Transcript Sorting Out Solubility Problems - PowerPoint Presentation

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:
First-a couple of things to ALWAYS
remember:
1. Compounds with Nitrate (NO3) are
ALWAYS soluble (NO3- is ALWAYS a
spectator in unit 3)!
First-a couple of things to ALWAYS
remember:
1. Compounds with Nitrate (NO3) are
ALWAYS soluble (NO3- 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 (NO3) are
ALWAYS soluble (NO3- 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 CaSO4
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 CaSO4 … 65.363 g
3. Volume of saturated CaSO4 evaporated …… 20.0 mL
Using the data from this experiment, calculate the molar solubility of CaSO4 .
2
-Experimental determination of solubility
An example problem is: An experiment was done in which 20.0 mL of saturated CaSO4
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 CaSO4 … 65.363 g
3. Volume of saturated CaSO4 evaporated …… 20.0 mL
Using the data from this experiment, calculate the molar solubility of CaSO4 .
1. Subtract # 1 from #2 to find the grams of solid CaSO4 which dissolved
2
-Experimental determination of solubility
An example problem is: An experiment was done in which 20.0 mL of saturated CaSO4
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 CaSO4 … 65.363 g
3. Volume of saturated CaSO4 evaporated …… 20.0 mL
Using the data from this experiment, calculate the molar solubility of CaSO4 .
1. Subtract # 1 from #2 to find the grams of solid CaSO4 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 CaSO4
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 CaSO4 … 65.363 g
3. Volume of saturated CaSO4 evaporated …… 20.0 mL
Using the data from this experiment, calculate the molar solubility of CaSO4 .
1. Subtract # 1 from #2 to find the grams of solid CaSO4 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 CaSO4
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 CaSO4 … 65.363 g
3. Volume of saturated CaSO4 evaporated …… 20.0 mL
Using the data from this experiment, calculate the molar solubility of CaSO4 .
1. Subtract # 1 from #2 to find the grams of solid CaSO4 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
mol to find Molar concentration or Molar Solubility.
4. Use
M
L
2
-Experimental determination of solubility
An example problem is: An experiment was done in which 20.0 mL of saturated CaSO4
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 CaSO4 … 65.363 g
3. Volume of saturated CaSO4 evaporated …… 20.0 mL
Using the data from this experiment, calculate the molar solubility of CaSO4 .
1. Subtract # 1 from #2 to find the grams of solid CaSO4 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
mol to find Molar concentration or Molar Solubility.
4. Use
M
L
THE ANSWER IS 0.00844 M or 8.44 x 10-3M
2
-Experimental determination of solubility
An example problem is: An experiment was done in which 20.0 mL of saturated CaSO4
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 CaSO4 … 65.363 g
3. Volume of saturated CaSO4 evaporated …… 20.0 mL
Using the data from this experiment, calculate the molar solubility of CaSO4 .
1. Subtract # 1 from #2 to find the grams of solid CaSO4 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
mol to find Molar concentration or Molar Solubility.
4. Use
M
L
THE ANSWER IS 0.00844 M or 8.44 x 10-3M
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 [IO3-] in saturated Cu(IO3)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 [IO3-] in saturated Cu(IO3)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 [IO3-] in saturated Cu(IO3)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 [IO3-] in saturated Cu(IO3)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(IO3)2(s)  Cu2+(aq) + 2 IO3-(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 [IO3-] in saturated Cu(IO3)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(IO3)2(s)  Cu2+(aq) + 2 IO3-(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 [IO3-] in saturated Cu(IO3)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)
-s
Cu(IO3)2(s)  Cu2+(aq) + 2 IO3-(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 [IO3-] in saturated Cu(IO3)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)
-s
Cu(IO3)2(s)  Cu2+(aq) + 2 IO3-(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”
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 [IO3-] in saturated Cu(IO3)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)
-s
+s
+2s
Cu(IO3)2(s)  Cu2+(aq) + 2 IO3-(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”
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 [IO3-] in saturated Cu(IO3)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)
-s
+s
+2s
Cu(IO3)2(s)  Cu2+(aq) + 2 IO3-(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”
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 [IO3-] in saturated Cu(IO3)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)
-s
+s
+2s
Cu(IO3)2(s)  Cu2+(aq) + 2 IO3-(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”
4. Write the Ksp expression. Ksp = [Cu2+] [IO3-]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 [IO3-] in saturated Cu(IO3)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)
-s
+s
+2s
Cu(IO3)2(s)  Cu2+(aq) + 2 IO3-(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”
