How it was expected the pH at equivalence point was 9.17 this is because of the domination of hydroxide ion in solution. The relationship between the pH and the amount of titrant added offered a better understanding of the equilibrium properties of the acid. Introduction Titrations are a convenient and common method of analysis. Generally titration is an experiment where a known property of one solution is used to infer an unknown property of another solution. There are several types of titrations: Acid-base titrations are based on the neutralization reaction between the analyte and an acidic or basic titrant.
In order to find the equivalent weight a titration of the unknown acid had to be conducted. Equivalent weight is the weight of the substance in grams divided by the average molarity of the sodium hydroxide solution times the volume from the titration in liters. If the molarity of your sodium hydroxide solution were too low then the equivalent weight would be off significantly. In order to calculate a decent molarity a short series of titrating was conducted using KHP. Find the molarity is essential to numerous amounts of future procedures such as equivalent weigh.
Objectives: The purpose of this lab is to observe the reaction of crystal violet and sodium hydroxide by looking at the relationship between concentration and time elapsed of the crystal violet. CV+ + OH- CVOH To quantitatively observe this reaction of crystal violet, the rate law is used. The rate law tells us that the rate is equal to a rate constant (k) multiplied by the concentration of crystal violet to the power of its reaction order ([CV+]p) and the concentration of hydroxide to the power of its reaction order ([OH-]q). Rate = k[CV+]p[OH-]q To fully understand the rate law, concentrations of the substances must be looked at first. The concentration is measured in molarity.
==> NaHCO3(aq.) + NaCl(aq.) We will standardize the HCl solution to use it in the titration. The standardization will come as a result of the 1:1 molar ratio above. Thus, the molarity of the HCl solution can be calculated by dividing the number of moles of HCl by the volume of HCl (in liters) used to neutralize the Na2CO3 .
Therefore the alkalinity of water samples is being calculated. In the second approach, the two volume readings for the respective amounts of sulfuric acid used are being determined an indicator based method. Congo red and bromocresol green are being used as the indicators. Procedure (Outline provided as pre-lab): A. The pH meter was calibrated using standard pH solutions provided.
This particular reaction is a strong acid and a strong base which means that when the reaction reaches the equivalence point, the moles of the acid and the base are equal and the solution is neutral so the pH should be around 7.0 depending on the final volume of each solution. To get this data, we will titrate an HCl solution with NaOH solution of which is a known concentration. We will record the initial and final reading of the NaOH while we record the pH of the titrated solution in the beaker. We will repeat this process with a solution of acetic acid which is a weak acid with NaOH and record the initial and final reading of NaOH and the pH of the solution in the beaker. Procedure Preview Calibrate the pH meter.
Ashley Peccatiello Experiment 7 – Dehydration of 2-Methylclyclohexanol, Tests for Unsaturation, and Gas Chromatography Date Performed: October 25, 2012 Date Written: October 29, 2012 Purpose: To dehydrate 2-methylclyclohexanol to obtain two isomers. To separate the products by simple distillation. To analyze the sample by introducing the technique of gas chromatography and unsaturation tests. Reaction: Figure 1. Overall reaction of the acid-catalyzed dehydration of 2-methylcyclohexanol Figure 2.
Write a hypothesis on what you think will happen when mixed. 3. Combine the substances; record observations. Hydrochloric Acid and Magnesium 1. Add hydrochloric acid to a test tube.
There are two parts to this lab. Part 1 which will be a known hydrate CoCl2 ∙6H2O or Cobalt (II) Chloride hexahydrate, and Part 2 which will be an unknown hydrate. Equations and Mechanisms * Moles of hydrate and water ratio: Moles of water Moles
3) Write equations to indicate what you consider to have happened in each case in which there was precipitate formed. Use ions to represent the species in the reacting solutions, but for those products that were precipitates write a formula for the compound. Place (aq) after those species in solution and (s) after the precipitates. Be sure to write the equations so that both atoms and charge are conserved. For example: Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) ( AgCl(s) + Na+(aq) + NO3-(aq) 4) Rewrite the equations, leaving out the ions not involved in the reaction (spectators).