solutions: 2001 d by: i schnizzle. 2001 d part a.1m pb(no3)2 /.1m nacl /.1m kmno4 /.1m c2h5oh /.1m...

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Solutions: Solutions: 2001 D 2001 D By: By: I Schnizzle I Schnizzle

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Page 1: Solutions: 2001 D By: I Schnizzle. 2001 D Part A.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH /.1M KC2H3O2.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH

Solutions:Solutions:2001 D2001 D

By:By:

I SchnizzleI Schnizzle

Page 2: Solutions: 2001 D By: I Schnizzle. 2001 D Part A.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH /.1M KC2H3O2.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH

2001 D Part A2001 D Part A

.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M .1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH / .1M KC2H3O2C2H5OH / .1M KC2H3O2A.) Which solution has the highest boiling point? A.) Which solution has the highest boiling point? Explain Explain The answer to this question can be found by The answer to this question can be found by referring back to the bond strengths within each referring back to the bond strengths within each of these solutions.of these solutions.The bond strength of any of these solutions The bond strength of any of these solutions determines how high or low the boiling point will determines how high or low the boiling point will be.be.

Page 3: Solutions: 2001 D By: I Schnizzle. 2001 D Part A.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH /.1M KC2H3O2.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH

2001 D Part A Cont’2001 D Part A Cont’

Since bonds within ionic compounds are Since bonds within ionic compounds are stronger than those within covalent stronger than those within covalent compounds the .1M of C2H5OH is out.compounds the .1M of C2H5OH is out.

Than looking at your other four solutions Than looking at your other four solutions which are ionic Pb(NO3)2 has the most which are ionic Pb(NO3)2 has the most ions of the remaining four solutions there ions of the remaining four solutions there for it will have the highest boiling point.for it will have the highest boiling point.

Page 4: Solutions: 2001 D By: I Schnizzle. 2001 D Part A.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH /.1M KC2H3O2.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH

2001 D Part B2001 D Part B

.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M .1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH / .1M KC2H3O2C2H5OH / .1M KC2H3O2

B.) Which solution has the highest pH? Explain B.) Which solution has the highest pH? Explain

The pH level of any solution is determined by The pH level of any solution is determined by how basic or acidic the solution is.how basic or acidic the solution is.

If a solution has a weak acid and a salt it will If a solution has a weak acid and a salt it will form a more basic solution. This goes for a salt form a more basic solution. This goes for a salt and weak base as well except the solution would and weak base as well except the solution would be more acidicbe more acidic

Page 5: Solutions: 2001 D By: I Schnizzle. 2001 D Part A.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH /.1M KC2H3O2.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH

2001 D Part B Cont’2001 D Part B Cont’

KC2H3O2 is the solution that will have the KC2H3O2 is the solution that will have the highest pH based on the weak acid and salt it highest pH based on the weak acid and salt it has.has.

When KC2H3O2 is added to H2O the compound When KC2H3O2 is added to H2O the compound will dissociate and the weak acid and salt, K in will dissociate and the weak acid and salt, K in this case will form new compounds.this case will form new compounds.

With the new compound formed from the weak With the new compound formed from the weak acid the new solution will be more basic, and in acid the new solution will be more basic, and in turn have higher pH.turn have higher pH.

Page 6: Solutions: 2001 D By: I Schnizzle. 2001 D Part A.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH /.1M KC2H3O2.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH

2001 D Part C 2001 D Part C

.1M Pb(NO3)2 /.1M NaCl /.1M .1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH / .1M KC2H3O2KMnO4 /.1M C2H5OH / .1M KC2H3O2C.) Identify a pair of solutions that would C.) Identify a pair of solutions that would produce a precipitate when mixed produce a precipitate when mixed together. Write the formula of the together. Write the formula of the precipitate.precipitate.In order to complete this question you In order to complete this question you must first have a general understanding of must first have a general understanding of your solubility rules.your solubility rules.

Page 7: Solutions: 2001 D By: I Schnizzle. 2001 D Part A.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH /.1M KC2H3O2.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH

2001 D Part C Cont’2001 D Part C Cont’

After looking over your solubility rules you After looking over your solubility rules you will find that Pb2+ does not tend to will find that Pb2+ does not tend to dissolve with most salts.dissolve with most salts.