4. Write the Ksp expression. Ksp = [Cu2+] [IO3-]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 [IO3-] in saturated Cu(IO3)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)
-s
+s
+2s
Cu(IO3)2(s)  Cu2+(aq) + 2 IO3-(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”
4. Write the Ksp expression. Ksp = [Cu2+] [IO3-]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 [IO3-] in saturated Cu(IO3)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)
-s
+s
+2s
Cu(IO3)2(s)  Cu2+(aq) + 2 IO3-(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”
4. Write the Ksp expression. Ksp = [Cu2+] [IO3-]2
5. Substitute concentrations into Ksp expression. Ksp = [s] [2s]2
Ksp
6. Solve for s ( Ksp = 4s3 so
) (look up Ksp on Ksp table)
3
s
4
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 [IO3-] in saturated Cu(IO3)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)
-s
+s
+2s
Cu(IO3)2(s)  Cu2+(aq) + 2 IO3-(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”
4. Write the Ksp expression. Ksp = [Cu2+] [IO3-]2
5. Substitute concentrations into Ksp expression. Ksp = [s] [2s]2
Ksp
6. Solve for s ( Ksp = 4s3 so
) (look up Ksp on Ksp table)
3
s
4
7. When you find “s”, the [IO3-] = 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 [IO3-] in saturated Cu(IO3)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)
-s
+s
+2s
Cu(IO3)2(s)  Cu2+(aq) + 2 IO3-(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”
4. Write the Ksp expression. Ksp = [Cu2+] [IO3-]2
5. Substitute concentrations into Ksp expression. Ksp = [s] [2s]2
Ksp
6. Solve for s ( Ksp = 4s3 so
) (look up Ksp on Ksp table)
3
s
4
7. When you find “s”, the [IO3-] = 2s. ANSWER [IO3-] = 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:
mol
M L
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:
mol
M L
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 PbCl2 will dissolve in 300.0 mL of water
to form a saturated solution. Use this information to calculate the Ksp for PbCl2.
4
-Ksp from solubility (one compound)
For example: It is found that 1.20 grams of PbCl2 will dissolve in 300.0 mL of water
to form a saturated solution. Use this information to calculate the Ksp for PbCl2.
1. First, we would go g  mol  M
4
-Ksp from solubility (one compound)
For example: It is found that 1.20 grams of PbCl2 will dissolve in 300.0 mL of water
to form a saturated solution. Use this information to calculate the Ksp for PbCl2.
1. First, we would go g  mol  M
1mol
1.20 gx
 0.004313mol
278.2 g
0.00431mol
M
 0.01438M
0.3000 L
4
-Ksp from solubility (one compound)
For example: It is found that 1.20 grams of PbCl2 will dissolve in 300.0 mL of water
to form a saturated solution. Use this information to calculate the Ksp for PbCl2.
1. First, we would go g  mol  M
1mol
1.20 gx
 0.004313mol
278.2 g
0.00431mol
M
 0.01438M
0.3000 L
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 PbCl2 will dissolve in 300.0 mL of water
to form a saturated solution. Use this information to calculate the Ksp for PbCl2.
1. First, we would go g  mol  M
1mol
1.20 gx
 0.004313mol
278.2 g
0.00431mol
M
 0.01438M
0.3000 L
2. Write the equilibrium equation and write the molar solubility on top of
the solid.
-0.01438M
PbCl2(s) 
Pb2+(aq)
+
2Cl-(aq)
4
-Ksp from solubility (one compound)
For example: It is found that 1.20 grams of PbCl2 will dissolve in 300.0 mL of water
to form a saturated solution. Use this information to calculate the Ksp for PbCl2.
1. First, we would go g  mol  M
1mol
1.20 gx
 0.004313mol
278.2 g
0.00431mol
M
 0.01438M
0.3000 L
2. Write the equilibrium equation and write the molar solubility on top of
the solid.
-0.01438M
PbCl2(s) 
Pb2+(aq)
+
2Cl-(aq)
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 PbCl2 will dissolve in 300.0 mL of water
to form a saturated solution. Use this information to calculate the Ksp for PbCl2.
1. First, we would go g  mol  M
1mol
1.20 gx
 0.004313mol
278.2 g
0.00431mol
M
 0.01438M
0.3000 L
2. Write the equilibrium equation and write the molar solubility on top of
the solid.
-0.01438M + 0.01438M +0.02876M
PbCl2(s)  Pb2+(aq) + 2Cl-(aq)
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 PbCl2 will dissolve in 300.0 mL of water
to form a saturated solution. Use this information to calculate the Ksp for PbCl2.
1. First, we would go g  mol  M
1mol
1.20 gx
 0.004313mol
278.2 g
0.00431mol
M
 0.01438M
0.3000 L
2. Write the equilibrium equation and write the molar solubility on top of
the solid.
-0.01438M + 0.01438M +0.02876M
PbCl2(s)  Pb2+(aq) + 2Cl-(aq)
3. Use coefficients to find concentrations of the ions. Write them above.
4. Write Ksp expression and substitute these values in to calculate Ksp.
4
-Ksp from solubility (one compound)
For example: It is found that 1.20 grams of PbCl2 will dissolve in 300.0 mL of water
to form a saturated solution. Use this information to calculate the Ksp for PbCl2.
1. First, we would go g  mol  M
1mol
1.20 gx
 0.004313mol
278.2 g
0.00431mol
M
 0.01438M
0.3000 L
2. Write the equilibrium equation and write the molar solubility on top of
the solid.
-0.01438M + 0.01438M +0.02876M
PbCl2(s)  Pb2+(aq) + 2Cl-(aq)
3. Use coefficients to find concentrations of the ions. Write them above.
4. Write Ksp expression and substitute these values in to calculate Ksp.

