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LABORATORY REPORTCHM 213(PHYSICAL CHEMISTRY)1. MUHAMMAD MIRZA HIZAMI BIN RAJIEI 2019289394GROUP : 1PROGRAMME : AS115 2ANO. AND TITLE OF EXPERIMENT :5 Acid-Base Equilibrium.DATE OF EXPERIMENT : 15 / 4 / 2020DATE OF REPORT SUBMISSION : 30 / 4 / 2020NAME OF LECTURER : DR. NUR ROYHAILA MOHAMADLaboratory...

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LABORATORY REPORT CHM 213 (PHYSICAL CHEMISTRY) 1.

MUHAMMAD MIRZA HIZAMI BIN RAJIEI

2019289394

GROUP

:1

PROGRAMME

: AS115 2A

NO. AND TITLE OF EXPERIMENT

:5 Acid-Base Equilibrium.

DATE OF EXPERIMENT

: 15 / 4 / 2020

DATE OF REPORT SUBMISSION

: 30 / 4 / 2020

NAME OF LECTURER

: DR. NUR ROYHAILA MOHAMAD

Laboratory Report Marking Scheme: Criteria Objective of the Experiment Introduction Experimental Procedure Results and Observations Calculations Discussion Conclusion Answers to Questions References Format Total Marks

Full Mark(s)

Marks (to be filled by lecturer)

OBJECTIVES: To measure the pH of solutions of different of acidity and basicity, to determine the ionisation constant of a weak acid, to study the effect of dilution on the degree of ionisation and to study the properties of buffer solutions.

INTRODUCTION: In discussing equilibrium chemical systems, we must consider as one of the most important of these topics reactions involving acids and bases. Most acid-base systems are equilibrium ones. Let's look at this field of chemistry, starting with some descriptive chemistry. Acids and bases are relative terms. A molecule cannot be called an acid in exclusion of all else. For example, the gaseous hydrogen chloride molecule HCl(g) is a very stable molecule and does not act as acid or base unless something else is introduced to the system. If water is introduced to the system, the HCl readily reacts with the water to form H3O+(aq) and Cl(aq). In other words, in the process of dissolving into the water, the HCl molecule gives up a proton to become Cl-(aq) and a water molecule accepts the proton and becomes H3O+(aq). According to the Bronsted-Lowrey definition, an acid is a molecule or ion that can easily give up (donate) a proton in a reaction with a base. Similarly, a base is a molecule or ion that can bond with (accept) a proton released by an acid in a reaction. Thus, in the reaction between hydrogen chloride (HCl) and water, the HCl is the acid and the water is a base. In the reaction between ammonia (NH3) and water, the water donates a proton to the ammonia molecule to make NH4+(aq) and OH-(aq). Thus, in this reaction, the ammonia is a base and the water is an acid. We have seen water as a Bronsted-Lowrey acid and as a Bronsted-Lowry base.

Arrhenius Bronsted-Lowry Lewis

Acid Substance that produce H3Oin water. Proton donor Substance that can accept a pair of electrons.

Base Substance that produce OHin water Proton acceptor Substance that can donate a pair of electrons.

Recall that we write concentration here but actually an equilibrium constant is properly written using relative activities. Relative activities are written as: For the pH scale in water solution we can easily find H3O+ and OH- concentrations in the range between 1 and 10-15. For reasons of convenience and other scientific reasons, concentrations are often expressed on a logarithmic scale. For this, the 'p' function is used. It was first proposed by Sørenson. (S1909). You will see this function used in numerous ways in Science. Don't get confused into thinking that each time you see it it is something new. In general: p X = -log X .Whatever X is. In this case, we use pH and pOH to measure concentrations. Hence, pH = -log [H+] and pOH = -log [OH-.We already know that in aqueous solution KW = [H+][OH-] = 1.0×10-14. Because of a law in mathematics that says :log (x×y) = log x + log y.We can write pKw = p([H+][OH-]) = pH + pOH = 14.We can also convert pH and pOH values back to [H+] and [OH-] values by using the inverse of the p function [H+] = antilog(-pH) or [H+] = 10-Ph .For buffer solutions in many chemical processes, both industrial and biological reactions can only take place in a narrow range of

[H3O+] concentrations (narrow pH range). To ensure that the desired reactions take place, buffers are used which hold the pH constant (within limits). Examples of such chemical processes are: • Sewage treatment: Bacteria only survive in a narrow pH range • Photographic developing • Electroplating • Most metabolic processes • Stomach pH ~ 1.5 • Saliva pH ~ 6.8 • Blood pH ~ 7.4 Buffers are used to maintain pH at some desired range. A Buffer is made up of approximately equimolar amounts of a weak acid and its conjugate base. To make these solutions, we could, for example add a weak acid (like Acetic acid) and the salt of the acid (like Sodium acetate) in equal proportions. The solution made up of CH3COOH and CH3COO- is considered in approximately equal amounts. CH3COO- + H3O+ CH3COOH + H2O According to La Châtelier's principal If we add a small amount of acid, the acetate ion will react with it and reduce the effective change in pH. If we add a small amount of base, the acetic acid will react with it and again reduce the effective change in pH. The amount of acid and base added must be smaller than the amount of buffer or it will simply use up all the buffer and then proceed to change the pH. To restate what we said in more mathematical terms: Here, we see that the equilibrium constant for the reaction of the acetate ion with acid is very large. Hence, any acid added to the buffer solution will be immediately used up in the buffer equilibrium. Similarly, the Kfor the reaction of the acetic acid with a base is large and the base will be used up. A buffer is most effective at resisting change in pH when [base] = [acid] pH = pKa + log(1) = pKa.

