Chem 110
Concentration Dilution Module
Instructions
This is optional for Chem 110.
This module is designed to provide students with the framework needed to solve a variety of
problems. Specifically, this should help students to successfully complete the Concentration
Dilution Skill Check Test. You can work on it alone or with a partner. If you have more than two
people working together, the value of the effort tends to diminish. We strongly encourage you to
do this alone or with only one partner.
NOTE: the estimated time need to complete this module is ~ 60-75 minutes. There is no “quick
and easy” way to learn how to do this. If you put in the time intentionally, it will save you a lot of
time in the long run!
Please do the module completely. If you skip sections of the module, it is more likely that you will
have problems understanding the intention in later sections. If you are not sure how to proceed,
please re read the instructions carefully. Working with another student may also be helpful in this
respect.
No answer key is available and it will not be made available. Please do NOT e-mail your instructor
or TA for the answers. If you need help or want to check your answers, do your best to complete
the questions and then bring your work to the Chemistry Resource Room (211 Whitmore) during
Chem 110 TA office hours. The TAs will be happy to go over the material with you and answer
your questions.
Finally, there are several fully worked out questions at the end. After working through the module,
they may be helpful in answering questions if you still have any. However, keep in mind that the
purpose of the worked out problems is to illustrate the method. READING or IMITATING the
solutions of problems will not help you.
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Chem 110
Concentration Dilution Module
Concentration and Dilution
Activity 1: Solvation at the Molecular Level
Explore the solvation process using a strong electrolyte and a nonelectrolyte as our examples.
Example 1: NaCl (Table Salt)
NaCl is an ionic compound that dissolves and completely dissociates in water.
Q1:
Based on the picture and your chemistry knowledge, is NaCl a strong electrolyte or
nonelectrolyte? ______________________________________
Q2:
What ions and/or molecules are present in a solution of NaCl and water?
___________________________________________________________
Write a balanced chemical equation for this dissolution process.
water
→
_____________________
2
_________________________
Chem 110
Concentration Dilution Module
There are several different acceptable ways a balanced chemical
equation can be written. For example, to express the process of
magnesium sulfate (MgSO4) dissolving in water, one could write
MgSO4 (s) + H2O (l)  Mg2+ (aq) + SO42(aq)
Here liquid water (as noted by the (l) following water) is included as
one of the reactants, or the components found on the left-hand side of
the arrow. Another acceptable way to write out this process is
MgSO4 (s)  Mg2+ (aq) + SO42(aq)
Notice the water is not shown in this case, but it is known to be
present because the magnesium sulfate is now present in the aqueous
form (indicated by the (aq) notation). Remember the definition of an
aqueous solution is a solute dissolved in water.
Typically we will use the second form when writing reactions that
correspond to the dissolution process.
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Chem 110
Concentration Dilution Module
Concentration and Dilution
Activity 1: Solvation at the Molecular Level
Example 2: C6H12O6 (glucose)
C6H12O6 is a molecular compound that completely dissolves but does not dissociate in water.
Q3:
Looking at the picture and the process represented, is glucose a strong electrolyte or
nonelectrolyte? ___________________________________________________
Q4:
What ions and/or molecules are present in a solution of glucose and water?
________________________________________________________________
.
Write a balanced chemical equation for this dissolution process.
water
_____________________________
→
4
_____________________________________
Chem 110
Concentration Dilution Module
Concentration and Dilution
Activity 1: Solvation at the Molecular Level
Let’s now compare the drawings for NaCl and C6H12O6.
Q5: Give one key similarity between these two compounds when water is added.
Give one key difference between these two compounds when water is added. (How are strong
electrolytes different from nonelectrolytes?)
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Chem 110
Concentration Dilution Module
Concentration and Dilution
Activity 2: Relative Ion Concentrations
Now that we are more familiar with what strong electrolytes and non-electrolytes look like on the
molecular level, let’s use this knowledge to investigate relative concentrations of both ionic and
molecular species in a solution.
GIVEN STOCK SOLUTION:
1 mole of calcium chloride (CaCl2) is dissolved in 1,000 mL of water.
Like NaCl from Activity 1, calcium chloride is a strong electrolyte, meaning it breaks up into
ions upon dissolving.
STEP 1: Write a balanced chemical equation for calcium chloride dissolving in water.
Use Activity 2 as a guide.