Ksp  Pb
2
Cl   0.004310.02876
 2
2
 1.19 x105
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 10-4M Pb(NO3)2 is mixed with 300.0 mL of
3.0 x 10-3M CaCl2. Use calculations to determine whether a precipitate will form.
Here’s an Example: 200.0 mL of 2.5 x 10-4M Pb(NO3)2 is mixed with 300.0 mL of
3.0 x 10-3M CaCl2. Use calculations to determine whether a precipitate will form.
Using the solubility table, the possible precipitate (compound with LOW
Solubility) here is PbCl2. So we calculate the [Pb2+] and the [Cl-] right
after mixing:
Here’s an Example: 200.0 mL of 2.5 x 10-4M Pb(NO3)2 is mixed with 300.0 mL of
3.0 x 10-3M CaCl2. Use calculations to determine whether a precipitate will form.
Using the solubility table, the possible precipitate (compound with LOW
Solubility) here is PbCl2. So we calculate the [Pb2+] and the [Cl-] right
after mixing:
Pb   2.5 x 10
2
-4
x
200.0mL
 1.0 x10  4 M
500.0mL
Here’s an Example: 200.0 mL of 2.5 x 10-4M Pb(NO3)2 is mixed with 300.0 mL of
3.0 x 10-3M CaCl2. Use calculations to determine whether a precipitate will form.
Using the solubility table, the possible precipitate (compound with LOW
Solubility) here is PbCl2. So we calculate the [Pb2+] and the [Cl-] right
after mixing:
Pb   2.5 x 10
2
-4
x
200.0mL
 1.0 x10  4 M
500.0mL
Since the Cl- comes from CaCl2 the initial [Cl-] from this solution will be
3.0 x 10-3M x 2 = 6.0 x 10-3M:
Here’s an Example: 200.0 mL of 2.5 x 10-4M Pb(NO3)2 is mixed with 300.0 mL of
3.0 x 10-3M CaCl2. Use calculations to determine whether a precipitate will form.
Using the solubility table, the possible precipitate (compound with LOW
Solubility) here is PbCl2. So we calculate the [Pb2+] and the [Cl-] right
after mixing:
Pb   2.5 x 10
2
-4
x
200.0mL
 1.0 x10  4 M
500.0mL
Since the Cl- comes from CaCl2 the initial [Cl-] from this solution will be
3.0 x 10-3M x 2 = 6.0 x 10-3M:
Cl   6.0 x 10

-3
x
300.0mL
 3.6 x10 3 M
500.0mL
Here’s an Example: 200.0 mL of 2.5 x 10-4M Pb(NO3)2 is mixed with 300.0 mL of
3.0 x 10-3M CaCl2. Use calculations to determine whether a precipitate will form.
Using the solubility table, the possible precipitate (compound with LOW
Solubility) here is PbCl2. So we calculate the [Pb2+] and the [Cl-] right
after mixing:
Pb   2.5 x 10
2
-4
x
200.0mL
 1.0 x10  4 M
500.0mL
Since the Cl- comes from CaCl2 the initial [Cl-] from this solution will be
3.0 x 10-3M x 2 = 6.0 x 10-3M:
Cl   6.0 x 10

-3
x
300.0mL
 3.6 x10 3 M
500.0mL
Now we write the equilibrium equation and the Ksp expression for the
PbCl2:
Here’s an Example: 200.0 mL of 2.5 x 10-4M Pb(NO3)2 is mixed with 300.0 mL of
3.0 x 10-3M CaCl2. Use calculations to determine whether a precipitate will form.
Using the solubility table, the possible precipitate (compound with LOW
Solubility) here is PbCl2. So we calculate the [Pb2+] and the [Cl-] right
after mixing:
Pb 
2
200.0mL
 2.5 x 10 x
 1.0 x10  4 M
500.0mL
-4
Since the Cl- comes from CaCl2 the initial [Cl-] from this solution will be
3.0 x 10-3M x 2 = 6.0 x 10-3M:
 
