Chemistry 12 – Sorting out Solubility Problems This will help you with the following types of Solubility Problems: 1-ion concentrations in mixtures (no.

Chemistry 12 – Sorting out Solubility Problems This will help you with the following types of Solubility Problems: 1-ion concentrations in mixtures (no ppts) 2-experimental determination of solubility 3-solubility (s) from Ksp (one compound) 4-Ksp from solubility (one compound) 5-predicting precipitates using Trial Ksp 6-finding maximum concentration of an ion in a solution in which another ion is present 7-finding minimum concentration of an ion necessary to just start precipitation 8-finding which precipitate will form first 9-using titrations to find unknown concentration

First-a couple of things to ALWAYS remember:

1. Compounds with Nitrate (NO 3 ) are ALWAYS soluble (NO 3 - is ALWAYS a spectator in unit 3)!

First-a couple of things to ALWAYS remember: 1. Compounds with Nitrate (NO 3 ) are ALWAYS soluble (NO 3 - is ALWAYS a spectator in unit 3)! 2.Compounds with Alkali ions (Na, K, Li etc.) are ALWAYS soluble. (Na +, K + etc. are ALWAYS spectators)!

First-a couple of things to ALWAYS remember: 1. Compounds with Nitrate (NO 3 ) are ALWAYS soluble (NO 3 - is ALWAYS a spectator in unit 3)! 2.Compounds with Alkali ions (Na, K, Li etc.) are ALWAYS soluble. (Na +, K + etc. are ALWAYS spectators)! 3.Always have your solubility and Ksp table handy when doing these problems!

1-Ion concentrations in mixtures (no ppts) In this type there are 1 or 2 compounds and they are BOTH soluble. (There are NO Low solubility Compounds!) An Example: Find the [OH - ] when 50.0 mL of 0.20M NaOH is mixed with 250.0 mL of 0.30M Sr(OH) 2.

1-Ion concentrations in mixtures (no ppts) In this type there are 1 or 2 compounds and they are BOTH soluble. (There are NO Low solubility Compounds!) An Example: Find the [OH - ] when 50.0 mL of 0.20M NaOH is mixed with 250.0 mL of 0.30M Sr(OH) 2. Both of these solutions are soluble (check sol. Table) There is NO compound of Low Solubility involved here. So no equilibrium and no Ksp!

1-Ion concentrations in mixtures (no ppts) In this type there are 1 or 2 compounds and they are BOTH soluble. (There are NO Low solubility Compounds!) An Example: Find the [OH - ] when 50.0 mL of 0.20M NaOH is mixed with 250.0 mL of 0.30M Sr(OH) 2. Both of these solutions are soluble (check sol. Table) There is NO compound of Low Solubility involved here. So no equilibrium and no Ksp! In this type of problem: -use the Dilution Formula to find the final [NaOH] in the mixture (final volume is 300.0 mL)

1-Ion concentrations in mixtures (no ppts) In this type there are 1 or 2 compounds and they are BOTH soluble. (There are NO Low solubility Compounds!) An Example: Find the [OH - ] when 50.0 mL of 0.20M NaOH is mixed with 250.0 mL of 0.30M Sr(OH) 2. Both of these solutions are soluble (check sol. Table) There is NO compound of Low Solubility involved here. So no equilibrium and no Ksp! In this type of problem: -use the Dilution Formula to find the final [NaOH] in the mixture (final volume is 300.0 mL) -dissociate the NaOH to find the [OH - ] in NaOH

1-Ion concentrations in mixtures (no ppts) In this type there are 1 or 2 compounds and they are BOTH soluble. (There are NO Low solubility Compounds!) An Example: Find the [OH - ] when 50.0 mL of 0.20M NaOH is mixed with 250.0 mL of 0.30M Sr(OH) 2. Both of these solutions are soluble (check sol. Table) There is NO compound of Low Solubility involved here. So no equilibrium and no Ksp! In this type of problem: -use the Dilution Formula to find the final [NaOH] in the mixture (final volume is 300.0 mL) -dissociate the NaOH to find the [OH - ] in NaOH -use the Dilution Formula to find the final [Sr(OH) 2 ]

1-Ion concentrations in mixtures (no ppts) In this type there are 1 or 2 compounds and they are BOTH soluble. (There are NO Low solubility Compounds!) An Example: Find the [OH - ] when 50.0 mL of 0.20M NaOH is mixed with 250.0 mL of 0.30M Sr(OH) 2. Both of these solutions are soluble (check sol. Table) There is NO compound of Low Solubility involved here. So no equilibrium and no Ksp! In this type of problem: -use the Dilution Formula to find the final [NaOH] in the mixture (final volume is 300.0 mL) -dissociate the NaOH to find the [OH - ] in NaOH -use the Dilution Formula to find the final [Sr(OH) 2 ] -dissociate the [Sr(OH) 2 ] to find the [OH - ] in [Sr(OH) 2 ]

1-Ion concentrations in mixtures (no ppts) In this type there are 1 or 2 compounds and they are BOTH soluble. (There are NO Low solubility Compounds!) An Example: Find the [OH - ] when 50.0 mL of 0.20M NaOH is mixed with 250.0 mL of 0.30M Sr(OH) 2. Both of these solutions are soluble (check sol. Table) There is NO compound of Low Solubility involved here. So no equilibrium and no Ksp! In this type of problem: -use the Dilution Formula to find the final [NaOH] in the mixture (final volume is 300.0 mL) -dissociate the NaOH to find the [OH - ] in NaOH -use the Dilution Formula to find the final [Sr(OH) 2 ] -dissociate the [Sr(OH) 2 ] to find the [OH - ] in [Sr(OH) 2 ] -add up the [OH - ] from NaOH and the [OH - ] from [Sr(OH) 2 ] to find the final [OH - ].

1-Ion concentrations in mixtures (no ppts) In this type there are 1 or 2 compounds and they are BOTH soluble. (There are NO Low solubility Compounds!) An Example: Find the [OH - ] when 50.0 mL of 0.20M NaOH is mixed with 250.0 mL of 0.30M Sr(OH) 2. Both of these solutions are soluble (check sol. Table) There is NO compound of Low Solubility involved here. So no equilibrium and no Ksp! In this type of problem: -use the Dilution Formula to find the final [NaOH] in the mixture (final volume is 300.0 mL) -dissociate the NaOH to find the [OH - ] in NaOH -use the Dilution Formula to find the final [Sr(OH) 2 ] -dissociate the [Sr(OH) 2 ] to find the [OH - ] in [Sr(OH) 2 ] -add up the [OH - ] from NaOH and the [OH - ] from [Sr(OH) 2 ] to find the final [OH - ]. -THE ANSWER IS [OH - ] = 0.53 M

2-Experimental determination of solubility An example problem is: An experiment was done in which 20.0 mL of saturated CaSO 4 were transferred to a weighed empty evaporating dish and were left to evaporate. The dish and the solid residue were then weighed. The data table is as follows: 1. Mass of empty evaporating dish ……………. 65.340 g 2. Mass of evaporating dish and solid CaSO 4 … 65.363 g 3. Volume of saturated CaSO 4 evaporated …… 20.0 mL Using the data from this experiment, calculate the molar solubility of CaSO 4.

2-Experimental determination of solubility An example problem is: An experiment was done in which 20.0 mL of saturated CaSO 4 were transferred to a weighed empty evaporating dish and were left to evaporate. The dish and the solid residue were then weighed. The data table is as follows: 1. Mass of empty evaporating dish ……………. 65.340 g 2. Mass of evaporating dish and solid CaSO 4 … 65.363 g 3. Volume of saturated CaSO 4 evaporated …… 20.0 mL Using the data from this experiment, calculate the molar solubility of CaSO 4. 1. Subtract # 1 from #2 to find the grams of solid CaSO 4 which dissolved

2-Experimental determination of solubility An example problem is: An experiment was done in which 20.0 mL of saturated CaSO 4 were transferred to a weighed empty evaporating dish and were left to evaporate. The dish and the solid residue were then weighed. The data table is as follows: 1. Mass of empty evaporating dish ……………. 65.340 g 2. Mass of evaporating dish and solid CaSO 4 … 65.363 g 3. Volume of saturated CaSO 4 evaporated …… 20.0 mL Using the data from this experiment, calculate the molar solubility of CaSO 4. 1. Subtract # 1 from #2 to find the grams of solid CaSO 4 which dissolved 2. Since you are asked for the molar solubility, change these grams to moles. (If you were asked for the solubility in g/L, you could leave them as grams.)