For 1 calcium chloride, how many ions are present after dissociation?
What are these ions? (Give chemical symbol, charge, and quantity)
STEP 2: What is the calcium chloride stock solution concentration?
STEP 3: (a) What is the concentration in molarity of Ca2+(aq) in this solution? (Show work)
(b) What is the concentration of Cl (aq)? (Show work)
(c) Do these concentration values differ? Why or why not?
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Chem 110
Concentration Dilution Module
Concentration and Dilution
Activity 3: Dilution Process
We have now seen how molecular and ionic concentrations can be determined from given
information of a stock solution. However, it is common to take a small portion of a stock solution
and dilute it to make a new solution of a different concentration. We shall explore this process
now.
GOAL: to find the concentration of a dilute solution
GIVEN:
A stock solution Kool Aid® (KA) can be prepared by dissolving 1 packet of Kool Aid® mix with 1
liter of water. (Let’s assume 1 packet contains 10 moles of Kool Aid®)
Stock Solution
1L
MS
VS
®
STEP 1: Write a balanced chemical equation for Kool Aid dissolving in water.
(Assume Kool Aid® is a nonelectrolyte and use KA to represent Kool Aid®.)
STEP 2: Calculate the concentration (MS) of the stock solution of Kool Aid®.
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Chem 110
Concentration Dilution Module
Concentration and Dilution
Activity 3: Dilution Process
125 mL of this Kool Aid® stock solution is removed and placed into another container. It is then
diluted with water to a final volume of 1 liter.
Solution 1
Solution 2
add water
M1=MS
M2
V1
V2
(a) How many moles of Kool Aid® are contained in 125 mL of stock solution (solution 1)?
(Show work including units)
(b) Using the symbols shown for molarity and volume, write a general equation for the calculation
of moles in a solution.
(c) How does the number of moles of Kool Aid® compare for solution 1 and solution 2?
You should have observed that the number of moles removed from a
stock solution equals the number of moles contained in the diluted
solution. This concept can be represented by a relationship you
should have seen before in your textbook:
M1·V1 = M2·V2
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Chem 110
Concentration Dilution Module
Concentration and Dilution
Activity 3: Dilution Process
Reminder: our goal is to find the concentration of the dilute solution (M2)
STEP 1 was to write the chemical equation for the process that was occurring
KA (s) + H2O (l) → KA (aq)
STEP 2 was to get the concentration of the stock solution MS = 10M
STEP 3: What information do we know about each Kool Aid® solution?
Organize your data from the previous steps below. Circle the variable that is unknown.
Variable
M1=MS
Value (Units)
10 M
Word Description
Concentration of the stock solution
V1
n1 =M1V1
M2
V2
n2=M2V2
STEP 4: Using your organized data from STEP 3, solve for the concentration of the dilute
solution. (Show all of your work before continuing.)
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Chem 110
Concentration Dilution Module
Concentration and Dilution
Activity 4: Putting it All Together
You have now seen questions relating to electrolyte strength, the solvation process, molecular
and ionic concentrations, and the dilution process. Now we shall apply several of these concepts
at the same time to answer a given question.
A 1-L stock solution of AlCl3 has a concentration of 20.0 M. When 100 mL of this stock
solution is diluted to 500 mL, what is the concentration of chloride ions (Cl(aq)) in the dilute
solution?
STEP 1: Write the balanced chemical equation for AlCl3 dissolving in water.
STEP 2: Draw how the diluted solution is being made (refer to the Kool Aid® example in
Activity 2 for help if necessary)
M1 =
M2 =
V1 =
V2 =
Concentration and volume
of stock soln = [AlCl3]
Concentration and volume
of dilute soln = [AlCl3]
STEP 3: What information about each solution do you know? Organize the information
underneath the pictures. Circle the unknown variable.
Did you include units?
STEP 4: Using information from STEP 1 through 3,
(a) Determine the [AlCl3] of the dilute solution. (Show ALL work)
Re-read the question: have you answered the question that was ASKED???
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Chem 110
Concentration Dilution Module
Concentration and Dilution
Activity 4: Putting it All Together
To get the [Cl−] the dissociation of AlCl3 must be considered (Step 1).
STEP 4 <continued>:
(b) Determine the [Al3+] of the dilute solution. (Show ALL work)
(c) Determine the [Cl−] of the dilute solution. (Show ALL work)
What if the question asked for the moles of Cl−(aq) in the dilute solution?