PROCEDURES: a. Measuring pH of solutions The pH values for each of the solutions listed in the Results and Discussion section predicted before any measurements are made. The pH values of the solution measured. The results filled in in the table provided b. Effect of dilution on a The solution prepared as required. The pH values predicted and the pH measured. The results filled in the table provided. c. Buffers The solution prepared as required. Solution nos.13 and 14 predicted either as buffers or not. The pH of all the solutions measured and the results filled in the table provided.

RESULTS: Chemical Solutions a) Hydrogen Ion Concentration: pH 1. Distilled water 2. tap water 3. vinegar 4. Milk 5. 0.1 M NH4OH 6. 0.1 M NH4NO3 7. 0.1 M NaCl 8. 0.1 M CH3COONa 9. 0.1 M CH3COONH4 b) Effect of Dilution 10. 0.1 M CH3COOH 11. 5 ml 0.1 M CH3COOH + 5 ml H2O 12. 1 ml 0.1 M CH3COOH + 99 ml H2O c) Buffers 13. 5 ml 0.1 M CH3COOH + 5 ml 0.1 M HCl 14. 50 ml 0.1 M NH4OH + 50 ml 0.1 M NH4NO3 15. 10 ml of (14) + 6 ml of H2O 16. 10 ml of (14) + 5 ml of H2O + 1ml 0.1 M HCl 17. 10 ml of (14) + 6ml 0.1 M HCl 18. 10 ml of (14 ) + 5 ml of H2O + 1 ml 0.1 M NaOH

Predicted pH

Measured pH

7 6.6 2 6.8 11 10 7 8.6 3

7 6.5 2.4 6.7 11.6 10.2 7 8.9 2.9

2.7 5.8 6.99

2.8 Dilution 6.99

2 10.1 9.8 9 8.3 10.3

Buffer 10.2 9.9 8.6 8 10.8

QUESTIONS: 1. Classify solutions nos 1-14 in the experiment into any of these group. Justify your answers. a. Strong acid - solutions no 3,9 b. Strong base - solutions no 5,6,14,18 c. Weak acid - solutions no 2,4,10,12 d. Weak base - solutions no 8,15,16,17 e. Buffer solutions - solutions no 13

2. Calculate the dissociation constants ( Ka ) and degree of ionization of CH 3COOH in solution nos 10-12 based on the pH values obtained. Compare Ka and degree of ionization with theoretical value. Solution nos 10: CH3COOH → H+ + CH3COOx= 10-2.8 x = 1.5849 × 10-3 Ka10 = ( 1.5849 × 10-3 )2 0.1 - 1.5849 × 10-3

Ionization =

𝑥 𝑀

× 100%

= 1.58%

= 2.5523 × 10-5 Solution nos 11 and 12 CH3COOH + H2O H3O+ + CH3COOx = 10-5.8 = 1.5849 × 10-6

Ionization =

𝑥 𝑀

× 100%

= 1.58 × 10-3 % -6

2

Ka11 = ( 1.5849 × 10 ) 0.1 × 1.5849 × 10-6 = 2.5119 × 10-11

x = 10-6.99 = 1.0233 × 10-7

Ionization =

𝑥 𝑀

× 100%

= 1.02 × 10-4 % -7

2

Ka12 = ( 1.0233 × 10 ) 0.1 × 1.0233 × 10-7 = 1.0471 × 10-13 3. Is the degree of ionisation affected by dilution? Explain Yes, On the increasing of dilution, the degree of dissociation increases. On dilution of the acid, it dissociates into ions because of the high dielectric constant of water. On dilution ions are readily available hence degree of dissociation increase.

4. What are the common ions that are found in solution no.14 ? NH4+

5. How does the pH change in solution no. 15 – 18 compared to solution no. 14 ? Explain your findings Solution no.15 – 17 have lower pH compared to solution no. 14 due to H2O and HCl added into the solution.Meanwhile solution no.18 have a higher pH compared to solution no. 14 due to NaOH is a strong base added into the solution.

DISCUSSION: The solution must be accurately measured otherwise there will be slight change in the results.

CONCLUSION: The purpose and learning objectives of this experiment were to measure pHs and understand how the different solutions created/mixed resulted in different pH, determine the ionisation of the weak acid, study the effect of dilution on the degree of ionisation and to examine how the buffer solutions functioned to prevent the pH from changing drastically....