300.0mL
Cl  6.0 x 10 x
 3.6 x10 3 M
500.0mL

-3
Now we write the equilibrium equation and the Trial Ksp expression for
the PbCl2:
PbCl2(s)  Pb2+(aq) + 2Cl-(aq)
Trial Ksp = [Pb2+] [Cl-]2
Here’s an Example: 200.0 mL of 2.5 x 10-4M Pb(NO3)2 is mixed with 300.0 mL of
3.0 x 10-3M CaCl2. Use calculations to determine whether a precipitate will form.
Using the solubility table, the possible precipitate (compound with LOW
Solubility) here is PbCl2. So we calculate the [Pb2+] and the [Cl-] right
after mixing:
Pb 
2
200.0mL
 2.5 x 10 x
 1.0 x10  4 M
500.0mL
-4
Since the Cl- comes from CaCl2 the initial [Cl-] from this solution will be
3.0 x 10-3M x 2 = 6.0 x 10-3M:
 
300.0mL
Cl  6.0 x 10 x
 3.6 x10 3 M
500.0mL

-3
Now we write the equilibrium equation and the Trial Ksp expression for
the PbCl2:
PbCl2(s)  Pb2+(aq) + 2Cl-(aq)
Trial Ksp = [Pb2+] [Cl-]2
Trial Ksp = [Pb2+] [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 10-4M Pb(NO3)2 is mixed with 300.0 mL of
3.0 x 10-3M CaCl2. Use calculations to determine whether a precipitate will form.
Using the solubility table, the possible precipitate (compound with LOW
Solubility) here is PbCl2. So we calculate the [Pb2+] and the [Cl-] right
after mixing:
Pb 
2
200.0mL
 2.5 x 10 x
 1.0 x10  4 M
500.0mL
-4
Since the Cl- comes from CaCl2 the initial [Cl-] from this solution will be
3.0 x 10-3M x 2 = 6.0 x 10-3M:
 
300.0mL
Cl  6.0 x 10 x
 3.6 x10 3 M
500.0mL

-3
Now we write the equilibrium equation and the Trial Ksp expression for
the PbCl2:
PbCl2(s)  Pb2+(aq) + 2Cl-(aq)
Trial Ksp = [Pb2+] [Cl-]2
Trial Ksp = [Pb2+] [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.
6
Finding 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 2nd and 3rd 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.
6
Finding 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 2nd and 3rd 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:
6
Finding 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 2nd and 3rd 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.
6
Finding 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 2nd and 3rd 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
6
Finding 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 2nd and 3rd 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.
6
Finding 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 2nd and 3rd 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 Na2SO4 which can be
added to 400.0 mL of a 0.0050M solution of Sr(NO3)2 without forming a precipitate.
Here is an example question: Calculate the maximum mass of Na2SO4 which can be
added to 400.0 mL of a 0.0050M solution of Sr(NO3)2 without forming a precipitate.
The compound of Low Solubility here is SrSO4 and the Sr2+ ions and the
SO42- ions are coming from DIFFERENT sources. (We are NOT trying to
find the solubility of SrSO4 here so DON’T solve for “s”)!
Here is an example question: Calculate the maximum mass of Na2SO4 which can be
added to 400.0 mL of a 0.0050M solution of Sr(NO3)2 without forming a precipitate.
The compound of Low Solubility here is SrSO4 and the Sr2+ ions and the
SO42- ions are coming from DIFFERENT sources. (We are NOT trying to
find the solubility of SrSO4 here so DON’T solve for “s”)!
We write the equilibrium equation and the Ksp expression for SrSO4:
Here is an example question: Calculate the maximum mass of Na2SO4 which can be
added to 400.0 mL of a 0.0050M solution of Sr(NO3)2 without forming a precipitate.
The compound of Low Solubility here is SrSO4 and the Sr2+ ions and the
SO42- ions are coming from DIFFERENT sources. (We are NOT trying to
find the solubility of SrSO4 here so DON’T solve for “s”)!
We write the equilibrium equation and the Ksp expression for SrSO4:
SrSO4(s)  Sr2+(aq) + SO42-(aq)
Ksp = [Sr2+] [SO42-]
Here is an example question: Calculate the maximum mass of Na2SO4 which can be
added to 400.0 mL of a 0.0050M solution of Sr(NO3)2 without forming a precipitate.
The compound of Low Solubility here is SrSO4 and the Sr2+ ions and the
SO42- ions are coming from DIFFERENT sources. (We are NOT trying to
find the solubility of SrSO4 here so DON’T solve for “s”)!
We write the equilibrium equation and the Ksp expression for SrSO4:
SrSO4(s)  Sr2+(aq) + SO42-(aq)
Ksp = [Sr2+] [SO42-]
We know the Sr(NO3)2 is soluble and the [Sr2+] is 0.0050M
Here is an example question: Calculate the maximum mass of Na2SO4 which can be
added to 400.0 mL of a 0.0050M solution of Sr(NO3)2 without forming a precipitate.
The compound of Low Solubility here is SrSO4 and the Sr2+ ions and the
SO42- ions are coming from DIFFERENT sources. (We are NOT trying to
find the solubility of SrSO4 here so DON’T solve for “s”)!
We write the equilibrium equation and the Ksp expression for SrSO4:
SrSO4(s)  Sr2+(aq) + SO42-(aq)
Ksp = [Sr2+] [SO42-]
We know the Sr(NO3)2 is soluble and the [Sr2+] is 0.0050M so we
rearrange the Ksp expression and solve for [SO42-]:
Here is an example question: Calculate the maximum mass of Na2SO4 which can be
added to 400.0 mL of a 0.0050M solution of Sr(NO3)2 without forming a precipitate.
The compound of Low Solubility here is SrSO4 and the Sr2+ ions and the
SO42- ions are coming from DIFFERENT sources. (We are NOT trying to
find the solubility of SrSO4 here so DON’T solve for “s”)!
We write the equilibrium equation and the Ksp expression for SrSO4:
SrSO4(s)  Sr2+(aq) + SO42-(aq)
Ksp = [Sr2+] [SO42-]
We know the Sr(NO3)2 is soluble and the [Sr2+] is 0.0050M so we
rearrange the Ksp expression and solve for [SO42-]:
SO   
2
4
Ksp 3.4 x107
5



