==> 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 .
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.
n (3) Solutions of aqueous sodium hydroxide and hydrochloric acid react to form water and aqueous sodium chloride. co NaOH(s) → Na+(aq) + OH–(aq) ∆H1 = ? Chemistry with Vernier py In this experiment, you will use a Styrofoam-cup calorimeter to measure the heat released by three reactions. One of the reactions is the same as the combination of the other two reactions. Therefore, according to Hess’s law, the heat of reaction of the one reaction should be equal to the sum of the heats of reaction for the other two.
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
Introduction In Chemistry 1211 lab the main objective was to identify the name of the unknown acid #2651145-PLF13 that was issued by the instructor. Melting point, titration of the unknown acid, calculation of equivalent weight, pKa, solubility tests, and properties of a hydrochloric acid; these procedures were used to characterize the unknown acid and then compared with known substances using GSU Chemistry department search engine. By doing this experiment he or she will develop an understanding of the properties of organic acids and differences between properties of a strong acid such hydrochloric acid. Experimental The first procedure to identify the unknown organic acid was finding the correct melting point. This was done by taking small amounts of the unknown acid.
Halides Lab: Background information: Halide ions are reactive and useful. Salts are positively charged ions (metals) combined with any negative ions (nonmetal), and when placed in a solution (water) it separates into the cations and anions that made it up. The Purpose of this lab is to find out how the Halides react with the indicators, and to determine the identity of the two unknown solutions (A and B). Color of solutions prior to experiment: NaF | NaCI | KBr | KI | Unknown A | Unknown B | clear | clear | clear | clear | clear | clear | Color of indicator prior to experiment: 5% Bleach (NaOCI) | 0.2 M Na2S2O3 | 0.1 M AgNO3 | 0.5 M Ca(NO3)2 | clear | clear | clear | clear | Halide solutions | NaF | NaCI | KBr | KI | unknown A | unknown B | Test 1: Ca(NO3)2 | Cloudy White (Nothing) | Clear | Nothing | light yellow (Nothing) | Nothing | Nothing | Test 2, Part A: AgNO3 | clear (Nothing) | Milky White | Gold (Cloudy yellow) | milky green (Cloudy yellow) | turned white, film developed on top layer | Milky | Test 2, Part B: add Na2S2O3 to test tube from part A | Dark Orange/brown | Clear | Dark Gold(precipitate yellow then clear) | milky green (no change) | white precipitation, settled on bottom | Milky | Test 3: NaOCI (Bleach) | Clear (Nothing) | Nothing | Nothing | Orange (Clear) | Nothing | Orange | Unknown A is identified as NaCI (Sodium Chloride), because in test#1 the solution turned a cloudy white color when Ca(NO3)2 (Calcium nitrate) was added. In the first part of test#2, when AgNO3 (Silver nitrate) is added, the solution turned white, with a thin layer of film developing on the surface.
The chemical reaction used to find this constant is as follows: MgC2O4 (s) ↔Mg(aq)2++ C2O4 (aq)2- Kc= Mg2+[C2O42-][MgC2O4] Ksp=Mg2+[C2O42-] The solid salt magnesium oxalate is prepared through the following precipitation reaction: Mg(SO4)(aq)+NaC2O4 (aq) → MgC2O4 (s)+NaSO4 (aq) Next, the concentration of the Mg2+ and C2O42- ions is found through a redox titration. This redox titration uses a standardized potassium permanganate solution. The potassium permanganate solution is standardized by titrating it with samples of iron(II)ammonium sulfate hexahydrate . The end point is reached when the solution has turned light purple which is a result of excess amounts of MNO4-. This reaction can be summed up using the following formula: 5Fe2++8H++MnO4- →5Fe3++Mn2++4H2O After standardization, the potassium permanganate solution is then titrated with 3 different magnesium oxalate solutions.
The purpose of the lab was to determine which reactant was the limiting reactant, and to see how much of the other reactant was used. The true molarity of a compound can be defined as the amount of moles per liter of that substance. The equation of this single displacement chemical reaction done during this lab is 2Al(s) + 3CuCl(aq) → 3Cu (s) + 2AlCl2 (aq). In the reaction, the solid Aluminum replaces the Copper in Copper (II) Chloride to produce solid copper, and Aluminum Chloride. In order to find which reactant is the limiting reactant, an equation based on the molarity of the Copper (II) Chloride may be used, or the products of the reaction may be observed.
3. To determine which copper chloride was being used (Copper (1) or Copper (II) we could look at the ratio of copper chloride to sodium hydroxide and if it is the same or closest, that is which chloride