2-Experimental determination of solubility An example problem is: An experiment was done in which 20.0 mL of saturated CaSO 4 were transferred to a weighed empty evaporating dish and were left to evaporate. The dish and the solid residue were then weighed. The data table is as follows: 1. Mass of empty evaporating dish ……………. 65.340 g 2. Mass of evaporating dish and solid CaSO 4 … 65.363 g 3. Volume of saturated CaSO 4 evaporated …… 20.0 mL Using the data from this experiment, calculate the molar solubility of CaSO 4. 1. Subtract # 1 from #2 to find the grams of solid CaSO 4 which dissolved 2. Since you are asked for the molar solubility, change these grams to moles. (If you were asked for the solubility in g/L, you could leave them as grams.) 3. Change 20.0 mL to 0.0200L

2-Experimental determination of solubility An example problem is: An experiment was done in which 20.0 mL of saturated CaSO 4 were transferred to a weighed empty evaporating dish and were left to evaporate. The dish and the solid residue were then weighed. The data table is as follows: 1. Mass of empty evaporating dish ……………. 65.340 g 2. Mass of evaporating dish and solid CaSO 4 … 65.363 g 3. Volume of saturated CaSO 4 evaporated …… 20.0 mL Using the data from this experiment, calculate the molar solubility of CaSO 4. 1. Subtract # 1 from #2 to find the grams of solid CaSO 4 which dissolved 2. Since you are asked for the molar solubility, change these grams to moles. (If you were asked for the solubility in g/L, you could leave them as grams.) 3. Change 20.0 mL to 0.0200L 4. Use to find Molar concentration or Molar Solubility.

2-Experimental determination of solubility An example problem is: An experiment was done in which 20.0 mL of saturated CaSO 4 were transferred to a weighed empty evaporating dish and were left to evaporate. The dish and the solid residue were then weighed. The data table is as follows: 1. Mass of empty evaporating dish ……………. 65.340 g 2. Mass of evaporating dish and solid CaSO 4 … 65.363 g 3. Volume of saturated CaSO 4 evaporated …… 20.0 mL Using the data from this experiment, calculate the molar solubility of CaSO 4. 1. Subtract # 1 from #2 to find the grams of solid CaSO 4 which dissolved 2. Since you are asked for the molar solubility, change these grams to moles. (If you were asked for the solubility in g/L, you could leave them as grams.) 3. Change 20.0 mL to 0.0200L 4. Use to find Molar concentration or Molar Solubility. THE ANSWER IS 0.00844 M or 8.44 x 10 -3 M

2-Experimental determination of solubility An example problem is: An experiment was done in which 20.0 mL of saturated CaSO 4 were transferred to a weighed empty evaporating dish and were left to evaporate. The dish and the solid residue were then weighed. The data table is as follows: 1. Mass of empty evaporating dish ……………. 65.340 g 2. Mass of evaporating dish and solid CaSO 4 … 65.363 g 3. Volume of saturated CaSO 4 evaporated …… 20.0 mL Using the data from this experiment, calculate the molar solubility of CaSO 4. 1. Subtract # 1 from #2 to find the grams of solid CaSO 4 which dissolved 2. Since you are asked for the molar solubility, change these grams to moles. (If you were asked for the solubility in g/L, you could leave them as grams.) 3. Change 20.0 mL to 0.0200L 4. Use to find Molar concentration or Molar Solubility. THE ANSWER IS 0.00844 M or 8.44 x 10 -3 M In a variation of this type of problem, you may be asked to calculate the Ksp. In this case, use this molar solubility and use it to find Ksp as outlined in #4, coming up…

3-Solubility (s) from Ksp (one compound) In this type of problem, you will have ONE COMPOUND and it will be a LOW SOLUBILITY compound. You are asked to find: The molar solubility of the compound or The solubility of the compound in g/L or The mass which will dissolve in a certain volume or The concentration of one of the ions in the compound

3-Solubility (s) from Ksp (one compound) In this type of problem, you will have ONE COMPOUND and it will be a LOW SOLUBILITY compound. You are asked to find: The molar solubility of the compound or The solubility of the compound in g/L or The mass which will dissolve in a certain volume or The concentration of one of the ions in the compound An example question could be: Find the [IO 3 - ] in saturated Cu(IO 3 ) 2.

3-Solubility (s) from Ksp (one compound) In this type of problem, you will have ONE COMPOUND and it will be a LOW SOLUBILITY compound. You are asked to find: The molar solubility of the compound or The solubility of the compound in g/L or The mass which will dissolve in a certain volume or The concentration of one of the ions in the compound An example question could be: Find the [IO 3 - ] in saturated Cu(IO 3 ) 2. The starting procedures for all problems of this type are the SAME:

3-Solubility (s) from Ksp (one compound) In this type of problem, you will have ONE COMPOUND and it will be a LOW SOLUBILITY compound. You are asked to find: The molar solubility of the compound or The solubility of the compound in g/L or The mass which will dissolve in a certain volume or The concentration of one of the ions in the compound An example question could be: Find the [IO 3 - ] in saturated Cu(IO 3 ) 2. The starting procedures for all problems of this type are the SAME: 1. Write the equilibrium equation. (solid on left, ions on right BALANCED)

3-Solubility (s) from Ksp (one compound) In this type of problem, you will have ONE COMPOUND and it will be a LOW SOLUBILITY compound. You are asked to find: The molar solubility of the compound or The solubility of the compound in g/L or The mass which will dissolve in a certain volume or The concentration of one of the ions in the compound An example question could be: Find the [IO 3 - ] in saturated Cu(IO 3 ) 2. The starting procedures for all problems of this type are the SAME: 1. Write the equilibrium equation. (solid on left, ions on right BALANCED) Cu(IO 3 ) 2(s)  Cu 2+ (aq) + 2 IO 3 - (aq)

3-Solubility (s) from Ksp (one compound) In this type of problem, you will have ONE COMPOUND and it will be a LOW SOLUBILITY compound. You are asked to find: The molar solubility of the compound or The solubility of the compound in g/L or The mass which will dissolve in a certain volume or The concentration of one of the ions in the compound An example question could be: Find the [IO 3 - ] in saturated Cu(IO 3 ) 2. The starting procedures for all problems of this type are the SAME: 1. Write the equilibrium equation. (solid on left, ions on right BALANCED) Cu(IO 3 ) 2(s)  Cu 2+ (aq) + 2 IO 3 - (aq) 2. Since you need to find the molar solubility (s), write “-s” above the solid

3-Solubility (s) from Ksp (one compound) In this type of problem, you will have ONE COMPOUND and it will be a LOW SOLUBILITY compound. You are asked to find: The molar solubility of the compound or The solubility of the compound in g/L or The mass which will dissolve in a certain volume or The concentration of one of the ions in the compound An example question could be: Find the [IO 3 - ] in saturated Cu(IO 3 ) 2. The starting procedures for all problems of this type are the SAME: 1. Write the equilibrium equation. (solid on left, ions on right BALANCED) Cu(IO 3 ) 2(s)  Cu 2+ (aq) + 2 IO 3 - (aq) 2. Since you need to find the molar solubility (s), write “-s” above the solid -s

3-Solubility (s) from Ksp (one compound) In this type of problem, you will have ONE COMPOUND and it will be a LOW SOLUBILITY compound. You are asked to find: The molar solubility of the compound or The solubility of the compound in g/L or The mass which will dissolve in a certain volume or The concentration of one of the ions in the compound An example question could be: Find the [IO 3 - ] in saturated Cu(IO 3 ) 2. The starting procedures for all problems of this type are the SAME: 1. Write the equilibrium equation. (solid on left, ions on right BALANCED) Cu(IO 3 ) 2(s)  Cu 2+ (aq) + 2 IO 3 - (aq) 2. Since you need to find the molar solubility (s), write “-s” above the solid 3. Use coefficients to find the concentrations of the ions in terms of “s” -s