6
.
8
x
10
M
2
Sr
0.0050
Here is an example question: Calculate the maximum mass of Na2SO4 which can be
added to 400.0 mL of a 0.0050M solution of Sr(NO3)2 without forming a precipitate.
The compound of Low Solubility here is SrSO4 and the Sr2+ ions and the
SO42- ions are coming from DIFFERENT sources. (We are NOT trying to
find the solubility of SrSO4 here so DON’T solve for “s”)!
We write the equilibrium equation and the Ksp expression for SrSO4:
SrSO4(s)  Sr2+(aq) + SO42-(aq)
Ksp = [Sr2+] [SO42-]
We know the Sr(NO3)2 is soluble and the [Sr2+] is 0.0050M so we
rearrange the Ksp expression and solve for [SO42-]:
SO   
2
4
Ksp 3.4 x107
5



6
.
8
x
10
M
2
Sr
0.0050
The Na2SO4 contains 1 SO4 so [Na2SO4] = [SO42-].
Here is an example question: Calculate the maximum mass of Na2SO4 which can be
added to 400.0 mL of a 0.0050M solution of Sr(NO3)2 without forming a precipitate.
The compound of Low Solubility here is SrSO4 and the Sr2+ ions and the
SO42- ions are coming from DIFFERENT sources. (We are NOT trying to
find the solubility of SrSO4 here so DON’T solve for “s”)!
We write the equilibrium equation and the Ksp expression for SrSO4:
SrSO4(s)  Sr2+(aq) + SO42-(aq)
Ksp = [Sr2+] [SO42-]
We know the Sr(NO3)2 is soluble and the [Sr2+] is 0.0050M so we
rearrange the Ksp expression and solve for [SO42-]:
SO   
2
4
Ksp 3.4 x107
5



6
.
8
x
10
M
2
Sr
0.0050
The Na2SO4 contains 1 SO4 so [Na2SO4] = [SO42-]. We go from Mmolg
Here is an example question: Calculate the maximum mass of Na2SO4 which can be
added to 400.0 mL of a 0.0050M solution of Sr(NO3)2 without forming a precipitate.
The compound of Low Solubility here is SrSO4 and the Sr2+ ions and the
SO42- ions are coming from DIFFERENT sources. (We are NOT trying to
find the solubility of SrSO4 here so DON’T solve for “s”)!
We write the equilibrium equation and the Ksp expression for SrSO4:
SrSO4(s)  Sr2+(aq) + SO42-(aq)
Ksp = [Sr2+] [SO42-]
We know the Sr(NO3)2 is soluble and the [Sr2+] is 0.0050M so we
rearrange the Ksp expression and solve for [SO42-]:
SO   
2
4
Ksp 3.4 x107
5