To answer this, an additional step (STEP 5) is necessary to calculate the
number of moles.
RECALL
moles = M×V
STEP 5: (a) How many moles of aluminum ion (Al3+) are present in the dilute solution?
(b) How many moles of chloride ion (Cl−) are present in the dilute solution?
(c) How does this amount of aluminum and chloride ions in the dilute solution compare
to the amount of aluminum and chloride ions removed from the stock solution?
(Recall Activity 3)
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Chem 110
Concentration Dilution Module
Concentration and Dilution
Activity 5: More Practice
There are several ways questions relating to concentrations and dilutions can be presented. The
same problem solving approach as outlined in Activities 3 and 4, however, still applies.
STEP 1 Write out the chemical process (what is happening on the atomic scale?)
STEP 2 –Draw pictures (what is happening on the macroscopic level?)
STEP 3 – Organize information given in the problem. (Put this information directly under
the pictures but be sure it is clear what M refers to, i.e., is M the concentration of
salt? Or the concentration of one of the ions?)
STEP 4 – Solve for unknowns by using M1V1 = M2V2 or other relationships
STEP 5 –What are you asked for? The process outlined in step 2 will help you clarify what
you want to know. Use the stoichiometry of the chemical process (step 1) or MV
= moles to obtain desired answer.
Let’s look at another example question:
10 mL of a stock solution of potassium sulfate (K2SO4) is removed and diluted with water to a
final volume of 2 liters. The dilute solution has 0.5 moles of K+ ions. What is the concentration
of the stock solution used to make the dilute solution?
STEP 1: Write out a balanced chemical equation for K2SO4 dissolving in water
STEP 2
Draw how the dilute solution is being made and label the pictures with the
information given.
V1 =
M1 =
Conc. [K2SO4]
V2 =
M2 =
dilute [K2SO4] BUT, also know
nK+= Moles of K+ in dilute solution
STEP 3: Organize all of the information given below the pictures. (be sure to include units).
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Chem 110
Concentration Dilution Module
Concentration and Dilution
Activity 5: More Practice
Reminder: this is the original problem we are trying to solve. Be sure to circle or otherwise
indicate what you want to know.
10 mL of a stock solution of potassium sulfate (K2SO4) is removed and diluted
with water to a final volume of 2 liters. The dilute solution has 0.5 moles of
K+ ions. What is the concentration of the stock solution used to make the
dilute solution?
Unlike the previous examples, you should notice that a concentration value for the stock solution
and the dilute solution are not given directly in the problem.
However, you ARE given the moles of potassium ions in the dilute solution. How can this be
used to find the concentration? Recall the definition of concentration from Activity 1:
M = moles per liter.
STEP 4: Using the information in STEP 1 through 3,
(a) Determine the [K+] for the dilute solution.
(b) Is this [K+] the same as the [K2SO4] of the dilute solution? (Hint: Look at STEP 1!)
Show why or why not below and calculate [K2SO4] for the dilute solution.
(c) Use M1·V1 = M2·V2 to determine the concentration of K2SO4 in the original solution.
(Show all work including units. Be sure to answer the question being asked)
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Chem 110
Concentration Dilution Module
Concentration and Dilution
Activity 6: Practice
The following questions apply several of the concepts you have learned from this module. You
should be able to answer these questions using both the skills you acquired in the previous
Activities as well as your previous chemistry knowledge.
When solving these problems, be sure to write out all of the STEPS as shown in the prior
Activities.
Also, when applicable, circle any “extra” information not needed to solve the problem.
Pract Q1: Calculate the concentration of the sodium ions (Na+) in a solution made by diluting 90.0
mL of a 0.750M solution of sodium sulfate (Na2SO4) to a total volume of 450.0 mL.
Pract Q2: Determine the number of moles of bromide ions (Br −) in 0.500 L of a 0.300 M solution of
potassium bromide (KBr) that has been diluted with water to a final volume of 1.500 L.
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Chem 110
Concentration Dilution Module
Concentration and Dilution
Worked Through Examples
The module you just completed attempted to cover the basic framework for concentration
and dilution problems. However, since there are many different ways you could see these
questions being asked, we have provided additional pages to demonstrating how some more
advanced problems are solved using the same step-by-step process shown in the module.