3-Solubility (s) from Ksp (one compound) In this type of problem, you will have ONE COMPOUND and it will be a LOW SOLUBILITY compound. You are asked to find: The molar solubility of the compound or The solubility of the compound in g/L or The mass which will dissolve in a certain volume or The concentration of one of the ions in the compound An example question could be: Find the [IO 3 - ] in saturated Cu(IO 3 ) 2. The starting procedures for all problems of this type are the SAME: 1. Write the equilibrium equation. (solid on left, ions on right BALANCED) Cu(IO 3 ) 2(s)  Cu 2+ (aq) + 2 IO 3 - (aq) 2. Since you need to find the molar solubility (s), write “-s” above the solid 3. Use coefficients to find the concentrations of the ions in terms of “s” -s +s +2s

3-Solubility (s) from Ksp (one compound) In this type of problem, you will have ONE COMPOUND and it will be a LOW SOLUBILITY compound. You are asked to find: The molar solubility of the compound or The solubility of the compound in g/L or The mass which will dissolve in a certain volume or The concentration of one of the ions in the compound An example question could be: Find the [IO 3 - ] in saturated Cu(IO 3 ) 2. The starting procedures for all problems of this type are the SAME: 1. Write the equilibrium equation. (solid on left, ions on right BALANCED) Cu(IO 3 ) 2(s)  Cu 2+ (aq) + 2 IO 3 - (aq) 2. Since you need to find the molar solubility (s), write “-s” above the solid 3. Use coefficients to find the concentrations of the ions in terms of “s” -s +s +2s 4. Write the Ksp expression.

3-Solubility (s) from Ksp (one compound) In this type of problem, you will have ONE COMPOUND and it will be a LOW SOLUBILITY compound. You are asked to find: The molar solubility of the compound or The solubility of the compound in g/L or The mass which will dissolve in a certain volume or The concentration of one of the ions in the compound An example question could be: Find the [IO 3 - ] in saturated Cu(IO 3 ) 2. The starting procedures for all problems of this type are the SAME: 1. Write the equilibrium equation. (solid on left, ions on right BALANCED) Cu(IO 3 ) 2(s)  Cu 2+ (aq) + 2 IO 3 - (aq) 2. Since you need to find the molar solubility (s), write “-s” above the solid 3. Use coefficients to find the concentrations of the ions in terms of “s” -s +s +2s 4. Write the Ksp expression. Ksp = [Cu 2+ ] [IO 3 - ] 2

3-Solubility (s) from Ksp (one compound) In this type of problem, you will have ONE COMPOUND and it will be a LOW SOLUBILITY compound. You are asked to find: The molar solubility of the compound or The solubility of the compound in g/L or The mass which will dissolve in a certain volume or The concentration of one of the ions in the compound An example question could be: Find the [IO 3 - ] in saturated Cu(IO 3 ) 2. The starting procedures for all problems of this type are the SAME: 1. Write the equilibrium equation. (solid on left, ions on right BALANCED) Cu(IO 3 ) 2(s)  Cu 2+ (aq) + 2 IO 3 - (aq) 2. Since you need to find the molar solubility (s), write “-s” above the solid 3. Use coefficients to find the concentrations of the ions in terms of “s” -s +s +2s 4. Write the Ksp expression. Ksp = [Cu 2+ ] [IO 3 - ] 2 5. Substitute concentrations into Ksp expression.

3-Solubility (s) from Ksp (one compound) In this type of problem, you will have ONE COMPOUND and it will be a LOW SOLUBILITY compound. You are asked to find: The molar solubility of the compound or The solubility of the compound in g/L or The mass which will dissolve in a certain volume or The concentration of one of the ions in the compound An example question could be: Find the [IO 3 - ] in saturated Cu(IO 3 ) 2. The starting procedures for all problems of this type are the SAME: 1. Write the equilibrium equation. (solid on left, ions on right BALANCED) Cu(IO 3 ) 2(s)  Cu 2+ (aq) + 2 IO 3 - (aq) 2. Since you need to find the molar solubility (s), write “-s” above the solid 3. Use coefficients to find the concentrations of the ions in terms of “s” -s +s +2s 4. Write the Ksp expression. Ksp = [Cu 2+ ] [IO 3 - ] 2 5. Substitute concentrations into Ksp expression. Ksp = [s] [2s] 2

3-Solubility (s) from Ksp (one compound) In this type of problem, you will have ONE COMPOUND and it will be a LOW SOLUBILITY compound. You are asked to find: The molar solubility of the compound or The solubility of the compound in g/L or The mass which will dissolve in a certain volume or The concentration of one of the ions in the compound An example question could be: Find the [IO 3 - ] in saturated Cu(IO 3 ) 2. The starting procedures for all problems of this type are the SAME: 1. Write the equilibrium equation. (solid on left, ions on right BALANCED) Cu(IO 3 ) 2(s)  Cu 2+ (aq) + 2 IO 3 - (aq) 2. Since you need to find the molar solubility (s), write “-s” above the solid 3. Use coefficients to find the concentrations of the ions in terms of “s” -s +s +2s 4. Write the Ksp expression. Ksp = [Cu 2+ ] [IO 3 - ] 2 5. Substitute concentrations into Ksp expression. Ksp = [s] [2s] 2 6. Solve for s ( Ksp = 4s 3 so ) (look up Ksp on Ksp table)

3-Solubility (s) from Ksp (one compound) In this type of problem, you will have ONE COMPOUND and it will be a LOW SOLUBILITY compound. You are asked to find: The molar solubility of the compound or The solubility of the compound in g/L or The mass which will dissolve in a certain volume or The concentration of one of the ions in the compound An example question could be: Find the [IO 3 - ] in saturated Cu(IO 3 ) 2. The starting procedures for all problems of this type are the SAME: 1. Write the equilibrium equation. (solid on left, ions on right BALANCED) Cu(IO 3 ) 2(s)  Cu 2+ (aq) + 2 IO 3 - (aq) 2. Since you need to find the molar solubility (s), write “-s” above the solid 3. Use coefficients to find the concentrations of the ions in terms of “s” -s +s +2s 4. Write the Ksp expression. Ksp = [Cu 2+ ] [IO 3 - ] 2 5. Substitute concentrations into Ksp expression. Ksp = [s] [2s] 2 6. Solve for s ( Ksp = 4s 3 so ) (look up Ksp on Ksp table) 7. When you find “s”, the [IO 3 - ] = 2s.

3-Solubility (s) from Ksp (one compound) In this type of problem, you will have ONE COMPOUND and it will be a LOW SOLUBILITY compound. You are asked to find: The molar solubility of the compound or The solubility of the compound in g/L or The mass which will dissolve in a certain volume or The concentration of one of the ions in the compound An example question could be: Find the [IO 3 - ] in saturated Cu(IO 3 ) 2. The starting procedures for all problems of this type are the SAME: 1. Write the equilibrium equation. (solid on left, ions on right BALANCED) Cu(IO 3 ) 2(s)  Cu 2+ (aq) + 2 IO 3 - (aq) 2. Since you need to find the molar solubility (s), write “-s” above the solid 3. Use coefficients to find the concentrations of the ions in terms of “s” -s +s +2s 4. Write the Ksp expression. Ksp = [Cu 2+ ] [IO 3 - ] 2 5. Substitute concentrations into Ksp expression. Ksp = [s] [2s] 2 6. Solve for s ( Ksp = 4s 3 so ) (look up Ksp on Ksp table) 7. When you find “s”, the [IO 3 - ] = 2s. ANSWER [IO 3 - ] = 5.2 x 10 -3 M

4-Ksp from solubility (one compound) In this type of question, you are dealing with ONE compound which has LOW SOLUBILITY and you are given either: -molar solubility -solubility in g/L -grams (or moles) which will dissolve in a certain volume of water (solution)

4-Ksp from solubility (one compound) In this type of question, you are dealing with ONE compound which has LOW SOLUBILITY and you are given either: -molar solubility -solubility in g/L -grams (or moles) which will dissolve in a certain volume of water (solution) Don’t forget, to find Molar Solubility (M) from grams, use the plan: grams  moles  M

4-Ksp from solubility (one compound) In this type of question, you are dealing with ONE compound which has LOW SOLUBILITY and you are given either: -molar solubility -solubility in g/L -grams (or moles) which will dissolve in a certain volume of water (solution) Don’t forget, to find Molar Solubility (M) from grams, use the plan: grams  moles  M In this type of problem, you will HAVE the Molar Solubility, so DON’T call it “s”, write the actual values on top of the equilibrium equation.