6
.
8
x
10
M
2
Sr
0.0050
The Na2SO4 contains 1 SO4 so [Na2SO4] = [SO42-]. We go from Mmolg
Mol Na2SO4 = M x L = 6.8 x 10-5M x 0.4000L = 2.72 x 10-5 mol Na2SO4
Here is an example question: Calculate the maximum mass of Na2SO4 which can be
added to 400.0 mL of a 0.0050M solution of Sr(NO3)2 without forming a precipitate.
The compound of Low Solubility here is SrSO4 and the Sr2+ ions and the
SO42- ions are coming from DIFFERENT sources. (We are NOT trying to
find the solubility of SrSO4 here so DON’T solve for “s”)!
We write the equilibrium equation and the Ksp expression for SrSO4:
SrSO4(s)  Sr2+(aq) + SO42-(aq)
Ksp = [Sr2+] [SO42-]
We know the Sr(NO3)2 is soluble and the [Sr2+] is 0.0050M so we
rearrange the Ksp expression and solve for [SO42-]:
SO   
2
4
Ksp 3.4 x107
5



6
.
8
x
10
M
2
Sr
0.0050
The Na2SO4 contains 1 SO4 so [Na2SO4] = [SO42]. We go from Mmolg
Mol Na2SO4 = M x L = 6.8 x 10-5M x 0.4000L = 2.72 x 10-5 mol Na2SO4
2.72 x 10-5 mol Na2SO4
x
142.1g
 3.9 x10 3 g Na2SO4
1mol
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 [CO32-] necessary to just start precipitation in a solution in
which [Ag+] = 3.6 x 10-6M
Example: Find the minimum [CO32-] necessary to just start precipitation in a solution in
which [Ag+] = 3.6 x 10-6M
The compound of low solubility here is Ag2CO3.
Example: Find the minimum [CO32-] necessary to just start precipitation in a solution in
which [Ag+] = 3.6 x 10-6M
The compound of low solubility here is Ag2CO3. We can look up the Ksp
and we are given the [Ag+] .
Example: Find the minimum [CO32-] necessary to just start precipitation in a solution in
which [Ag+] = 3.6 x 10-6M
The compound of low solubility here is Ag2CO3. We can look up the Ksp
and we are given the [Ag+] .
We write the equilibrium equation and the Ksp expression for Ag2CO3.
Example: Find the minimum [CO32-] necessary to just start precipitation in a solution in
which [Ag+] = 3.6 x 10-6M
The compound of low solubility here is Ag2CO3. We can look up the Ksp
and we are given the [Ag+] .
We write the equilibrium equation and the Ksp expression for Ag2CO3.
Ag2CO3(s)  2Ag+(aq) + CO32-(aq)
Ksp = [Ag+]2[CO32-]
Example: Find the minimum [CO32-] necessary to just start precipitation in a solution in
which [Ag+] = 3.6 x 10-6M
The compound of low solubility here is Ag2CO3. We can look up the Ksp
and we are given the [Ag+] .
We write the equilibrium equation and the Ksp expression for Ag2CO3.
Ag2CO3(s)  2Ag+(aq) + CO32-(aq)
Ksp = [Ag+]2[CO32-]
Solve the Ksp expression for [CO32-]
Example: Find the minimum [CO32-] necessary to just start precipitation in a solution in
which [Ag+] = 3.6 x 10-6M
The compound of low solubility here is Ag2CO3. We can look up the Ksp
and we are given the [Ag+] .
We write the equilibrium equation and the Ksp expression for Ag2CO3.
Ag2CO3(s)  2Ag+(aq) + CO32-(aq)
Ksp = [Ag+]2[CO32-]
Solve the Ksp expression for [CO32-]
[CO3
2
Ksp
]
[ Ag  ]2
Example: Find the minimum [CO32-] necessary to just start precipitation in a solution in
which [Ag+] = 3.6 x 10-6M
The compound of low solubility here is Ag2CO3. We can look up the Ksp
and we are given the [Ag+] .
We write the equilibrium equation and the Ksp expression for Ag2CO3.
Ag2CO3(s)  2Ag+(aq) + CO32-(aq)
Ksp = [Ag+]2[CO32-]
Solve the Ksp expression for [CO32-]
[CO3
2
Ksp
]
[ Ag  ]2
Look up the Ksp and substitute 3.6 x 10-6 in for [Ag+]:
Example: Find the minimum [CO32-] necessary to just start precipitation in a solution in
which [Ag+] = 3.6 x 10-6M
The compound of low solubility here is Ag2CO3. We can look up the Ksp
and we are given the [Ag+] .
We write the equilibrium equation and the Ksp expression for Ag2CO3.
Ag2CO3(s)  2Ag+(aq) + CO32-(aq)
Ksp = [Ag+]2[CO32-]
Solve the Ksp expression for [CO32-]
[CO3
2
Ksp
]
[ Ag  ]2
Look up the Ksp and substitute 3.6 x 10-6 in for [Ag+]:
12
Ksp
8
.
5
x
10
2
[CO3 ] 

