Dissapering Salts Lab Report 8 PDF

Preview

Summary

Lab 8 assignment...

Description

Disappearing/Repapering Salts

Results For this experiment, 5mL of each of the following solutions were obtained in 50mL beakers: 0.1 M BaCl2, 1.0 M Na2CO3, 0.3 M H2SO4, 1.0 M NaCl, 0.1 M MgSO4 M, and 0.15M NaOH. Each were placed in wells in designated spots per Table 1. Three drops per each solution was added to the corresponding solution. Some of the solutions formed precipitates and some stayed in solution. The solutions that did not form precipitates was because the ions that were in the solution are all soluble. Table 1: Solution Combination for Precipitation Experiment and Observations for Precipitated Solution BaCl2 Na2CO3 H2SO4 NaCl MgSO4 BaCl2 X X X X X Na2CO3 White X X X X precipitate immediately formed No X X X H2SO4 White precipitate precipitate formed immediately formed X X No No NaCl No precipitate precipitate precipitate formed formed formed No No X Cloudy, MgSO4 White precipitate precipitate white precipitate formed formed precipitate immediately formed formed NaOH Very small No Very No White amount of precipitate minimal, precipitate precipitate white formed cloudy color formed formed at precipitate precipitate bottom formed, took a few minutes. H2SO4 Color- cloudy Rate of changequick Color- clear Rate of changeSlow Color- clear

Rate of change- slow Color- clear Rate of changequick Observationsreaction fizzed a little X NaOH Color- not as cloudy as other reactions Rate of change- slow Color- somewhat cloudy Rate of changequick Color- clear Rate of change- slow Color- clear Rate of change- slow Color- clear Rate of change- slo Part II of the experiment was to further analyze two different precipitates. Eight test tubes were set up with four containing 2.5mL BaCl2 and 2.5mL Na2CO3 and four containing 2.5 mL MgSO4 and 2.5mL NaOH. Next, four different methods of re-dissolving the precipitates were done. The results of each trial are shown in Table 2.

Table 2: Results of Different Ways to Re-dissolve the Precipitate.

Heating

Test tube 1: 2.5mL BaCl2 and 2.5mL Na2CO3

Test tube 2: 2.5mL BaCl2 and 2.5mL Na2CO3

Test tube 3: 2.5mL BaCl2 and 2.5mL Na2CO3

Test tube 4: 2.5mL BaCl2 and 2.5mL Na2CO3

Test tube 5: 2.5 mL MgSO4 and 2.5mL NaOH

Test tube 6: 2.5 mL MgSO4 and 2.5mL NaOH

Test tube 7: 2.5 mL MgSO4 and 2.5mL NaOH

Test tube 8: 2.5 m MgSO4 and 2.5mL NaOH

Helped separate precipitate quicker from the solution, white precipitate remained at bottom of test

X

X

X

Dispersed precipitate throughout the solution, white precipitate.

X

X

X

Dilution

tube. X

Adding acid (H2SO4)

X

Gave off clean odor, did not change color or precipitate, only volume. X

Adding base (NaOH)

X

X

X

X

X

Did not change precipitate, slight color change.

X

X

No change observed

X

X

X

X

X

Turned the solution an orange color at the very top of solution, no change in precipitate

