Data See Graphs Calculations The calculations consisted of the deviations to create the pH curve graph. The Ka was calculated by taking the –log of the amount of acetic acid added and squaring it then dividing it by the amount of solution added. The moles for the NaOH was calculated by taking .1 moles and dividing it by 1,000 mL and then multiplying it by the amount of NaOH added. For the Molarity, the mole-mole ratio is used to convert it. Conclusion The pH curve came out zig zagged.
This is a neutralization reaction between a strong acid and strong base. Therefore the heat of reaction (∆H2) is called as the heat of neutralization of HCl and NaOH solutions. The ∆H2 calculated from this experiment is -6.6944KJ/mol. This is because the enthalpy changes when one mole of H+ ions from an acid (HCl) reacts with one mole of OH- from an alkali (NaOH) to form one mole of water molecules under the stated conditions of the experiment. In the final reaction of the experiment (Part C), solid NaOH will react with an aqueous solution of HCl.
From these equilibrium concentrations, the equilibrium constant for the reaction can be determined. The purpose of this experiment is to determine the equilibrium constant for the following hydrolysis of an ester reaction: CH3COOCH2CH3 (aq) + H2O (l) CH3CH2OH (aq) + CH3COOH (aq) Ethyl Acetate Water Ethanol Acetic Acid (EtAc) (EtOH) (HAc) The equilibrium constant, Kc, for the reaction will have the following expression: Several reaction mixtures will be prepared with different initial amounts of ethyl
At 50C our results indicated a solubility of 89 g/100mL of H2O which was close to the known solubility of 80 g/100mL. Introduction: When a salt, such as potassium nitrate or sodium chloride, is placed in water a dissolving reaction will occur. At first, the positive and negative ions of the salt compound are only attracted to each other. In order for the salt to dissolve, these bonds must be broken so that the ions disassociate from each other. In the water molecules, hydrogen is slightly positive and oxygen slightly negative so they are attracted to ions of the opposite charge, known as dipole attraction.
Also the concentration of the unknown can be determined by observing the slope of the graph of Beer’s Law because the graph is linear. The equation for Beer’s Law is A=E * l * c where A is the absorbance, E is the extinction which is the constant based on what you are working with, l is the length in centimeters, and c is the concentration of the solution (this will change) the units for this is (mol/L). Data/ Results: Table 1: Concentrations of cobalt chloride hexahydrate, with there absorbance found using a calorimeter. With this information we then calculated the concentration into moles. Trial | Concentration (mg/mL) | Concentration, calc.
The acids are determined on how much H they have and bases are determined on how much they mess with the hydrogen concentration. When looking at the acid and base used in this lab HCl is an acid and NaOH is a base. To determine a pH value the molarity of the acids and bases must be below one, so using .1 M NaOH and HCl is ideal. To measure the pH the mathematical equation is used: pH = -log [H+] and pOH = -log [OH-]. When adding these up the end result has to be 14 because acids go from 1-7 and basic from 14-7.
Introduction The purpose of this lab is to determine the normality of an unknown base using a volumetric titration. The volumetric titration used for this experiment consisted of a standard acid called potassium hydrogen phthalate (KHC8H4O4) titrated with a weak unknown base. An acid-base indicator called phenolphthalein is used in order to view the reaction proceeding to completion. The indicator allows visualization of the acid changing colour when the solution has reached the end point. The normality of the unknown base is calculated after the solution has reached the end point.
The changes in the pH can be followed using a pH sensor. Acids and bases neutralize, or reverse, the action of one another. By adding a known amount of acid to a basic solution, until it completely reacts with it, the amount of the base can be determined. During neutralization, acids and bases react with each other to produce ionic substances, called salts. In this titration the concentration of the NaOH was determined by titrating it with a solution acidic acid.
Unknown acid buret | | | | Trial #1 | Trial #2 | Initial buret reading | 20.5 mL | 21.8 mL | Final buret reading | 40.5 mL | 45.1 mL | Volume of acid added | 20.0 mL | 23.3 mL | This second table shows the calculation of the pH at half equivalence point, average pKa and the average of the unknown acid. The volume of equivalence point was found by picking the steepest portion of the curve or point of inflection from the graph; and tracing the point to the volume NaOH added. One-half the equivalence point was determined by dividing by ½ of the volume of equivalence point. Once the one-half of the equivalence point was found, trace the point until you hit the graph line and trace it to the pH-axis. Results | | | | Trial #1 | Trial #2 | Volume of NaOH at equivalence point | 9.5 mL | 13.7 mL | Volume of NaOH at one-half the equivalence point | 4.75 mL | 6.9 mL | pH at half equivalence point | 5.42 | 5.3 | pKa of unknown acid | 5.42 | 5.3 | Average pKa | 5.36 | 5.36 | Average Ka | 4.4 * 10-6 | 4.4 * 10-6 | Moles of unknown acid | 1.30 * 10-3 moles | 1.87 * 10-3 moles | M of unknown acid | 0.0650 M | 0.0805 M | Average M of unknown acid | 0.07275 M | 0.07275 M | 2.
Introduction High performance liquid chromatography (HPLC) is used to separate compounds in a sample, identify compounds and can even be used to deduce the relative amounts of different compounds in a mixture. HPLC works under the same principle as thin layer chromatography using both a stationary and mobile phase. The mobile phase carries the mixture across the stationary phase which is used to separate the compounds. Although in HPLC mobile phase is tailor made to suit the polarity of the analytes. The mobile phase used in this particular experiment was Methanol and 0.1M sodium dihydrogen phosphate at a ratio of 30:70 and a pH of 4.5, slightly acidic.