Titration

INTRODUCTION

Differences between acids and bases
An acid-base reaction is based on the reaction involving the ionization of water H2O -> H+ + OH-
This means that water can break apart into a hydrogen ion and a hydroxide ion. These two ions can also join together to form a water molecule. When a strong acid is placed in water, it will ionize completely, and break down into its constituent ions in which one of it a hydrogen ion. When a strong base is placed in water, it will ionize completely and dissociate into its constituent ions in which one of them is a hydroxide ion. Based on their reactions in water, we can effectively define what an acid and a base is and distinguish between the two.
The first modern definition of an acid and a base was proposed by Arrhenius. According to his theory, Arrhenius stated that an acid is a substance that dissociates in water to form hydrogen ions (H+) and increases the concentration of H+ ions in an aqueous solution and that a base is a substance that dissociates in water to form hydroxide ions (OH-) and increases the concentration of OH- ions in an aqueous solution. His definitions were however limited to aqueous solutions.
A major problem with the Arrhenius acid-base concept is that certain substances such as ammonia, (NH3), produce basic solutions and react with acids, yet contain no hydroxide. In 1923, J. N. Bronsted and T. M. Lowry dependently proposed a new way of defining acids and bases in aqueous solutions. According to the Bronsted-Lowry concept of acids and bases, acids are hydrogen ion donors and bases are hydrogen ion acceptors. An acid can donate an H+ ion (proton) to another substance while a base can accept an H+ ion from another substance. The term proton means the species H+ (the nucleus of the hydrogen atom) rather than the actual hydrogen ions that occur in various solutions. The definition thus shows that reaction is independent of the solvent. The use of the word species rather than substance or molecule implies that the terms acid and base are not restricted to uncharged molecules but apply also to positively or negatively charged ions. This extension is one of the important features of the Bronsted-Lowry definition.
Based on these definitions, the properties of an acid can be distinguished from that of a base. An acid react with most metals to form hydrogen gas, have a sour taste, frequently feel sticky and are usually gases or liquids. Vinegar and grapefruit juice are too harmless acids. Hydrochloric acid is much more dangerous and may cause severe burns. A base on the other hand feels slippery because the skin dissolves a little when it comes in contact with them, have a bitter taste, react with grease and oil, and frequently solids.

Discussion of indicators
An indicator is a chemical compound that is added in small amounts to an aqueous solution so that the pH of that solution can be determined visually. An acid-base indicator is either a weak acid or a weak base. An indicator does not change color from pure acid to pure alkaline at specific hydrogen ion concentration, but rather, the color change occurs over a range of hydrogen ion concentrations. This range is termed the color change interval. It is expressed as a pH range. Some examples of indicators include thymol blue, methyl yellow, methyl orange, bromphenol blue, bromcresol green, methyl red, bromthymol blue, phenol red, neutral red, phenolphthalein, thymolphthalein, alizarin yellow, nitramine, and trinitrobenzoic acid. In this experiment phenolphthalein is the indicator used in the titration process.

Discussion of terms
Titration is the process of determining the quantity of a substance A which is the acid by adding measured increments of substance B which is the base, with which it reacts (almost always as a standardized solution called the titrant) with provision for some means of recognizing (indicating) the end point at which essentially all of the acid has reacted using an indicator. If the end point coincides with the addition of the exact chemical equivalence, it is called the equivalence point or stoichiometric or theoretical end point, thus allowing the amount of the acid to be found from known amount of the base added up to this point, the reacting weight ratio of the acid to the base being known from stoichiometry or otherwise. The titrant solution, containing the active agent with which a titration is made, is standardized, that is the concentration of the active agent is determined, usually by titration with a standard solution of accurately known concentration. Standard solutions are prepared using standard substances in one of several ways. A primary standard is a substance of known high purity which may be dissolved in a known volume of solvent to give a primary standard solution. If stoichiometry is used to establish the strength of a titrant, it is called a secondary standard solution. Titration error is the difference in the amount of titrant, or the corresponding difference in the amount of substance being titrated, represented by the expression:
(End-point value) - (Equivalent-point value)
Titration curve is a plot of the volume of the base used against the pH of each respective volume recorded. In this experiment, the pH of the solution is plotted against the respective volume of the NaOH solution at each pH value.