X

X

Dissolved precipitate turned translucent. X

Data Analysis 1. The observations of the precipitates that formed were that they were are similar to each other. The common between the different precipitates was that they were white and cloudy. The difference between the precipitates was how the precipitates looked in each solution. Some were formed only at the bottom of the test tube, and some formed throughout the entire solution. The precipitates that formed were fairly similar to each other. For instance, each precipitate that was formed was white and cloudy. The common difference between the precipitates was how they were formed. Some of the precipitates formed at the bottom of the test tube, while others formed throughout the entire test tube. In some reactions, no precipitate formed and there was no color change, so the solution remained clear. 2. A. The combination of ions that resulted in formation of a precipitate in this experiment, was: BaCl2 and MgSO4, BaCl2 and Na2CO3, BaCl2 and H2SO4, BaCl2 and NaOH, MgSO4 and Na2CO3, MgSO4 and NaOH, and NaOH and H2SO4. B. There were no common ions present in all precipitates. Multiple precipitates contained Na atoms. The solutions that precipitated the most were the ones that were mixed with BaCl2. C. The combination of ions that did not form precipitates was BaCl2 and NaCl, MgSO4 and NaCl, MgSO4 and H2SO4, NaCl and NaCO3, NaCl and H2SO4, NaCl and NaOH, Na2CO3 and NaOH and Na2CO3 and H2SO4. D. Na2CO3 + MgSO4 = Na2SO4(aq) + MgCO3(s)

No change observed

Discussion 1.       

BaCl2 + MgSO4 = BaSO4+MgCl2 BaCl2 + Na2CO3 = BaCO3 + 2 NaCl BaCl2 + H2SO4 = BaSO4 + 2 HCl BaCl2 + NaOH = Ba(OH)2 + 2 NaCl MgSO4 + Na2CO3 = MgCO3 + Na2SO4 MgSO4 + 2 NaOH= Mg(OH)2 + Na2SO4 H2SO4 + 2 NaOH = 2 H2O + Na2SO4

2. As the temperature of a solution is increased, the average kinetic energy of the molecules that make up the solution also increases1. This increase in kinetic energy allows the solvent molecules to more effectively break apart the solute molecules that are held together by intermolecular attractions1. When heating the mixtures that produced a precipitate, the precipitate moved differently in each trail. As trial one test tube was heated, the precipitate separated from the solution to the bottom of the test tube very quickly. When trial two was heated, the precipitated dispersed throughout the test tube in the solution. When the temperature of the solution was increased, the solubility of the precipitate increased. No phase changes occurred because the precipitate remained as a solid. If the precipitate would have completely dissolved, then a phase change from liquid to solid. 3. The effect of diluting the precipitate did not affect the precipitate. In trial one, there was no effect at all. The volume of the solution just increased. In trial two, the precipitated thinned out a little bit. Dilution is the process where a solution is added more of the solvent to decrease the concentration of the solute2. In dilution, the amount of solute does not change; the number of moles is the same before and after dilution2. 4. In one of the experiments, H2SO4 was added to the precipitated solution. After adding the acid, the precipitate dissolved. This occurred due to the process of equilibrium. Equilibrium is when the rate of the forward reaction equals the rate of the reverse reaction; all reactant and product concentrations are constant at equilibrium3. An equation that could help clarify this would be expressed as YX(s) ←> Y+(aq) + X-(aq). 5. A. LiOH (aq) + NaCl (aq)  LiCl(aq) + NaOH (aq). This combination of compounds would stay in solution because each compound is soluble so that no precipitate would form. B. 3 BaCl2 (aq) + 2 Na3PO4 (aq)  Ba3(PO4)2 (s) + 6 NaCl (aq). This combination of compounds would precipitate because Ba3(PO4)2 is insoluble, which would form a precipitate. C. MgSO4 (aq) + KOH (aq)  MgOH (aq) + KSO4 (aq). This combination of compounds would stay in solution because each compound is soluble so that no precipitate would form.

Reference [1] Boundless. Boundless Chemistry. Lumen. https://courses.lumenlearning.com/boundlesschemistry/chapter/factors-affecting-solubility/ [2] Solutions, molarity and dilution. Engineering ToolBox. https://www.engineeringtoolbox.com/solution-solute-solvent-molar-molarity-concentrationdilution-d_1964.html [3] The equilibrium constant K. Khan Academy. Khan Academy. https://www.khanacademy.org/science/chemistry/chemical-equilibrium/equilibriumconstant/a/the-equilibrium-constant-k...