4-Ksp from solubility (one compound) For example: It is found that 1.20 grams of PbCl 2 will dissolve in 300.0 mL of water to form a saturated solution. Use this information to calculate the Ksp for PbCl 2.

4-Ksp from solubility (one compound) For example: It is found that 1.20 grams of PbCl 2 will dissolve in 300.0 mL of water to form a saturated solution. Use this information to calculate the Ksp for PbCl 2. 1. First, we would go g  mol  M

4-Ksp from solubility (one compound) For example: It is found that 1.20 grams of PbCl 2 will dissolve in 300.0 mL of water to form a saturated solution. Use this information to calculate the Ksp for PbCl 2. 1. First, we would go g  mol  M 2. Write the equilibrium equation and write the molar solubility on top of the solid.

4-Ksp from solubility (one compound) For example: It is found that 1.20 grams of PbCl 2 will dissolve in 300.0 mL of water to form a saturated solution. Use this information to calculate the Ksp for PbCl 2. 1. First, we would go g  mol  M 2. Write the equilibrium equation and write the molar solubility on top of the solid. PbCl 2(s)  Pb 2+ (aq) + 2Cl - (aq) -0.01438M

4-Ksp from solubility (one compound) For example: It is found that 1.20 grams of PbCl 2 will dissolve in 300.0 mL of water to form a saturated solution. Use this information to calculate the Ksp for PbCl 2. 1. First, we would go g  mol  M 2. Write the equilibrium equation and write the molar solubility on top of the solid. PbCl 2(s)  Pb 2+ (aq) + 2Cl - (aq) -0.01438M 3. Use coefficients to find concentrations of the ions. Write them above.

4-Ksp from solubility (one compound) For example: It is found that 1.20 grams of PbCl 2 will dissolve in 300.0 mL of water to form a saturated solution. Use this information to calculate the Ksp for PbCl 2. 1. First, we would go g  mol  M 2. Write the equilibrium equation and write the molar solubility on top of the solid. PbCl 2(s)  Pb 2+ (aq) + 2Cl - (aq) -0.01438M + 0.01438M +0.02876M 3. Use coefficients to find concentrations of the ions. Write them above.

4-Ksp from solubility (one compound) For example: It is found that 1.20 grams of PbCl 2 will dissolve in 300.0 mL of water to form a saturated solution. Use this information to calculate the Ksp for PbCl 2. 1. First, we would go g  mol  M 2. Write the equilibrium equation and write the molar solubility on top of the solid. PbCl 2(s)  Pb 2+ (aq) + 2Cl - (aq) -0.01438M + 0.01438M +0.02876M 3. Use coefficients to find concentrations of the ions. Write them above. 4. Write Ksp expression and substitute these values in to calculate Ksp.

5-Predicting Precipitates using Trial Ksp In this type of problem you are given 2 SOLUBLE SOLUTIONS, which when mixed will give one compound of LOW SOLUBILITY. You are asked to say whether a ppt will form or not using calculations.

5-Predicting Precipitates using Trial Ksp In this type of problem you are given 2 SOLUBLE SOLUTIONS, which when mixed will give one compound of LOW SOLUBILITY. You are asked to say whether a ppt will form or not using calculations. 1. First, you would find which combination of the 4 ions gives you a compound of low solubility (use your solubility table or Ksp table)

5-Predicting Precipitates using Trial Ksp In this type of problem you are given 2 SOLUBLE SOLUTIONS, which when mixed will give one compound of LOW SOLUBILITY. You are asked to say whether a ppt will form or not using calculations. 1. First, you would find which combination of the 4 ions gives you a compound of low solubility (use your solubility table or Ksp table) 2. Next, use the DILUTION FORMULA to find the concentration of each of the ions of that compound right after mixing.

5-Predicting Precipitates using Trial Ksp In this type of problem you are given 2 SOLUBLE SOLUTIONS, which when mixed will give one compound of LOW SOLUBILITY. You are asked to say whether a ppt will form or not using calculations. 1. First, you would find which combination of the 4 ions gives you a compound of low solubility (use your solubility table or Ksp table) 2. Next, use the DILUTION FORMULA to find the concentration of each of the ions of that compound right after mixing. 3. Write the equilibrium equation and Ksp expression for the compound in question.

5-Predicting Precipitates using Trial Ksp In this type of problem you are given 2 SOLUBLE SOLUTIONS, which when mixed will give one compound of LOW SOLUBILITY. You are asked to say whether a ppt will form or not using calculations. 1. First, you would find which combination of the 4 ions gives you a compound of low solubility (use your solubility table or Ksp table) 2. Next, use the DILUTION FORMULA to find the concentration of each of the ions of that compound right after mixing. 3. Write the equilibrium equation and Ksp expression for the compound in question. 4. Remember, these ions are coming from DIFFERENT SOURCES, so the coefficients in the equilibrium equation have NOTHING TO DO WITH their concentrations.(Coefficients ARE exponents in the Ksp expression)

5-Predicting Precipitates using Trial Ksp In this type of problem you are given 2 SOLUBLE SOLUTIONS, which when mixed will give one compound of LOW SOLUBILITY. You are asked to say whether a ppt will form or not using calculations. 1. First, you would find which combination of the 4 ions gives you a compound of low solubility (use your solubility table or Ksp table) 2. Next, use the DILUTION FORMULA to find the concentration of each of the ions of that compound right after mixing. 3. Write the equilibrium equation and Ksp expression for the compound in question. 4. Remember, these ions are coming from DIFFERENT SOURCES, so the coefficients in the equilibrium equation have NOTHING TO DO WITH their concentrations.(Coefficients ARE exponents in the Ksp expression) 5. In your Ksp expression, write “Trial” in front of Ksp. Plug in the values of the ion concentrations from step 2 and calculate the “Trial Ksp”

5-Predicting Precipitates using Trial Ksp In this type of problem you are given 2 SOLUBLE SOLUTIONS, which when mixed will give one compound of LOW SOLUBILITY. You are asked to say whether a ppt will form or not using calculations. 1. First, you would find which combination of the 4 ions gives you a compound of low solubility (use your solubility table or Ksp table) 2. Next, use the DILUTION FORMULA to find the concentration of each of the ions of that compound right after mixing. 3. Write the equilibrium equation and Ksp expression for the compound in question. 4. Remember, these ions are coming from DIFFERENT SOURCES, so the coefficients in the equilibrium equation have NOTHING TO DO WITH their concentrations.(Coefficients ARE exponents in the Ksp expression) 5. In your Ksp expression, write “Trial” in front of Ksp. Plug in the values of the ion concentrations from step 2 and calculate the “Trial Ksp” 6. Compare the “Trial Ksp” to the Actual Ksp (look it up) If Trial Ksp > Ksp, a ppt WILL form. If Trial Ksp < Ksp there is NO ppt.

Here’s an Example: 200.0 mL of 2.5 x 10 -4 M Pb(NO 3 ) 2 is mixed with 300.0 mL of 3.0 x 10 -3 M CaCl 2. Use calculations to determine whether a precipitate will form.