Discussion of calculations
Using the constructed titration curve from pH (y-axis) and volume of NaOH (x-axis) data, the moles of NaOH added the molar mass of CH3COOH acid, the percentage of the CH3COOH acid in the vinegar and the mass of vinegar used will be calculated. Also the pKa and Ka values of the acid will be determined from the data.

EXPERIMENTAL DATA AND CALCULATIONS
Recorded Data
The starting volume which was 0.30 ml was subtracted from the volumes recorded to get the actual volume delivered by the burette.

Volume of NaOH (ml) | pH | 0 | 2.75 | 2.95 | 3.30 | 6.00 | 3.68 | 8.95 | 3.93 | 11.90 | 4.12 | 14.95 | 4.29 | 17.90 | 4.45 | 20.95 | 4.60 | 24.10 | 4.77 | 26.90 | 4.94 | 29.95 | 5.16 | 33.00 | 5.49 | 36.05 | 6.2 | 36.60 | 6.54 | 37.00 | 6.95 | 37.20 | 7.23 | 37.30 | 8.39 | 37.70 | 10.34 | 37.90 | 10.62 | 38.00 | 10.69 | 38.20 | 10.93 |

Calculation of Moles of NaOH/CH3COOH
The chemical reaction between CH3COOH and NaOH is shown below: NaOH + CH3COOH -> NaCH3COO + H2O
To calculate the moles of NaOH added which is equivalent to the moles of CH3COOH in the vinegar: moles CH3COOH = moles NaOH = (Volume in litres)(Molarity of NaOH solution) volume of NaOH = 37.5 volume in litres = 37.5/1000 = 0.0375 molarity of NaOH solution = 0.09476 M moles of NaOH = 0.0375 × 0.09476 = 0.00355 moles moles of CH3COOH = moles of NaOH = 0.00355 moles

Calculation of mass of CH3COOH
Mass of CH3COOH = no. of moles × molar mass
No. of moles = 0.00355 moles
Molar mass of CH3COOH = 12 + (1×3) + 12 +16 +16 +1 = 60 g/mol
Mass of CH3COOH = 0.00355 × 60 = 0.213 g

Calculation of percentage CH3COOH percentage (%) of CH3COOH = mass of CH3COOH × 100% mass of vinegar mass of CH3COOH = 0.213 g mass of vinegar = 4.071 g % CH3COOH = 0.213 g × 100% 4.071 g = 5.23%

Determination of equivalence point
The equivalence point is the volume where the solution is completely neutralised and the indicator changes colour to pink from the colourless state it was in. From our experiment the equivalence point was 37.50 ml.

Calculation of pKa
To calculate the pKa of the acid, The half-volume of the equivalence point is taken = 37.5ml/2 = 18.75 ml
Using the titration curve, the pH at this volume = 4.52
Therefore the pKa of the acid = 4.52
Ka = 10-pKa = 10-(4.52) = 3.02 × 10-5

CONCLUSION.
Summary of results
From my data and calculations, there were some titration errors that occurred during the experiment that affected the final results. The percentage of the CH3COOH solution from my calculations is 5.23%. Vinegar contains 5% of acetic acid and the value calculated is slightly higher but approximately 5% but with an error of +0.23. The equivalence point was identified at 37.5 ml and used to calculate the number of moles of CH3COOH as 0.00355 moles. The experimental Ka value of CH3COOH was very different from the theoretical value. In this experiment, Ka = 3.02 × 10-5 but the theoretical value is 1.8 × 10-5. The values are different due to titration errors that occurred in the course of the experiment.

What was learned
In this experiment, I learnt how to interpolate values. I also learnt how to find the number of moles of acids and bases and how to find the mass, molarity and mass concentration. I was able to recognize the differences between acids and bases and how they react to achieve neutralization. I also learnt how to find the acid content in vinegar which would help do so for other substances/solutions. I also learnt how to find the endpoint of a titration. I also learnt how to titrate properly and obtain the amount of acids or base.