For ease of understanding, each step to solving the problem has been made a different
color. Also, since these problems are more advanced, additional steps beyond Step 5 may
be necessary.
STEP
1&2
–
Writing
out
the
chemical
process
and
drawing
STEP 3 – Organizing information given in the problem
STEP 4 – Solving for unknowns by using M1V1 = M2V2 or other relationships
STEP 5 – Manipulating information solved for to obtain desired answer
ADDITIONAL STEPS NOT ADDRESSED IN THE MODULE
FEEDBACK FOR CLARIFICATION
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pictures
Chem 110
Concentration Dilution Module
Concentration and Dilution
Worked Through Examples
EX1: Determine the number of moles of chloride ions (Cl–) in 1.800 L of a 0.650 M solution
of magnesium chloride (MgCl2) that has been diluted with water to a final volume of
2.500 L.
STEP 1
MgCl2 (s) + H2O (l)  Mg2+ (aq) + 2Cl– (aq)
STEP 2
Add water
V1 = 1.800L
M1 = 0.650M
MgCl2 solution
n (Cl− )= ???
V2 = 2.500L
M2 = ????
MgCl2 solution
STEP 3
M1 = 0.650 M MgCl2
V1 = 1.800 L
V2 = 2.500 L
M1 = ???
Moles of Cl− (aq) = ???
STEP 4
M1V1 = n1
0.650 moles MgCl2 × 1.800 L = 1.17 moles MgCl2
1L
STEP 5
1.17 moles MgCl2 × 2 moles Cl–
1 mole MgCl2
= 2.34 moles Cl–
The final volume of 2.500 L the solution was diluted to was NOT necessary to answer the
question. This dilution did not change the number of moles present in the 1.800-L of 0.650
M solution. If the question had asked for the new CONCENTRATION (or the amount of
moles per unit volume), then this volume would have been important.
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Chem 110
Concentration Dilution Module
Concentration and Dilution
Worked Through Examples
EX2:(a)
STEP 1
What is the concentration of potassium ions (K+) present
980 mL of a 0.45 M solution of potassium sulfate (K2SO4) diluted to 2.0 L?
in
K2SO4 (s) + H2O  2K+ (aq) + SO42– (aq)
STEP 2
K2SO4 solution
Add water
V1 = 980mL
M1 = 0.450M
K2SO4 solution
V2 = 2.00L
M2 = ????
K2SO4 solution
[K+]= ???
STEP 3
M1 = 0.45 M K2SO4
V1 = 980 mL (0.980L)
V2 = 2,000 mL (2.0L)
STEP 4
M1V1=M2V2
(0.45M) (0.980L) = M2 (2.0L)
M2 = (0.45M) (0.980L)
2.0L
M2 = 0.2205 M K2SO4
STEP 5
0.2205 moles K2SO4 × 2 moles K+ ions = 0.4410 moles K+ = 0.4410 M K+
1L
1 mole K2SO4
1L
The final volume of 2.0 L that the solution was diluted to WAS necessary to answer this
question because a concentration was asked for. Recall that concentration is dependent on
the volume.
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Chem 110
Concentration Dilution Module
Concentration and Dilution
Worked Through Examples
(b) How many sulfate ions (SO42–) ions are present in 500 mL of the dilute solution?
STEP 1
K2SO4 (s) + H2O  2K+ (aq) + SO42– (aq)
STEP 2 shown on previous page
STEPS 3 and 4 (above)
M2 = 0.2205M K2SO4
STEP 5 To get number of sulfate ions, first we need moles of sulfate ions in the dilute
solution. Use reaction shown in step 1
0.2205 moles K2SO4 × 1 mole SO42– = 0.2205 moles SO42– = 0.2205 M SO42–
1L
1 mole K2SO4
1L
This is the concentration of the SO42– ion for the ENTIRE dilute solution. However, the
question is asking for the number of SO42– ions in a specified volume, NOT the entire dilute
solution.
Moles = M×V
= 0.2205 moles SO42– × 0.500L
1L
= 0.1103 moles SO42–
We still have NOT answered the question! We need to use our known relationship between
a mole and the number of objects in a mole (Avogadro’s Number = 6.02 × 1023).