 0.66M
 2
6 2
[ Ag ]
(3.6 x10 )
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(NO3)3 is added dropwise to a solution containing 0.10M KF and
0.10M Na2SO4. 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(NO3)3 is added dropwise to a solution containing 0.10M KF and
0.10M Na2SO4. 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(NO3)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(NO3)3 is added dropwise to a solution containing 0.10M KF and
0.10M Na2SO4. 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(NO3)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(NO3)3 is added dropwise to a solution containing 0.10M KF and
0.10M Na2SO4. 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(NO3)3 is irrelevant since it is
added dropwise to a solution with a larger volume. So we can re-state it
like this:
Sr2+ is added dropwise to a solution containing 0.10M F- and 0.10M SO42-.
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(NO3)3 is added dropwise to a solution containing 0.10M KF and
0.10M Na2SO4. 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(NO3)3 is irrelevant since it is
added dropwise to a solution with a larger volume. So we can re-state it
like this:
Sr2+ is added dropwise to a solution containing 0.10M F- and 0.10M SO42-.
Which precipitate will form first?
We calculate the [Sr2+] necessary to start precipitation of each
precipitate (SrF2 and SrSO4)
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(NO3)3 is added dropwise to a solution containing 0.10M KF and
0.10M Na2SO4. 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(NO3)3 is irrelevant since it is
added dropwise to a solution with a larger volume. So we can re-state it
like this:
Sr2+ is added dropwise to a solution containing 0.10M F- and 0.10M SO42-.
Which precipitate will form first?
We calculate the [Sr2+] necessary to start precipitation of each
precipitate (SrF2 and SrSO4)
Whichever precipitate requires the LESSER [Sr2+] will form first (as
[Sr2+] gradually increases as it is added.)
Sr2+ is added dropwise to a solution containing 0.10M F- and 0.10M SO42-.
Which precipitate will form first?
First we find [Sr2+] necessary to start formation of SrF2:
Sr2+ is added dropwise to a solution containing 0.10M F- and 0.10M SO42-.
Which precipitate will form first?
First we find [Sr2+] necessary to start formation of SrF2:
We write the equilibrium equation and Ksp expression for SrF2:
Sr2+ is added dropwise to a solution containing 0.10M F- and 0.10M SO42-.
Which precipitate will form first?
First we find [Sr2+] necessary to start formation of SrF2:
We write the equilibrium equation and Ksp expression for SrF2:
SrF2(s)  Sr2+(aq) + 2F-(aq)
Sr2+ is added dropwise to a solution containing 0.10M F- and 0.10M SO42-.
Which precipitate will form first?
First we find [Sr2+] necessary to start formation of SrF2:
We write the equilibrium equation and Ksp expression for SrF2:
SrF2(s)  Sr2+(aq) + 2F-(aq)
Ksp = [Sr2+] [F-]2
Sr2+ is added dropwise to a solution containing 0.10M F- and 0.10M SO42-.
Which precipitate will form first?
First we find [Sr2+] necessary to start formation of SrF2:
We write the equilibrium equation and Ksp expression for SrF2:
SrF2(s)  Sr2+(aq) + 2F-(aq)
We now solve for [Sr2+] :
[ Sr 2 ] 
Ksp

[F ]
2

Ksp = [Sr2+] [F-]2
Sr2+ is added dropwise to a solution containing 0.10M F- and 0.10M SO42-.
Which precipitate will form first?
First we find [Sr2+] necessary to start formation of SrF2:
We write the equilibrium equation and Ksp expression for SrF2:
SrF2(s)  Sr2+(aq) + 2F-(aq)
We now solve for [Sr2+] :
Ksp = [Sr2+] [F-]2
4.3 x10 9
7
[ Sr ]   2 

4
.
3
x
10
M
2
(0.10)
[F ]
2
Ksp
Sr2+ is added dropwise to a solution containing 0.10M F- and 0.10M SO42-.
Which precipitate will form first?
First we find [Sr2+] necessary to start formation of SrF2:
We write the equilibrium equation and Ksp expression for SrF2:
SrF2(s)  Sr2+(aq) + 2F-(aq)
Ksp = [Sr2+] [F-]2
We now solve for [Sr2+] :
4.3 x10 9
7
[ Sr ]   2 

4
.
3
x
10
M
2
(0.10)
[F ]
2
Ksp
Next we find [Sr2+] necessary to start formation of SrSO4:
SrSO4(s)  Sr2+(aq) + SO42-(aq)
Ksp = [Sr2+] [SO42-]
Sr2+ is added dropwise to a solution containing 0.10M F- and 0.10M SO42-.
Which precipitate will form first?
First we find [Sr2+] necessary to start formation of SrF2:
We write the equilibrium equation and Ksp expression for SrF2:
SrF2(s)  Sr2+(aq) + 2F-(aq)
Ksp = [Sr2+] [F-]2
We now solve for [Sr2+] :
4.3 x10 9
7
[ Sr ]   2 