Therefore the solutions of Pb(NO3)2 and Therefore the solutions of Pb(NO3)2 and NaCl are good solutions that will form a NaCl are good solutions that will form a precipitate since the ions will form new precipitate since the ions will form new compounds one of which is PbCl2 which is compounds one of which is PbCl2 which is insoluble.insoluble.

Page 8: Solutions: 2001 D By: I Schnizzle. 2001 D Part A.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH /.1M KC2H3O2.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH

2001 D Part C Cont’2001 D Part C Cont’

The equation for this new precipitate The equation for this new precipitate formed would be as follows.formed would be as follows.

Pb(NO3)2 + NaCl yields NaNO3 + PbCl2 Pb(NO3)2 + NaCl yields NaNO3 + PbCl2

After balancing this equation the final After balancing this equation the final balanced equation would be Pb(NO3)2 + balanced equation would be Pb(NO3)2 + 2NaCl yields 2NaNO3 + PbCl22NaCl yields 2NaNO3 + PbCl2

Where the PbCl2 is the precipitate formed Where the PbCl2 is the precipitate formed and is also the formula of the precipitate.and is also the formula of the precipitate.

Page 9: Solutions: 2001 D By: I Schnizzle. 2001 D Part A.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH /.1M KC2H3O2.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH

2001 D Part D2001 D Part D

.1M Pb(NO3)2 /.1M NaCl /.1M .1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH / .1M KC2H3O2KMnO4 /.1M C2H5OH / .1M KC2H3O2Which solution could be used to oxidize Which solution could be used to oxidize the Cl-(aq) ion? Identify the product of the the Cl-(aq) ion? Identify the product of the oxidation.oxidation.In order to oxidize something the charge In order to oxidize something the charge on the ion becomes more positive so the on the ion becomes more positive so the goal is to find a solution that allows for goal is to find a solution that allows for Cl-’s charge to become more positive.Cl-’s charge to become more positive.

Page 10: Solutions: 2001 D By: I Schnizzle. 2001 D Part A.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH /.1M KC2H3O2.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH

2001 D Part D Cont’2001 D Part D Cont’

KMnO4 is a perfect solution for this KMnO4 is a perfect solution for this oxidation to occur because once the ions oxidation to occur because once the ions of the KMnO4 are broken down they will of the KMnO4 are broken down they will oxidize the Cl- the most.oxidize the Cl- the most.

This oxidation occurs and forms the This oxidation occurs and forms the product of ClO3- which gives the Cl- a product of ClO3- which gives the Cl- a more positive charge of 5+more positive charge of 5+

Page 11: Solutions: 2001 D By: I Schnizzle. 2001 D Part A.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH /.1M KC2H3O2.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH

2001D Part E2001D Part E

.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M .1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH / .1M KC2H3O2C2H5OH / .1M KC2H3O2E.) Which solution would be the least effective E.) Which solution would be the least effective conductor of electricity? Explain.conductor of electricity? Explain.First of all good conductors of electricity are First of all good conductors of electricity are compounds that have ionic bonds within them as compounds that have ionic bonds within them as opposed to covalent bonds.opposed to covalent bonds.So we are looking for a covalent bonded So we are looking for a covalent bonded compound since that would be the least effective compound since that would be the least effective solution to conduct electricity.solution to conduct electricity.

Page 12: Solutions: 2001 D By: I Schnizzle. 2001 D Part A.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH /.1M KC2H3O2.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH

2001D Part E Cont’2001D Part E Cont’

C2H5OH is the only compound that has C2H5OH is the only compound that has covalent bonds and therefore does not covalent bonds and therefore does not form any ions in H2O.form any ions in H2O.

Since H2O is also covalently bonded the Since H2O is also covalently bonded the solution would be the least likely to be a solution would be the least likely to be a good conductor of electricity.good conductor of electricity.

Page 13: Solutions: 2001 D By: I Schnizzle. 2001 D Part A.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH /.1M KC2H3O2.1M Pb(NO3)2 /.1M NaCl /.1M KMnO4 /.1M C2H5OH

BAM!BAM!

The EndThe End