.0 = 0.096 mol/dm3 Equation for titration of sodium hydroxide and ethanoic acid: NaOH(aq) + HCOOH(aq)

HCOONa(aq) + H2O(l)

Volume of HCOOH, V1 = 10.0cm3 Concentration of NaOH, M2 = 0.1 mol/dm3 Volume of NaOH, V2 = 11.5 cm3 1 mol of NaOH reacted with 1 mol of HCOOH Therefore, concentration of HCl, M1 = [(M2V2)]/V1 = (0.1 x 11.5)/10.0 = 0.115 mol/dm3

Discussion: Part (A) Ammonium chloride 1. Do you expect an ammonium chloride solution to be neutral? No. 2. On what previously obtained experimental evidence do you do you base your above answer? Ammonium chloride solution was not expected to be neutral, but acidic. Ammonium chloride is the salt resulted from the reaction between strong acid and weak base. The dissociation of ammonium chloride in water will produce hydroxonium ions (H3O+) in excess, leading pH < 7. 3. Measure the pH of a 1 mol/dm3 solution of ammonium chloride, pH = 5.27.

4. Is the solution more acidic or basic than pure water? More acidic than pure water. 5. Write an equation for the reaction causing this. NH4+(aq) + H2O(l)

NH3(aq) + H3O+(aq)

6. Bronsted defined an acid as a proton donor and a base as a proton acceptor. What

is water in this reaction. Water in this reaction received protons from ammonium ions; therefore water was base in the reaction. 7. Write an expression for the hydrolysis constant Ka(NH4+).

Ka(NH4+) = [NH3][H3O+]/[NH4+] 8. What are the concentration of the following ions in the solution? a) [H3O+] = 5.37 x 10-6 mol/dm3 b) [NH4+] = (1.0 – 5.37 x 10-6 ) mol/dm3

≈ 1.0 mol/dm3 c) [Cl-] = 1.0 mol/dm3 d) [OH-] = 1.86 x 10-9 mol/dm3 e) [NH3] = 1.86 x 10-9 mol/dm3 9. Now calculate Ka(NH4+).

Ka(NH4+) = [NH3][H3O+]/[NH4+] = 2.88 x 10-11 mol/dm3 10. What is the degree of hydrolysis of NH4+?

5.37 x 10-6 11. Realizing that you have a solution of ammonium chloride at the end point of the titration of the strong acid hydrochloric acid and weak base ammonia, what indicator would you use to detect this end point in such a titration of approximately molar solution? Ammonium chloride is the salt produced at the end point of titration between a strong acid hydrochloric acid and weak base ammonia. At the end point, pH < 7. Any indicator that change colour between pH 3 – pH 7 can be used to determine the end point of the titration, for example methyl red that has pH range of 3.1 – 4.4. Part (B) sodium salts 1. Measure the pH of 1M sodium acetate, 1M sodium formate, and 0.5M sodium carbonate solution and complete the following table. pH *[H3O+] *[OH-] *[Na+]

CH3COONa 8.95 1.12 x 10-9 8.91 x 10-6 1.00

HCOONa 8.26 5.50 x 10-9 1.82 x 10-6 1.00

Na2CO3 11.48 3.31 x 10-12 3.02 x 10-3 1.00

*Concentration of ions in mol/dm3 CH3COONa: Sodium acetate HCOONa: Sodium formate Na2CO3: Sodium carbonate 2. What are the concentration of the following ions in NaAc solution? a) [CH3COOH] = 8.91 x 10-6 mol/dm3 b) [CH3COO-] = (1 – 8.91 x 10-6 ) mol/dm3

3. What is the percentage of hydrolysis of acetate ion? Percentage of hydrolysis = degree of hydrolysis x 100%

= 8.93 x 10-6 x 100% = 8.93 x 10-4% 4. What indicator would you use for the titration of acetic acid with strong base?

Why? Phenolphthalein. Any indicator that change colour within the range of pH 7 – pH 11 can be used to determine the end point of the titration because at end point, pH > 7. Phenolphthalein has the pH range of 8 – 10, therefore suitable. 5. Write the equation for the reaction of formate ion with water. HCOO-(aq) + H2O(l)

HCOOH(aq) + OH-(aq)

6. Consider the second protonation is very small, find the following in Na2CO3

solution. a) [H3O+] = 3.31 x 10-12 mol/dm3 b) [HCO3-] = 3.02 x 10-3 mol/dm3 c) [OH-] = 3.02 x 10-3 mol/dm3 d) [CO32-] = 0.497 mol/dm3

7. Comment on the relative basic strength of acetate, formate, and carbonate ions. CO32- > CH3COO- > HCOO8. Does this agree with the fact that formic acid is a stronger acid than acetic acid and acetic acid is stronger acid than bicarbonate ion? Yes. 9. Consequently, what is the relationship between the strength of an acid and the strength of its conjugate base? The stronger the acid is, the weaker the conjugate base will be.

Part (C) Acid-base Titration with Different Indicator For the titration of sodium hydroxide and hydrochloric acid with different indicators, the results obtained were quite similar with minor difference. This was a strong acid and strong base titration, before and after the end point was reached, there was a sharp

increase in pH from pH 3 – pH 11. Any indicator that change colour within the range can be used as the indicator of the titration. Phenolphthalein has pH range of 8 – 10 while Bromophenol blue has pH range of 3.0 – 4.6. Therefore both of the indicators were suitable for the titration. For the titration of sodium hydroxide and ethanoic acid with different indicator, the results obtained were totally different. For weak acid and strong base titration, the pH range before and after the end point was 7 – 11. Any indicators that change colour within this range can be used as indicator for this titration. Only phenolphthalein was suitable in the titration.

Conclusion: 1. Ammonium chloride solution is an acidic solution. 2. Formic acid is a stronger acid than acetic acid and acetic acid is stronger acid than bicarbonate ion. 3. Stronger acid will have weaker conjugate base. 4. For strong acid and strong base titration, both Bromophenol blue and phenolphthalein are suitable as indicators. 5. For weak acid and strong base titration, only phenolphthalein is suitable as

indicator in this experiment. References: Lim Y. S., Yip K. H. (2006). Pre-U Text STPM Chemistry, Longman, Pearson Malaysia Sdn.Bhd.