4
.
3
x
10
M
2
(0.10)
[F ]
2
Ksp
Next we find [Sr2+] necessary to start formation of SrSO4:
SrSO4(s)  Sr2+(aq) + SO42-(aq)
7
Ksp
3
.
4
x
10
6
[ Sr 2 ] 


3
.
4
x
10
M
2
(0.10)
[ SO4 ]
Ksp = [Sr2+] [SO42-]
Sr2+ is added dropwise to a solution containing 0.10M F- and 0.10M SO42-.
Which precipitate will form first?
First we find [Sr2+] necessary to start formation of SrF2:
We write the equilibrium equation and Ksp expression for SrF2:
SrF2(s)  Sr2+(aq) + 2F-(aq)
Ksp = [Sr2+] [F-]2
We now solve for [Sr2+] :
4.3 x10 9
7
[ Sr ]   2 

4
.
3
x
10
M
2
(0.10)
[F ]
2
Ksp
Next we find [Sr2+] necessary to start formation of SrSO4:
SrSO4(s)  Sr2+(aq) + SO42-(aq)
Ksp = [Sr2+] [SO42-]
7
Ksp
3
.
4
x
10
6
[ Sr 2 ] 


3
.
4
x
10
M
2
(0.10)
[ SO4 ]
So, since the SrF2 requires the LESSER concentration of Sr2+ (4.3 x 107M)
the precipitate SrF2 will form first.
9 -Using Titrations to find Unknown Concentration
9 -Using Titrations to find Unknown Concentration
The following diagram applies to Titration questions in unit 3:
Mole bridge
M x L of
Standard
Moles of
Standard
Moles of
Sample
M or L of
Sample
9 -Using Titrations to find Unknown Concentration
The following diagram applies to Titration questions in unit 3:
Mole bridge
M x L of
Standard
Moles of
Standard
Moles of
Sample
M or L of
Sample
For Example: It takes 16.45 mL of 0.100M AgNO3 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:
Mole bridge
M x L of
Standard
Moles of
Standard
Moles of
Sample
M or L of
Sample
For Example: It takes 16.45 mL of 0.100M AgNO3 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:
Mole bridge
M x L of
Standard
Moles of
Standard
Moles of
Sample
M or L of
Sample
For Example: It takes 16.45 mL of 0.100M AgNO3 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 AgNO3 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:
Mole bridge
M x L of
Standard
Moles of
Standard
Moles of
Sample
M or L of
Sample
For Example: It takes 16.45 mL of 0.100M AgNO3 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 AgNO3 because you are given the
volume and the concentration.)
So we calculate the MOLES of AgNO3 (or Ag+)
9 -Using Titrations to find Unknown Concentration
The following diagram applies to Titration questions in unit 3:
Mole bridge
M x L of
Standard
Moles of
Standard
Moles of
Sample
M or L of
Sample
For Example: It takes 16.45 mL of 0.100M AgNO3 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 AgNO3 because you are given the
volume and the concentration.)
So we calculate the MOLES of AgNO3 (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:
Mole bridge
M x L of
Standard
Moles of
Standard
Moles of
Sample
M or L of
Sample
For Example: It takes 16.45 mL of 0.100M AgNO3 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 AgNO3 because you are given the
volume and the concentration.)
So we calculate the MOLES of AgNO3 (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:

1
molBr

0.001645 mol Ag+ x

0
.
001645
molBr
1molAg 
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:

1
molBr

0.001645 mol Ag+ x

0
.
001645
molBr
1molAg 
mol
Now we use M 
to calculate the [Br-]. BE CAREFUL TO MAKE
L
SURE THE “L” WE USE ARE OF THE Br- SOLUTION AND NOT THE
AgNO3 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:

1
molBr

0.001645 mol Ag+ x

0
.
001645
molBr
1molAg 
mol
Now we use M 
to calculate the [Br-]. BE CAREFUL TO MAKE
L
SURE THE “L” WE USE ARE OF THE Br- SOLUTION AND NOT THE
AgNO3 SOLUTION!!!
 

0
.
001645
molBr

Br 
 0.0329M

0.0500 LofBr
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:

1
molBr

0.001645 mol Ag+ x

0
.
001645
molBr
1molAg 
mol
Now we use M 
to calculate the [Br-]. BE CAREFUL TO MAKE
L
SURE THE “L” WE USE ARE OF THE Br- SOLUTION AND NOT THE
AgNO3 SOLUTION!!!
 

0
.
001645
molBr

Br 
 0.0329M

0.0500 LofBr
The Volume of the
Sample with Brwas 50.0 mL