Using the solubility table, the possible precipitate (compound with LOW Solubility) here is PbCl 2. So we calculate the [Pb 2+ ] and the [Cl - ] right after mixing:

Here’s an Example: 200.0 mL of 2.5 x 10 -4 M Pb(NO 3 ) 2 is mixed with 300.0 mL of 3.0 x 10 -3 M CaCl 2. Use calculations to determine whether a precipitate will form. Using the solubility table, the possible precipitate (compound with LOW Solubility) here is PbCl 2. So we calculate the [Pb 2+ ] and the [Cl - ] right after mixing:

Here’s an Example: 200.0 mL of 2.5 x 10 -4 M Pb(NO 3 ) 2 is mixed with 300.0 mL of 3.0 x 10 -3 M CaCl 2. Use calculations to determine whether a precipitate will form. Using the solubility table, the possible precipitate (compound with LOW Solubility) here is PbCl 2. So we calculate the [Pb 2+ ] and the [Cl - ] right after mixing: Since the Cl - comes from CaCl 2 the initial [Cl - ] from this solution will be 3.0 x 10 -3 M x 2 = 6.0 x 10 -3 M:

Here’s an Example: 200.0 mL of 2.5 x 10 -4 M Pb(NO 3 ) 2 is mixed with 300.0 mL of 3.0 x 10 -3 M CaCl 2. Use calculations to determine whether a precipitate will form. Using the solubility table, the possible precipitate (compound with LOW Solubility) here is PbCl 2. So we calculate the [Pb 2+ ] and the [Cl - ] right after mixing: Since the Cl - comes from CaCl 2 the initial [Cl - ] from this solution will be 3.0 x 10 -3 M x 2 = 6.0 x 10 -3 M: Now we write the equilibrium equation and the Ksp expression for the PbCl 2 :

Here’s an Example: 200.0 mL of 2.5 x 10 -4 M Pb(NO 3 ) 2 is mixed with 300.0 mL of 3.0 x 10 -3 M CaCl 2. Use calculations to determine whether a precipitate will form. Using the solubility table, the possible precipitate (compound with LOW Solubility) here is PbCl 2. So we calculate the [Pb 2+ ] and the [Cl - ] right after mixing: Since the Cl - comes from CaCl 2 the initial [Cl - ] from this solution will be 3.0 x 10 -3 M x 2 = 6.0 x 10 -3 M: Now we write the equilibrium equation and the Trial Ksp expression for the PbCl 2 : PbCl 2(s)  Pb 2+ (aq) + 2Cl - (aq) Trial Ksp = [Pb 2+ ] [Cl - ] 2

Here’s an Example: 200.0 mL of 2.5 x 10 -4 M Pb(NO 3 ) 2 is mixed with 300.0 mL of 3.0 x 10 -3 M CaCl 2. Use calculations to determine whether a precipitate will form. Using the solubility table, the possible precipitate (compound with LOW Solubility) here is PbCl 2. So we calculate the [Pb 2+ ] and the [Cl - ] right after mixing: Since the Cl - comes from CaCl 2 the initial [Cl - ] from this solution will be 3.0 x 10 -3 M x 2 = 6.0 x 10 -3 M: Now we write the equilibrium equation and the Trial Ksp expression for the PbCl 2 : PbCl 2(s)  Pb 2+ (aq) + 2Cl - (aq) Trial Ksp = [Pb 2+ ] [Cl - ] 2 Trial Ksp = [Pb 2+ ] [Cl - ] 2 = (1.0 x 10 -4 )(3.6 x 10 -3)2 = 1.3 x 10 -9

Here’s an Example: 200.0 mL of 2.5 x 10 -4 M Pb(NO 3 ) 2 is mixed with 300.0 mL of 3.0 x 10 -3 M CaCl 2. Use calculations to determine whether a precipitate will form. Using the solubility table, the possible precipitate (compound with LOW Solubility) here is PbCl 2. So we calculate the [Pb 2+ ] and the [Cl - ] right after mixing: Since the Cl - comes from CaCl 2 the initial [Cl - ] from this solution will be 3.0 x 10 -3 M x 2 = 6.0 x 10 -3 M: Now we write the equilibrium equation and the Trial Ksp expression for the PbCl 2 : PbCl 2(s)  Pb 2+ (aq) + 2Cl - (aq) Trial Ksp = [Pb 2+ ] [Cl - ] 2 Trial Ksp = [Pb 2+ ] [Cl - ] 2 = (1.0 x 10 -4 )(3.6 x 10 -3)2 = 1.3 x 10 -9 The actual Ksp = 1.2 x 10 -5 so Trial Ksp < Ksp and a Precipitate will NOT form.

6Finding Maximum Concentration of an ion in a solution in which another ion is present. When you read this type of question, it should be apparent that the ions of the Low Solubility Compound are coming from DIFFERENT SOURCES, and it is NOT just about the one compound (as it was in the 2 nd and 3 rd type we looked at). You will be given some combination of the following information: -the concentration of one ion or of a soluble compound containing that ion. You will be asked: -the maximum concentration possible of another ion (this ion and the first one would form a LOW SOLUBILITY compound) or -the maximum concentration of a soluble compound containing the other ion or -the maximum mass of a compound containing the other ion that can be added without forming a ppt.

6Finding Maximum Concentration of an ion in a solution in which another ion is present. When you read this type of question, it should be apparent that the ions of the Low Solubility Compound are coming from DIFFERENT SOURCES, and it is NOT just about the one compound (as it was in the 2 nd and 3 rd type we looked at). You will be given some combination of the following information: -the concentration of one ion or of a soluble compound containing that ion. You will be asked: -the maximum concentration possible of another ion (this ion and the first one would form a LOW SOLUBILITY compound) or -the maximum concentration of a soluble compound containing the other ion or -the maximum mass of a compound containing the other ion that can be added without forming a ppt. The solution for this type of problem involves:

6Finding Maximum Concentration of an ion in a solution in which another ion is present. When you read this type of question, it should be apparent that the ions of the Low Solubility Compound are coming from DIFFERENT SOURCES, and it is NOT just about the one compound (as it was in the 2 nd and 3 rd type we looked at). You will be given some combination of the following information: -the concentration of one ion or of a soluble compound containing that ion. You will be asked: -the maximum concentration possible of another ion (this ion and the first one would form a LOW SOLUBILITY compound) or -the maximum concentration of a soluble compound containing the other ion or -the maximum mass of a compound containing the other ion that can be added without forming a ppt. The solution for this type of problem involves: 1. Determining the compound of LOW SOLUBILITY which could be formed.

6Finding Maximum Concentration of an ion in a solution in which another ion is present. When you read this type of question, it should be apparent that the ions of the Low Solubility Compound are coming from DIFFERENT SOURCES, and it is NOT just about the one compound (as it was in the 2 nd and 3 rd type we looked at). You will be given some combination of the following information: -the concentration of one ion or of a soluble compound containing that ion. You will be asked: -the maximum concentration possible of another ion (this ion and the first one would form a LOW SOLUBILITY compound) or -the maximum concentration of a soluble compound containing the other ion or -the maximum mass of a compound containing the other ion that can be added without forming a ppt. The solution for this type of problem involves: 1. Determining the compound of LOW SOLUBILITY which could be formed. 2. Writing the equilibrium and the Ksp expression for that compound

6Finding Maximum Concentration of an ion in a solution in which another ion is present. When you read this type of question, it should be apparent that the ions of the Low Solubility Compound are coming from DIFFERENT SOURCES, and it is NOT just about the one compound (as it was in the 2 nd and 3 rd type we looked at). You will be given some combination of the following information: -the concentration of one ion or of a soluble compound containing that ion. You will be asked: -the maximum concentration possible of another ion (this ion and the first one would form a LOW SOLUBILITY compound) or -the maximum concentration of a soluble compound containing the other ion or -the maximum mass of a compound containing the other ion that can be added without forming a ppt. The solution for this type of problem involves: 1. Determining the compound of LOW SOLUBILITY which could be formed. 2. Writing the equilibrium and the Ksp expression for that compound. 3. Looking up the Ksp, plugging in the conc. of the ion you know.

