Experiment 5 (1) Tanvver Singh CHEM PDF

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EXPERIMEWNT 5...

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Experiment 5: Double Displacement Reactions/Precipitation Lab Required reading: Ebbing, 11th Edition Chapters 2 and 4. -Naming compounds -Balancing equations -Solubility rules -Molecular and Ionic equations -Displacement reactions Learning Goals:  To be able to write IUPAC names and formulas of ionic compounds using crossover rule.  Understand the principle of metathesis reactions and be able to write formulas of products for metathesis reactions.  Balance chemical equations for ionic compounds.  To be able to understand and use solubility rules to predict the formation of precipitate.  To be able to write ionic and net ionic equations and determine spectator ions in a given reaction. Background information and theory: When two aqueous solutions of ionic compounds react to form a new ionic substance that is not soluble in water, the presence of a solid (or a precipitate) is observed in the mixture. This type of reaction is called a precipitation reaction. A metathesis reaction is a reaction between compounds that, when written as a molecular equation, it looks like there is an exchange between cations and anions. If one of these products is not soluble, according to solubility rules (table 4.1 of your textbook), then the reaction will proceed and the formation a solid will be observed. If both products are soluble in water then the ions will remain in solution and no reaction will take place; the formation of a solid substance is the driving force for these reactions. The following reaction is an example of a metathesis reaction: 3 CoCl2(aq) +

2 Na3PO4(aq) 

Co3(PO4)2(s)

+

6 NaCl(aq)

however, the following reaction: KCl(aq) +

NaOH (aq) 

N.R. (no reaction)

won’t proceed since both potential products KOH and NaCl are soluble in water. In this experiment, you will be mixing aqueous solutions and observing the formation of precipitates. A table of ions can be used to write formulas of ionic compound: Refer to table 2.4, 2.5 and 2.6 of your textbook. In the next page there is a copy of the solubility table provided in your textbook.

The solubility of chromates is similar to the solubility of carbonates.

Materials (listed per student): Equipment: Spot plate, glass stirring rod, 10 small test tubes and test tube rack Chemicals: 0.1 M solutions of the following chemicals: Na3PO4 K2CrO4 AgNO3

Na2CO3 KI Ni(NO3)2

Na2SO4 Ba(NO3)2 Pb(NO3)2

NaOH BaCl2 CuSO4

NaCl CaCl2 FeCl2

Lab Notebook and Lab Reports:  Your instructor may have you record your lab work directly in the lab manual for this experiment rather than in your lab notebook. Follow your instructor’s directions.  A separate formal lab report may not be assigned by your instructor for this experiment. Safety concerns to note with this lab: All strong acids should be elaborated upon, as well as dangerous or toxic chemicals. AgNO3 must be handled carefully. If AgNO3 comes in direct contact with skin it stains skin and may even cause burns. To prevent this use gloves while handling the chemical. Prevent cross contamination while mixing the chemicals. Refer to the SDS for silver nitrate and sodium hydroxide. To do in the lab Procedure: Work in groups of 2. Part 1: Avoid excessive usage of chemicals. 1) In labeled disposable test tubes place about 3 mL of each solution (use plastic pipets). 2) Obtain a clean spot plate and using a dropper add 3-5 drops of the first reactant in reaction 1 into a cavity in your plate (use one dropper for each solution to avoid contamination)

3) Add 3-5 drops of the second reactant from the same reaction into the same cavity using dropper. NOTE: while adding the second chemical, the tip of the dropper should not touch the chemical in the spot plate. If required, stir solution by using a stirring rod. 4) Write down all your observations including appearance of reactants and products. 5) If you cannot see a clear change perform the experiment in the small test tube as follows: Take 0.5 to 1 ml of each reactant and mix well. Hold it against the light to see formation of any precipitate. 6) Reactions which involve formation of molecular compounds like water or gases may not show any visual changes. 7) Repeat the experiment with the remaining reactions. All reactions in this lab should provide a precipitate, if you don’t see one, please repeat the reaction.

Waste Disposal and Clean-Up: 1) Any excess chemicals and products should be disposed of in designated labelled waste containers. 2) Clean glassware thoroughly.