0.1103 moles SO42– × 6.02 × 1023 SO42– ions = 6.64 × 1022 SO42– ions
1 mole SO42–
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Chem 110
Concentration Dilution Module
Concentration and Dilution
Worked Through Examples
EX3: (a) 13 mL of a dilute solution of calcium nitrate (Ca(NO3)2) has 1.61× 1023 NO3– ions. If
0.450 L of a stock solution of calcium nitrate was used to prepare a 1.0 L dilute
solution, what is the concentration of the stock solution?
STEP 1
Ca(NO3)2 (s) + H2O (l)  Ca2+ (aq) + 2 NO3– (aq)
STEP 2
Ca(NO3)2 solution
Add water
V1 = 0.450L
M1 = ???
Ca(NO3)2 solution
STEP 3
M1
=
M2
=
# NO3– ions =
remove some
of dilute soln
V2 =1.0L
M2 = ????
Ca(NO3)2 solution
? M Ca(NO3)2
? M Ca(NO3)2
1.61× 1023
V1
V2
V3
V3 = 13 mL
M3 = M2
Ca(NO3)2 solution
# of NO3− = 1.61× 1023
=
0.450 L
=
1.0 L
= V of dilute examined = 13 mL (0.013L)
Unlike previous examples, this question only gives you 2 of the variables in the expression
M1V1 = M2V2 . We must use other pieces of information in order to solve for one of the
concentrations (either the dilute solution or the stock solution) before we can use this
expression.
Since we know the number of nitrate ions present in a given volume of dilute solution, we can
first determine the number of moles of nitrate ions present in this volume.
1.61× 1023 NO3– ions × 1 mole
= 0.2674 moles NO3– ions
23
–
6.02× 10 NO3
Using the balanced chemical equation, we can relate the number of moles of nitrate ions to
the number of moles of calcium nitrate present in the specified volume of dilute solution.
0.2674 moles NO3– × 1 mole Ca(NO3)2 = 0.1337 moles Ca(NO3)2
2 moles NO3–
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Chem 110
Concentration Dilution Module
Concentration and Dilution
Worked Through Examples
Reminder: Problem being solved
EX3: (a) 13 mL of a dilute solution of calcium nitrate (Ca(NO3)2) has 1.61× 1023 NO3– ions. If
0.450 L of a stock solution of calcium nitrate was used to prepare a 1.0 L dilute
solution, what is the concentration of the stock solution?
Goal is to get M1 = concentration of the stock solution
STEP 1
Ca(NO3)2 (s) + H2O (l)  Ca2+ (aq) + 2 NO3– (aq)
STEP 2
Ca(NO3)2 solution
dilute soln
Add water
V1 = 0.450L
M1 = ???
soln 3
remove some
of dilute soln
V2 =1.0L
M2 = ????
V3 = 13 mL
M3 = M2
# of NO3− = 1.61× 1023
We used # of NO3− to find moles of Ca(NO3)2 in solution 3 =0.1337 moles Ca(NO3)2
Now we can use the definition of concentration to calculate M3 (concentration of solution 3)
M3 = n3 = 0.1337 moles Ca(NO3)2= 10.2862 M Ca(NO3)2
V3
0.0013L
Note: M2 = M3 so
M1V1 = M2V2 can be applied.
M1V1 = M2V2
M1 (0.450L) = (10.2862M) (1.000L)
M1 = (10.2862M) (1.000L)
0.450 L
M1 = 22.9 M Ca(NO3)2
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Chem 110
Concentration Dilution Module
Concentration and Dilution
Worked Through Examples
(b) What is the total ion concentration of 50 mL of the dilute solution of calcium nitrate?
Given/Determined Previously:
M1
=
22.9 M Ca(NO3)2
M2
=
10.3 M Ca(NO3)2
V1
=
0.450 L
V2
=
1.0L
# NO3– ions =
1.61× 1023
V3
= V of dilute examined = 13 mL (0.013L)
Using the information we have, the number of moles of calcium nitrate in 50 mL of the dilute
solution can be determined.
Concentration of 50mL of dilute solution is the same as the concentration of the full liter of
solution (Concentration is an intensive property.)
M of 50 mL = M2 = 10.3M
Based on the balanced chemical equation, for every 1 mole of Ca(NO3)2, 3 ions (1 Ca2+ and 2
NO3–) result.
[ions] = 10.3M Ca(NO3)2 ×
3 ions
1 mole Ca(NO3)2
21
= 30.9 M ion concentration