6Finding Maximum Concentration of an ion in a solution in which another ion is present. When you read this type of question, it should be apparent that the ions of the Low Solubility Compound are coming from DIFFERENT SOURCES, and it is NOT just about the one compound (as it was in the 2 nd and 3 rd type we looked at). You will be given some combination of the following information: -the concentration of one ion or of a soluble compound containing that ion. You will be asked: -the maximum concentration possible of another ion (this ion and the first one would form a LOW SOLUBILITY compound) or -the maximum concentration of a soluble compound containing the other ion or -the maximum mass of a compound containing the other ion that can be added without forming a ppt. The solution for this type of problem involves: 1. Determining the compound of LOW SOLUBILITY which could be formed. 2. Writing the equilibrium and the Ksp expression for that compound. 3. Looking up the Ksp, plugging in the conc. of the ion you know. 4. Solving for the conc. of the other ion and doing anything else necessary to answer the question.

Here is an example question: Calculate the maximum mass of Na 2 SO 4 which can be added to 400.0 mL of a 0.0050M solution of Sr(NO 3 ) 2 without forming a precipitate.

The compound of Low Solubility here is SrSO 4 and the Sr 2+ ions and the SO 4 2- ions are coming from DIFFERENT sources. (We are NOT trying to find the solubility of SrSO 4 here so DON’T solve for “s”)!

Here is an example question: Calculate the maximum mass of Na 2 SO 4 which can be added to 400.0 mL of a 0.0050M solution of Sr(NO 3 ) 2 without forming a precipitate. The compound of Low Solubility here is SrSO 4 and the Sr 2+ ions and the SO 4 2- ions are coming from DIFFERENT sources. (We are NOT trying to find the solubility of SrSO 4 here so DON’T solve for “s”)! We write the equilibrium equation and the Ksp expression for SrSO 4 :

Here is an example question: Calculate the maximum mass of Na 2 SO 4 which can be added to 400.0 mL of a 0.0050M solution of Sr(NO 3 ) 2 without forming a precipitate. The compound of Low Solubility here is SrSO 4 and the Sr 2+ ions and the SO 4 2- ions are coming from DIFFERENT sources. (We are NOT trying to find the solubility of SrSO 4 here so DON’T solve for “s”)! We write the equilibrium equation and the Ksp expression for SrSO 4 : SrSO 4(s)  Sr 2+ (aq) + SO 4 2- (aq) Ksp = [Sr 2+ ] [SO 4 2- ]

Here is an example question: Calculate the maximum mass of Na 2 SO 4 which can be added to 400.0 mL of a 0.0050M solution of Sr(NO 3 ) 2 without forming a precipitate. The compound of Low Solubility here is SrSO 4 and the Sr 2+ ions and the SO 4 2- ions are coming from DIFFERENT sources. (We are NOT trying to find the solubility of SrSO 4 here so DON’T solve for “s”)! We write the equilibrium equation and the Ksp expression for SrSO 4 : SrSO 4(s)  Sr 2+ (aq) + SO 4 2- (aq) Ksp = [Sr 2+ ] [SO 4 2- ] We know the Sr(NO 3 ) 2 is soluble and the [Sr 2+ ] is 0.0050M

Here is an example question: Calculate the maximum mass of Na 2 SO 4 which can be added to 400.0 mL of a 0.0050M solution of Sr(NO 3 ) 2 without forming a precipitate. The compound of Low Solubility here is SrSO 4 and the Sr 2+ ions and the SO 4 2- ions are coming from DIFFERENT sources. (We are NOT trying to find the solubility of SrSO 4 here so DON’T solve for “s”)! We write the equilibrium equation and the Ksp expression for SrSO 4 : SrSO 4(s)  Sr 2+ (aq) + SO 4 2- (aq) Ksp = [Sr 2+ ] [SO 4 2- ] We know the Sr(NO 3 ) 2 is soluble and the [Sr 2+ ] is 0.0050M so we rearrange the Ksp expression and solve for [SO 4 2- ]:

Here is an example question: Calculate the maximum mass of Na 2 SO 4 which can be added to 400.0 mL of a 0.0050M solution of Sr(NO 3 ) 2 without forming a precipitate. The compound of Low Solubility here is SrSO 4 and the Sr 2+ ions and the SO 4 2- ions are coming from DIFFERENT sources. (We are NOT trying to find the solubility of SrSO 4 here so DON’T solve for “s”)! We write the equilibrium equation and the Ksp expression for SrSO 4 : SrSO 4(s)  Sr 2+ (aq) + SO 4 2- (aq) Ksp = [Sr 2+ ] [SO 4 2- ] We know the Sr(NO 3 ) 2 is soluble and the [Sr 2+ ] is 0.0050M so we rearrange the Ksp expression and solve for [SO 4 2- ]: The Na 2 SO 4 contains 1 SO 4 so [Na 2 SO 4 ] = [SO 4 2- ].

Here is an example question: Calculate the maximum mass of Na 2 SO 4 which can be added to 400.0 mL of a 0.0050M solution of Sr(NO 3 ) 2 without forming a precipitate. The compound of Low Solubility here is SrSO 4 and the Sr 2+ ions and the SO 4 2- ions are coming from DIFFERENT sources. (We are NOT trying to find the solubility of SrSO 4 here so DON’T solve for “s”)! We write the equilibrium equation and the Ksp expression for SrSO 4 : SrSO 4(s)  Sr 2+ (aq) + SO 4 2- (aq) Ksp = [Sr 2+ ] [SO 4 2- ] We know the Sr(NO 3 ) 2 is soluble and the [Sr 2+ ] is 0.0050M so we rearrange the Ksp expression and solve for [SO 4 2- ]: The Na 2 SO 4 contains 1 SO 4 so [Na 2 SO 4 ] = [SO 4 2- ]. We go from M  mol  g

Here is an example question: Calculate the maximum mass of Na 2 SO 4 which can be added to 400.0 mL of a 0.0050M solution of Sr(NO 3 ) 2 without forming a precipitate. The compound of Low Solubility here is SrSO 4 and the Sr 2+ ions and the SO 4 2- ions are coming from DIFFERENT sources. (We are NOT trying to find the solubility of SrSO 4 here so DON’T solve for “s”)! We write the equilibrium equation and the Ksp expression for SrSO 4 : SrSO 4(s)  Sr 2+ (aq) + SO 4 2- (aq) Ksp = [Sr 2+ ] [SO 4 2- ] We know the Sr(NO 3 ) 2 is soluble and the [Sr 2+ ] is 0.0050M so we rearrange the Ksp expression and solve for [SO 4 2- ]: The Na 2 SO 4 contains 1 SO 4 so [Na 2 SO 4 ] = [SO 4 2- ]. We go from M  mol  g Mol Na 2 SO 4 = M x L = 6.8 x 10 -5 M x 0.4000L = 2.72 x 10 -5 mol Na 2 SO 4

Here is an example question: Calculate the maximum mass of Na 2 SO 4 which can be added to 400.0 mL of a 0.0050M solution of Sr(NO 3 ) 2 without forming a precipitate. The compound of Low Solubility here is SrSO 4 and the Sr 2+ ions and the SO 4 2- ions are coming from DIFFERENT sources. (We are NOT trying to find the solubility of SrSO 4 here so DON’T solve for “s”)! We write the equilibrium equation and the Ksp expression for SrSO 4 : SrSO 4(s)  Sr 2+ (aq) + SO 4 2- (aq) Ksp = [Sr 2+ ] [SO 4 2- ] We know the Sr(NO 3 ) 2 is soluble and the [Sr 2+ ] is 0.0050M so we rearrange the Ksp expression and solve for [SO 4 2- ]: The Na 2 SO 4 contains 1 SO 4 so [Na 2 SO 4 ] = [SO 4 2 ]. We go from M  mol  g Mol Na 2 SO 4 = M x L = 6.8 x 10 -5 M x 0.4000L = 2.72 x 10 -5 mol Na 2 SO 4 2.72 x 10 -5 mol Na 2 SO 4 x Na 2 SO 4

7-Finding Minimum Concentration of an ion necessary to just start precipitation This is exactly the same type of problem as type 6 (Finding Maximum Concentration of an ion in a solution in which another ion is present.) The calculations are done exactly the same way.