To do at home (online) Part 2: Theoretical work: Complete the data sheet independently using chapters 2 and 4 as follows. Do these steps for all reactions provided. Step 1: Write IUPAC names of products expected Step 2: Write formulas of reactants and products for the double displacement reaction. Step 3: Balance your equation Step 4: Use solubility rules to predict the physical states of products Note: All reactants are aqueous solutions Step 5: Write ionic, net ionic equations and determine spectator ions All equations must be balanced and must indicate the physical states of reactants and products. Please as practice write the molecular and ionic equation even for those reaction that will not work and you can cross with a line the net ionic equation as everything will cancel in these cases. Watch the following video for an example/explanation on how to write molecular, ionic and net ionic equations: https://lonestar.techsmithrelay.com/1sq7

Name: Tanveer Singh Lab Partner’s names: ____________________________________ Double Displacement reactions/Precipitations Lab Data Sheet For each reaction performed you must write: observations, balanced chemical equation, ionic equation, net ionic equation, spectator ions. Use the following example as a guide. Cobalt (II) chloride + sodium phosphate  Cobalt (II) phosphate + sodium chloride Balanced chemical equation (Molecular equation) 3 CoCl2(aq) + 2 Na3PO4 (aq)  Co3(PO4)2(s)

+

6 NaCl(aq)

Ionic equation 3 Co+2(aq) + 6 Cl-(aq) + 6Na+(aq) + 2 PO43-(aq)  Co3(PO4)2(s) + 6 Na+(aq) + 6Cl- (aq) Net ionic equation 3Co+2(aq) + 2 PO4 3-(aq)  Co3(PO4)2 (s) Spectator ions: Cl-(aq) and Na+(aq) Observations: -One of the initial solutions was colorless and the other was light brown. -Upon combining both solutions, the mixture turned dark blue. -A new blue solid is observed in the mixture (a precipitate was formed). 1: silver nitrate + sodium chloride  Silver chloride + Sodium Nitrate Observations: When solution of silver nitrate(colorless) is added to the solution of sodium chloride(colorless), the silver ions combine with the chlorine ions to form the precipitate of silver chloride (Light brown in color) Balanced chemical equation (Molecular equation) AgNo3 (aq) + NaCl (aq)  AgCl (s) +NaNO3 (Aq) Ionic equation Na (aq) + Cl (aq) + Ag (aq) + NO (aq) AgCl (s) + Na (aq) +NO (aq) Net ionic equation Cl (aq) + Ag (aq)  AgCl (s) Spectator ions: Na , NO

2: silver nitrate + potassium iodide  Silver iodide + Potassium nitrate Observations: When solution of silver nitrate and a solution of potassium iodide is mixed together they changed from being clear to yellow. Balanced chemical equation (Molecular equation) AgNO (aq) + KI (aq)  AgI (S) + KNO (Aq) Ionic equation Ag (aq) + NO (aq) + K (aq) + I (aq)  AgI (s) + K (aq) + NO (aq) Net ionic equation Ag (aq) + I (aq)  Agl (s) Spectator ions: K (aq) , NO (aq)

3: lead(II) nitrate + potassium iodide  Lead (II) iodide + Potassium nitrate Observations: Potassium iodide and lead(II) nitrate are combined and undergo a double replacement reaction. Potassium iodide reacts with lead(II) nitrate and produces lead(II) iodide and potassium nitrate. ... The lead ions and iodide ions will eventually meet to form a yellow precipitate.

Balanced chemical equation (Molecular equation) Pb(NO ) (aq) + 2KI (aq)  PBI (s) + 2KNO (aq) Ionic equation Pb (aq) + 2NO (aq)+ 2k (aq) + 2I (aq)  PBI (s) + 2K (aq) + 2NO (aq) Net ionic equation PB (aq) + 2I (aq)  PBI (s) Spectator ions: K (aq) , NO (aq)

4: copper (II) sulfate + sodium hydroxide sulphate

 copper hydroixide + Sodium

Observations: Copper sulphate reacts with sodium hydroxide to form a blue precipitate of copper hydroxide and sodium sulphate Balanced chemical equation (Molecular equation) CuSo (aq) + 2NaOH (aq)  Cu(OH) (s) + Na So (aq) Ionic equation Cu (aq) + SO (aq) + 2Na (aq) + 2OH (aq)  Cu(OH) (s)+ 2Na (aq) + SO (aq) Net ionic equation Cu (aq) + 2OH (aq)  Cu(OH) (s) Spectator ions: Na (aq), SO (aq)

5: iron (II) chloride + sodium hydroxide  Sodium chloride + iron(II)hydroxide Observations: iron (II) chloride reacts with Sodium hydroxide to form a reddish brown precipitate. Balanced chemical equation (Molecular equation) 2NaOH (aq) + FeCl (aq)  2NaCl (aq) + Fe(OH) (s) Ionic equation 2Na (aq) + 2OH (aq) + Fe (aq) + 2Cl (aq)  2Na (aq) + 2Cl (aq) + Fe(OH) (s) Net ionic equation 2OH (aq) + Fe (aq)  Fe(OH) (s) Spectator ions: Na (aq) , Cl (aq)

