Modbury High School SACE Stage 1 Chemistry Topic 5 Mole Concept and Stoichiometry Assignment 5: Volumetric analysis (titrations), stoichiometry SOLUTIONS Note: Write answers neatly and legibly in your exercise book or on pad paper. ALWAYS include a title and name for your work and clearly indicate each answer. 1. a) 23.08 and 23.00 mL are concordant titre values. Average titre = (23.08 + 23.00) = 23.04 mL 2 b) Ca(OH)2 + 2HCl → CaCl2 + 2H2O V = 23.04 mL C = 0.0934 M = 0.02304 L V = 20.0 mL C = ? = 0.0200 L …show more content…
systematic errors, which affect the accuracy of a set of measurements As only one titre was produced, random errors also affect the accuracy, such as: Technique of the
2. Write a statement to explain the molecular composition of the unknown solution based on the results obtained during testing with the Biuret solution and each sample solution.
2. To titrate a hydrochloric acid solution of “known” concentration with standardized 0.5M sodium hydroxide.
In six different 50mL volumetric flasks there was 0.00 mL, 2.00 mL, 4.00 mL, 6.00 mL, 8.00 mL, and 10.00mL of the 5.00 g/L Cu^(2+)stock solution. 3mL of 15M ammonia was added into each flask, once the ammonia was added each flask was diluted with water to 50 mL then were all thoroughly mixed.
Stoichiometry is a very important part of chemistry. Stoichiometry refers to calculating the masses of molecules and their products . The reactants are usually given and stoichiometry is used to find the products of the equations as well balancing the equation. An example of this would be sodium chloride (NaCl). Stoichiometry will say that if there are ten thousand atoms of sodium and one atom of chlorine, only one molecule of sodium chloride can be made and that fact can never be changed.
Can I apply this somehow to volume? Well at standard temperature and pressure (STP) a mole of a gas will occupy 22.4 liters. So If we keep our units straight we should be able calculate a given volume of gas from moles. Check it out…… http://www.sciencegeek.net/Chemistry/Video/Unit4/GMV4.shtml
In the ADI Molarity Lab, the primary tasks was to use different values of moles of solute, volume of solvent, and molarity to find the mathematical relationships between them. To find these relationships, our group had to change the quantities of each of the variables and visually observe the molarity within the solution. For instance when using Cobalt (II) Nitrate to find the relationship between volume of the solution and the molarity of the solution; the group kept the amount of moles of the solute at a constant of 1.00 moles because if it would have changed it would have caused inaccurate data. We first set the volume of the solution to 0.2 liters. The molarity of the solution was 5.00 mol/L. Then we changed the volume of the solution
Question of the day: What is the stoichiometric ratio of reactions in the chemical synthesis of the (2, 4-pentanedianato) iron (III) complex ion?
c. A lab exercise in BIO156 required 300 ml of water that was poured from a two-liter container. How many milliliters were left in the original container? (4 points)
The problem that was trying to be solved in this study deals with analyzing unknown solutions. In this particular case, a chemical company has several unknown solutions and to correctly dispose of them they need to know their properties. To figure out the properties several qualitative tests were performed throughout the study (Cooper 2012).
After the twenty minutes elapsed, the flask was cooled to room temperature and then titrated with the remaining NaOH until the colorless solution remained pink. The final volume was then recorded. While solution #1 was heating the same process was repeated with solution#2 and the second burette
9. How many moles of NaOH would be needed to completely react with all of the excess HCl determined in problem 8?
3. How could a conductometric titration be used to determine the molarity of either reacting solutions, assuming the concentration of one solution was known? HINT: Consider the variables needed to calculate molarity, and how can these values be obtained from the titration. The symbol M stands for molarity with units of mol/L. Molarity is a measure of the concentration of a solution. If you knew the concentration of one of the solutions, the molarity could be found in this way. The concept plan would go as follows: Volume of Titrant (in liters) x Moles of Titrant (mol) = moles of the unknown, and then take Moles of the unknown / Liters of the unknown to get the Molarity of the unknown in mol/L. 4
ii. The second part of the titration series involves titration of NaOH with Hydrochloric acid (HCL). Again, three reps of titration and a blank titration have to be completed. A volumetric pipet is used to measure 10.00mL of HCL into three labeled conical flasks. Then the flasks are filled with deionized water until about the 50mL mark. A buret is
Mol of Cu = 0.035983 mol Fe * 1 mol Cu1 mol Fe = 0.035983 mol [V]
A mock trial was performed to approximate the 1st and 2nd equivalence point regions. The HCl titrant was added into the soda ash solution in increments of 1.0 mL until the pH was close to ~3.0-2.0 then, the HCl solution was dispensed in increment of 0.1 mL until the pH was exactly 2.0. The volume of the titrant added and the pH was recorded. With the aid of the instructor the 1st and 2nd equivalence was determined based on the pH/mL change. This method was repeated once more; as instructed although the experimental procedure stated to do perform four trials. The calculated amount of soda ash was weighted and the mass was recorded. The soda ash was then transferred into a 250 mL beaker and dissolved in 70 mL of water. 1.0 mL increments of HCl solution was dispensed until 2 mL before equivalence point 1 then HCl was added in 0.3