Example: Find the minimum [CO 3 2- ] necessary to just start precipitation in a solution in which [Ag + ] = 3.6 x 10 -6 M

The compound of low solubility here is Ag 2 CO 3.

Example: Find the minimum [CO 3 2- ] necessary to just start precipitation in a solution in which [Ag + ] = 3.6 x 10 -6 M The compound of low solubility here is Ag 2 CO 3. We can look up the Ksp and we are given the [Ag + ].

Example: Find the minimum [CO 3 2- ] necessary to just start precipitation in a solution in which [Ag + ] = 3.6 x 10 -6 M The compound of low solubility here is Ag 2 CO 3. We can look up the Ksp and we are given the [Ag + ]. We write the equilibrium equation and the Ksp expression for Ag 2 CO 3.

Example: Find the minimum [CO 3 2- ] necessary to just start precipitation in a solution in which [Ag + ] = 3.6 x 10 -6 M The compound of low solubility here is Ag 2 CO 3. We can look up the Ksp and we are given the [Ag + ]. We write the equilibrium equation and the Ksp expression for Ag 2 CO 3. Ag 2 CO 3(s)  2Ag + (aq) + CO 3 2- (aq) Ksp = [Ag + ] 2 [CO 3 2- ]

Example: Find the minimum [CO 3 2- ] necessary to just start precipitation in a solution in which [Ag + ] = 3.6 x 10 -6 M The compound of low solubility here is Ag 2 CO 3. We can look up the Ksp and we are given the [Ag + ]. We write the equilibrium equation and the Ksp expression for Ag 2 CO 3. Ag 2 CO 3(s)  2Ag + (aq) + CO 3 2- (aq) Ksp = [Ag + ] 2 [CO 3 2- ] Solve the Ksp expression for [CO 3 2- ]

Example: Find the minimum [CO 3 2- ] necessary to just start precipitation in a solution in which [Ag + ] = 3.6 x 10 -6 M The compound of low solubility here is Ag 2 CO 3. We can look up the Ksp and we are given the [Ag + ]. We write the equilibrium equation and the Ksp expression for Ag 2 CO 3. Ag 2 CO 3(s)  2Ag + (aq) + CO 3 2- (aq) Ksp = [Ag + ] 2 [CO 3 2- ] Solve the Ksp expression for [CO 3 2- ] Look up the Ksp and substitute 3.6 x 10 -6 in for [Ag + ]:

8-Finding Which Precipitate will form First In this type of problem, an ion, or a solution containing a certain ion is added slowly to another solution which contains 2 ions that will precipitate with the added one. You are asked to determine which precipitate forms first.

8-Finding Which Precipitate will form First In this type of problem, an ion, or a solution containing a certain ion is added slowly to another solution which contains 2 ions that will precipitate with the added one. You are asked to determine which precipitate forms first. In this type of problem, it’s best to use an example:

8-Finding Which Precipitate will form First In this type of problem, an ion, or a solution containing a certain ion is added slowly to another solution which contains 2 ions that will precipitate with the added one. You are asked to determine which precipitate forms first. In this type of problem, it’s best to use an example: A solution of 1.0M Sr(NO 3 ) 3 is added dropwise to a solution containing 0.10M KF and 0.10M Na 2 SO 4. Which precipitate will form first?

8-Finding Which Precipitate will form First In this type of problem, an ion, or a solution containing a certain ion is added slowly to another solution which contains 2 ions that will precipitate with the added one. You are asked to determine which precipitate forms first. In this type of problem, it’s best to use an example: A solution of 1.0M Sr(NO 3 ) 3 is added dropwise to a solution containing 0.10M KF and 0.10M Na 2 SO 4. Which precipitate will form first? First, we eliminate all spectators and look at the actual ions we’re dealing with. Also the 1.0M of the Sr(NO 3 ) 3 is irrelevant since it is added dropwise to a solution with a larger volume.

8-Finding Which Precipitate will form First In this type of problem, an ion, or a solution containing a certain ion is added slowly to another solution which contains 2 ions that will precipitate with the added one. You are asked to determine which precipitate forms first. In this type of problem, it’s best to use an example: A solution of 1.0M Sr(NO 3 ) 3 is added dropwise to a solution containing 0.10M KF and 0.10M Na 2 SO 4. Which precipitate will form first? First, we eliminate all spectators and look at the actual ions we’re dealing with. Also the 1.0M of the Sr(NO 3 ) 3 is irrelevant since it is added dropwise to a solution with a larger volume. So we can re-state it like this:

8-Finding Which Precipitate will form First In this type of problem, an ion, or a solution containing a certain ion is added slowly to another solution which contains 2 ions that will precipitate with the added one. You are asked to determine which precipitate forms first. In this type of problem, it’s best to use an example: A solution of 1.0M Sr(NO 3 ) 3 is added dropwise to a solution containing 0.10M KF and 0.10M Na 2 SO 4. Which precipitate will form first? First, we eliminate all spectators and look at the actual ions we’re dealing with. Also the 1.0M of the Sr(NO 3 ) 3 is irrelevant since it is added dropwise to a solution with a larger volume. So we can re-state it like this: Sr 2+ is added dropwise to a solution containing 0.10M F - and 0.10M SO 4 2-. Which precipitate will form first?

8-Finding Which Precipitate will form First In this type of problem, an ion, or a solution containing a certain ion is added slowly to another solution which contains 2 ions that will precipitate with the added one. You are asked to determine which precipitate forms first. In this type of problem, it’s best to use an example: A solution of 1.0M Sr(NO 3 ) 3 is added dropwise to a solution containing 0.10M KF and 0.10M Na 2 SO 4. Which precipitate will form first? First, we eliminate all spectators and look at the actual ions we’re dealing with. Also the 1.0M of the Sr(NO 3 ) 3 is irrelevant since it is added dropwise to a solution with a larger volume. So we can re-state it like this: Sr 2+ is added dropwise to a solution containing 0.10M F - and 0.10M SO 4 2-. Which precipitate will form first? We calculate the [Sr 2+ ] necessary to start precipitation of each precipitate (SrF 2 and SrSO 4 )

8-Finding Which Precipitate will form First In this type of problem, an ion, or a solution containing a certain ion is added slowly to another solution which contains 2 ions that will precipitate with the added one. You are asked to determine which precipitate forms first. In this type of problem, it’s best to use an example: A solution of 1.0M Sr(NO 3 ) 3 is added dropwise to a solution containing 0.10M KF and 0.10M Na 2 SO 4. Which precipitate will form first? First, we eliminate all spectators and look at the actual ions we’re dealing with. Also the 1.0M of the Sr(NO 3 ) 3 is irrelevant since it is added dropwise to a solution with a larger volume. So we can re-state it like this: Sr 2+ is added dropwise to a solution containing 0.10M F - and 0.10M SO 4 2-. Which precipitate will form first? We calculate the [Sr 2+ ] necessary to start precipitation of each precipitate (SrF 2 and SrSO 4 ) Whichever precipitate requires the LESSER [Sr 2+ ] will form first (as [Sr 2+ ] gradually increases as it is added.)