6: nickel (II) nitrate + sodium hydroxide  sodium nitrate + nickel hydroxide

Observations: sodium hydroxide (colorless) reacts with nickel(II) nitrate(green) produces a light green precipitate. Balanced chemical equation (Molecular equation) 2NaOH (aq) + Ni(NO ) (aq)  2NaNO (aq) + Ni(OH) (s) Ionic equation 2Na (aq) + 2OH (aq) + Ni (aq) + 2NO (aq)  2Na (aq) + 2No (aq) + Ni(OH) (s) Net ionic equation 2OH (aq) + Ni (aq)  Ni(OH) (s) Spectator ions:

Na (aq), NO (aq)

7: nickel (II) nitrate + sodium phosphate  nickel phosphate + sodium nitrate Observations: Nickel(II) nitrate reacts with sodium phosphate the solution turns milky and forms a white precipitate. Balanced chemical equation (Molecular equation) 3Ni(NO ) (aq) + 2Na Po (aq)  Ni (PO ) (s) + 6NaNO (aq) Ionic equation 3Ni (aq) + 6No (aq) + 6Na (aq) + 2PO (aq)  NI (PO ) (s) + 6Na + 6NO Net ionic equation 3Ni (aq) + 2PO (aq)  Ni (PO ) (s) Spectator ions: NO (aq), Na (aq)

8: barium nitrate + sodium sulfate  Observations: Barium nitrate reacts with sodium sulphate to form a white solution and a white precipitate.

Balanced chemical equation (Molecular equation) Ba(NO ) (aq) + Na SO (aq)  BaSO (s) + 2NaNO (aq) Ionic equation Ba (aq) + 2NO (aq) + 2Na (aq) + SO (aq)  BaSo (s) + 2Na (aq) + 2NO (aq) Net ionic equation Ba (aq) + SO (aq)  BaSO (s) Spectator ions: NO (aq) , Na (aq)

9: calcium chloride + sodium phosphate  calcium phosphate + sodium chloride Observations: Two solutions of sodium phosphate and calcium chloride combine to form calcium phosphate precipitate and aqueous sodium chloride. The solution turns cloudy white. Balanced chemical equation (Molecular equation) 3Cacl (aq) + 2Na PO (aq) Ca (PO ) (s) + 6NaCl (aq) Ionic equation 3Ca + 6Cl + 6Na + 2PO  Ca (PO ) (s) + 6Na (aq) + 6Cl (aq) Net ionic equation 3Ca (aq) + 2PO (aq)  Ca (PO ) (s) Spectator ions: Cl (aq), Na (aq)

10: lead(II) nitrate + sodium phosphate  lead phospahate + sodium nitrate . Observations: lead(II) nitrate racts with sodium phosphate to form a cloudy white solution and a precipitate.

Balanced chemical equation (Molecular equation) 3Pb(NO ) (aq) + 2Na PO (aq)  Pb (PO ) (s) + 6NaNO (aq) Ionic equation 3Pb (aq) + 6NO (aq) + 6Na (aq) + 2PO (aq)  Pb (PO ) (s) + 6Na (aq) + 6NO (aq) Net ionic equation 3Pb (aq) + 2PO (aq)  Pb (PO ) (s) Spectator ions: NO (aq), Na (aq)

11: copper (II) sulfate + sodium carbonate  copper carbonate + sodium sulfate Observations: Copper(II) Sulphate reacts with sodium carbonate to produce a light blue precipitates.

Balanced chemical equation (Molecular equation) Na CO (aq) + CuSO (aq)  Na SO (aq) + CuCO (s) Ionic equation 2Na (aq) + CO (aq) + Cu (aq) + SO (aq)  Na (aq) + SO (aq) + CuCO (s) Net ionic equation CO (aq) + Cu (aq)  CuCO (s) Spectator ions: Na (aq), SO (aq)

12: nickel (II) nitrate + sodium carbonate  nickel carbonate + sodium nitrate

Observations: nickel(II) nitrate reacts with sodium carbonate to form a light green precipitate. Balanced chemical equation (Molecular equation) Ni(NO ) (aq) + Na CO (aq)  NiCO (s) + 2NaNO (aq) Ionic equation Ni (aq) + 2NO (aq) + 2Na (aq) + CO (aq)  NiCO (s) + 2Na (aq) + 2NO (aq) Net ionic equation Ni (aq) + CO (aq)  NiCO (s) Spectator ions: NO (aq), Na (aq)

13: barium chloride + potassium chromate  barium chromate + potassium chloride Observations: When mixed, solutions of barium chloride, and potassium chromate, , form a yellow precipitate of barium chromate.