Sr 2+ is added dropwise to a solution containing 0.10M F - and 0.10M SO 4 2-. Which precipitate will form first? First we find [Sr 2+ ] necessary to start formation of SrF 2 :

Sr 2+ is added dropwise to a solution containing 0.10M F - and 0.10M SO 4 2-. Which precipitate will form first? First we find [Sr 2+ ] necessary to start formation of SrF 2 : We write the equilibrium equation and Ksp expression for SrF 2 :

Sr 2+ is added dropwise to a solution containing 0.10M F - and 0.10M SO 4 2-. Which precipitate will form first? First we find [Sr 2+ ] necessary to start formation of SrF 2 : We write the equilibrium equation and Ksp expression for SrF 2 : SrF 2(s)  Sr 2+ (aq) + 2F - (aq)

Sr 2+ is added dropwise to a solution containing 0.10M F - and 0.10M SO 4 2-. Which precipitate will form first? First we find [Sr 2+ ] necessary to start formation of SrF 2 : We write the equilibrium equation and Ksp expression for SrF 2 : SrF 2(s)  Sr 2+ (aq) + 2F - (aq) Ksp = [Sr 2+ ] [F - ] 2

Sr 2+ is added dropwise to a solution containing 0.10M F - and 0.10M SO 4 2-. Which precipitate will form first? First we find [Sr 2+ ] necessary to start formation of SrF 2 : We write the equilibrium equation and Ksp expression for SrF 2 : SrF 2(s)  Sr 2+ (aq) + 2F - (aq) Ksp = [Sr 2+ ] [F - ] 2 We now solve for [Sr 2+ ] :

Sr 2+ is added dropwise to a solution containing 0.10M F - and 0.10M SO 4 2-. Which precipitate will form first? First we find [Sr 2+ ] necessary to start formation of SrF 2 : We write the equilibrium equation and Ksp expression for SrF 2 : SrF 2(s)  Sr 2+ (aq) + 2F - (aq) Ksp = [Sr 2+ ] [F - ] 2 We now solve for [Sr 2+ ] : Next we find [Sr 2+ ] necessary to start formation of SrSO 4 : SrSO 4(s)  Sr 2+ (aq) + SO 4 2- (aq) Ksp = [Sr 2+ ] [SO 4 2- ]

Sr 2+ is added dropwise to a solution containing 0.10M F - and 0.10M SO 4 2-. Which precipitate will form first? First we find [Sr 2+ ] necessary to start formation of SrF 2 : We write the equilibrium equation and Ksp expression for SrF 2 : SrF 2(s)  Sr 2+ (aq) + 2F - (aq) Ksp = [Sr 2+ ] [F - ] 2 We now solve for [Sr 2+ ] : Next we find [Sr 2+ ] necessary to start formation of SrSO 4 : SrSO 4(s)  Sr 2+ (aq) + SO 4 2- (aq) Ksp = [Sr 2+ ] [SO 4 2- ] So, since the SrF 2 requires the LESSER concentration of Sr 2+ (4.3 x 10 - 7 M) the precipitate SrF 2 will form first.

9-Using Titrations to find Unknown Concentration

The following diagram applies to Titration questions in unit 3: Moles of Standard Moles of Sample Mole bridge M x L of Standard M or L of Sample

9-Using Titrations to find Unknown Concentration The following diagram applies to Titration questions in unit 3: Moles of Standard Moles of Sample Mole bridge M x L of Standard M or L of Sample For Example: It takes 16.45 mL of 0.100M AgNO 3 to titrate a 50.0 mL sample containing Br - ions. Calculate the [ Br - ] in the sample.

9-Using Titrations to find Unknown Concentration The following diagram applies to Titration questions in unit 3: Moles of Standard Moles of Sample Mole bridge M x L of Standard M or L of Sample For Example: It takes 16.45 mL of 0.100M AgNO 3 to titrate a 50.0 mL sample containing Br - ions. Calculate the [ Br - ] in the sample. DON’T use anything silly like the dilution formula or Ksp’s here! Simply think MOLES!!!

9-Using Titrations to find Unknown Concentration The following diagram applies to Titration questions in unit 3: Moles of Standard Moles of Sample Mole bridge M x L of Standard M or L of Sample For Example: It takes 16.45 mL of 0.100M AgNO 3 to titrate a 50.0 mL sample containing Br - ions. Calculate the [ Br - ] in the sample. DON’T use anything silly like the dilution formula or Ksp’s here! Simply think MOLES!!! The solution that you can find the moles of is the Standard Solution (in this case the AgNO 3 because you are given the volume and the concentration.)

9-Using Titrations to find Unknown Concentration The following diagram applies to Titration questions in unit 3: Moles of Standard Moles of Sample Mole bridge M x L of Standard M or L of Sample For Example: It takes 16.45 mL of 0.100M AgNO 3 to titrate a 50.0 mL sample containing Br - ions. Calculate the [ Br - ] in the sample. DON’T use anything silly like the dilution formula or Ksp’s here! Simply think MOLES!!! The solution that you can find the moles of is the Standard Solution (in this case the AgNO 3 because you are given the volume and the concentration.) So we calculate the MOLES of AgNO 3 (or Ag + )

9-Using Titrations to find Unknown Concentration The following diagram applies to Titration questions in unit 3: Moles of Standard Moles of Sample Mole bridge M x L of Standard M or L of Sample For Example: It takes 16.45 mL of 0.100M AgNO 3 to titrate a 50.0 mL sample containing Br - ions. Calculate the [ Br - ] in the sample. DON’T use anything silly like the dilution formula or Ksp’s here! Simply think MOLES!!! The solution that you can find the moles of is the Standard Solution (in this case the AgNO 3 because you are given the volume and the concentration.) So we calculate the MOLES of AgNO 3 (or Ag + ) Moles Ag + = M x L = 0.100M x 0.01645L = 0.001645 mol Ag +

9-Using Titrations to find Unknown Concentration The following diagram applies to Titration questions in unit 3: Moles of Standard Moles of Sample Mole bridge M x L of Standard M or L of Sample For Example: It takes 16.45 mL of 0.100M AgNO 3 to titrate a 50.0 mL sample containing Br - ions. Calculate the [ Br - ] in the sample. DON’T use anything silly like the dilution formula or Ksp’s here! Simply think MOLES!!! The solution that you can find the moles of is the Standard Solution (in this case the AgNO 3 because you are given the volume and the concentration.) So we calculate the MOLES of AgNO 3 (or Ag + ) Moles Ag + = M x L = 0.100M x 0.01645L = 0.001645 mol Ag + Next, we write the balanced Net-Ionic Equation for the precipitation reaction: (see the next slide..)

Moles Ag + = M x L = 0.100M x 0.01645L = 0.001645 mol Ag + Next, we write the balanced Net-Ionic Equation for the precipitation reaction:

Moles Ag + = M x L = 0.100M x 0.01645L = 0.001645 mol Ag + Next, we write the balanced Net-Ionic Equation for the precipitation reaction: Ag + + Br -  AgBr (s)

Moles Ag + = M x L = 0.100M x 0.01645L = 0.001645 mol Ag + Next, we write the balanced Net-Ionic Equation for the precipitation reaction: Ag + + Br -  AgBr (s) We use the coefficient ratio to find the MOLES of Br - in the sample:

Moles Ag + = M x L = 0.100M x 0.01645L = 0.001645 mol Ag + Next, we write the balanced Net-Ionic Equation for the precipitation reaction: Ag + + Br -  AgBr (s) We use the coefficient ratio to find the MOLES of Br - in the sample: 0.001645 mol Ag + x

Moles Ag + = M x L = 0.100M x 0.01645L = 0.001645 mol Ag + Next, we write the balanced Net-Ionic Equation for the precipitation reaction: Ag + + Br -  AgBr (s) We use the coefficient ratio to find the MOLES of Br - in the sample: 0.001645 mol Ag + x Now we use to calculate the [Br - ]. BE CAREFUL TO MAKE SURE THE “L” WE USE ARE OF THE Br - SOLUTION AND NOT THE AgNO 3 SOLUTION!!!

Moles Ag + = M x L = 0.100M x 0.01645L = 0.001645 mol Ag + Next, we write the balanced Net-Ionic Equation for the precipitation reaction: Ag + + Br -  AgBr (s) We use the coefficient ratio to find the MOLES of Br - in the sample: 0.001645 mol Ag + x Now we use to calculate the [Br - ]. BE CAREFUL TO MAKE SURE THE “L” WE USE ARE OF THE Br - SOLUTION AND NOT THE AgNO 3 SOLUTION!!! The Volume of the Sample with Br - was 50.0 mL

Chemistry 12 – Sorting out Solubility Problems This will help you with the following types of Solubility Problems: 1-ion concentrations in mixtures (no.

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Chemistry 12 – Sorting out Solubility Problems This will help you with the following types of Solubility Problems: 1-ion concentrations in mixtures (no.

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Presentation on theme: "Chemistry 12 – Sorting out Solubility Problems This will help you with the following types of Solubility Problems: 1-ion concentrations in mixtures (no."— Presentation transcript:

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