Balanced chemical equation (Molecular equation) BaCl (aq) + K CrO (aq)  BaCrO (s) + 2KCl (aq) Ionic equation Ba (aq) + 2Cl (aq) + 2k (aq) + CrO (aq)  BaCrO (s) + 2k (aq) + 2Cl (aq) Net ionic equation Ba (aq) + CrO (aq)  BaCrO (s) Spectator ions: Cl (aq), K (aq)

14: lead (II) nitrate + potassium chromate lead chromate + potassium nitrate.

Observations: lead(II) nitrate reacts with potassium chromate to form a light yellow precipitate.

Balanced chemical equation (Molecular equation) Pb(NO ) (aq) + K CrO (aq)  PbCrO (s) + 2KNO (aq) Ionic equation Pb (aq) + 2NO (aq) + 2K (aq) + CrO (aq)  PbCrO (s) + 2K (aq) + 2NO (aq) Net ionic equation Pb (aq) + CrO (aq)  PbCrO (s) Spectator ions: NO (aq), K (aq)

Name: Tanveer Singh _ Double Displacement reactions/Precipitations Lab Prelab questions: Your instructor may ask you to answer these in your lab notebook in a “prelab” section, online or they may ask you to write your answers in your lab manual and submit them before starting your experiment. Follow your instructor’s directions. 1: What is a metathesis reaction? Metathesis reactions are chemical reactions in which two hydrocarbons (alkanes, alkenes or alkynes) are converted to two new hydrocarbons by the exchange of carbon–carbon single, double or triple bonds 2a: What is a strong electrolyte? Give two examples A strong electrolyte is a solution/solute that completely, or almost completely, ionizes or dissociates in a solution. Example: HCl, Hbr

2b: What is a weak electrolyte? Give two examples Weak electrolytes are partially dissociate into ions in solution and are weak conductors of electricity. Examples: H2CO3, NH3 2c: What is a non-electrolyte? Give two examples A non-electrolyte is a chemical compound which does not conduct electricity in any state. It does not provide ions in a solution and therefore current does not flow through such solution. Examples of non-electrolytes are latex, and sugar solution.

3: Using solubility rules determine whether following five compounds are soluble or insoluble in water. Hg2I2 (Insoluble)

Al2S3 (Insoluble)

Na2CO3 (soluble)

KOH (soluble)

Ca3(PO4)2 (Insoluble)

4: Predict the names of the products, write balanced chemical equation, ionic, net ionic equations and spectator ions for the following two metathesis reactions: a) aluminum nitrate + sodium hydroxide  aluminium hydroxide + sodium nitrate Molecular equation 3NaOH (aq) + Al(NO ) (aq)  3NaNO (aq) + Al(OH) (s)

Complete ionic equation 3Na (aq) + 3OH (aq) + Al (aq) + 3NO (aq)  3Na (aq) + 3NO (aq) + Al(OH) (s) Net ionic equation Al (aq) + 3OH (aq)  Al(OH) (s) Spectator ions

Na (aq), NO (aq)

b) silver acetate + sodium chloride silver chloride + sodium acetate Molecular equation NaCl (aq) + AgC H O (aq)  NaC H O (aq) + AgCl (s) Complete ionic equation Na (aq) + Cl (aq) + Ag (aq) + C H O  Na (aq) + C H O + AgCl (s) Net ionic equation Cl (aq) + Ag (aq)  AgCl (s) Spectator ions are Na (aq), C H O (Aq)

Name: _Tanveer Singh _ Lab partner’s names: ____________________________________ Double Displacement reactions/Precipitations Lab Postlab questions: Your instructor may ask you to answer these in your lab notebook, or to answer directly on this page and turn it in, or to include these answers in a formal lab report. Follow your instructor’s directions. 1.- Predict the name of the products, write molecular, ionic and net ionic equations for the following reactions: A) barium chloride + sodium sulfate  barium sulphate + sodium chloride Balanced chemical equation (Molecular equation) BaCl (aq) + Na SO (aq) BaSO (s) + 2NaCl (aq) Ionic equation Ba (aq) + 2Cl (aq) + 2Na (aq) + So (aq)  BaSO (s) + 2Na (aq) + 2Cl (aq) Net ionic equation Ba (aq) + So (aq)  BaSO (s) B) lithium chloride + potassium hydroxide  lithium hydroxide + potassium chloride Balanced chemical equation (Molecular equation) LiCl (aq) + KOH (aq)  LiOH (s) +KCl (aq) Ionic equation Li (aq) + Cl (aq) + K (aq) + OH (aq)  LiOH (s) + K (aq) + Cl (aq) Net ionic equation Li (aq) + OH  Li(OH)...