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1. The “Concentration”
[ ] of a solution is a
measure of how
much solute is
dissolved in the
solvent and can be
measured in units
like ppm, g/L , %.= Solution!!
2. The unit of concentration most often used in
chemistry is Molarity: mol/L.(M)
Molarity is defined as:
So a 1M solution of NaCl contains 1 mole
of NaCl in 1L of solution.
How would you make 1 M NaCl…
1 mol of NaCl = 58.44g
Therefore, add 58.44 g of NaCl to a
container, and add H2O to make up 1L of
solution. = 1 M NaCl
solution
solute
V
molsM
Molarity = Mol
L
M = 2.5mol
1.5L
M = 1.67 mol/L
You would call this a 1.67 “Molar” solution… Or 1.67M solution!
Molarity = Mol/L
We have to get grams to mols, and ml to L…
𝟓. 𝟏𝟐𝒈𝑵𝒂𝑪𝒍 ×𝟏𝒎𝒐𝒍
𝟓𝟖.𝟓𝒈= 0.0875mol NaCl
𝟐𝟓𝟎. 𝟎𝒎𝒍 ×𝟏𝑳
𝟏𝟎𝟎𝟎𝒎𝒍= 0.2500L
M=0.0875mol
0.250L
= 0.350 M NaCl
Eg.3 What mass of NaCl would make up
2.0L of a 0.25M(mol/L) NaCl solution?
1. Find the moles of NaCl needed.
2. Change moles to mass.
3. Answer: Add 29.22 g of NaCl to a container then add enough water to make up 2.0L of solution.
mol5.0
gNaClmol
gNaClmol 22.29
1
44.585.0
L
molL
1
25.00.2
Eg. 4. What mass of NaOH is
contained in 3.50L of 0.200 M NaOH?
1. Find the moles of NaCl needed.
2. Change moles to mass.
3. Answer: 28 g of NaOH is present in the solution
mol70.0
gNaOHmol
gNaOHmol 28
1
0.407.0
L
molL
1
200.05.3
Molarity = mol/L
So we need to get the g to moles, and the ml to L
𝟏. 𝟖𝟗𝟑𝒈 ×𝟏𝒎𝒐𝒍
𝟗𝟖. 𝟏𝒈= 𝟎. 𝟎𝟏𝟗𝟑𝒎𝒐𝒍
𝟏𝒎𝒍 ×𝟏𝑳
𝟏𝟎𝟎𝟎𝒎𝒍= 𝟎. 𝟎𝟎𝟏𝑳
Molarity = 0.0193mol/0.01L
= 18.7M H2SO4
Hebden: Pg 98 #59a,e,
#60a,e, #61, #62, #63, #66,
#68, #69
Does the amount of the
solute change?
Do the mols of the solute
change?
What do you think
happens to the Molarity?
1. When H2O is ADDED or REMOVED (evaporated)
from a solution OR if two solutions are mixed
together:
1. The concentration or Molarity (M) changes
2. The moles/grams of solute does NOT change
So if moles don’t change… then…
Moles of initial solution = Moles of final solution
And we know that…
Mol x L = Mol
L
Molarity x Volume = Mole
1. The new concentration of the solution can be found using the formula:
where [ ]= concentration or molarity (M)
Mi = Molarity of initial
Vi= Volume of initial
Mf = Molarity of final(diluted)
Vf= Volume of final(diluted)
ffii VMVM
MM
L
LM
L
mol
LLMLM
VMVM
f
f
f
ffii
0.1
0.3
)0.3(
0.3
0.3
)0.20.1()0.1)(0.3(
(Q#1) A student has 500.mL of a 0.4M NaCl
solution. How much water is needed to make
this solution 0.10M?
LV
VMmol
VMLM
VMVM
f
f
f
ffii
0.2
)10.0(20.0
)10.0()500.0)(40.0(
(Q#2) A chemist adds water to 120 mL of a
6.0M solution of NaOH until the final volume is
2.0L. What is the molarity (M) of the resulting
solution?
MM
LMmol
LMLM
VMVM
f
f
f
ffii
36.0
)0.2(720.0
)0.2()120.0)(0.6(
(Q#3) What volume of 12M HCl acid must be
added to water to make 5.0L of 0.10M HCl
solution?
LV
molVM
LMVM
VMVM
i
i
i
ffii
042.0
50.0)12(
)0.5)(10.0()12(
(Q#4)What concentration results when 150. mL
of a 0.36M solution of magnesium sulphate
are added to 750. mL of water?
MM
LMmol
LLMLM
VMVM
f
f
f
ffii
060.0
)900.0(054.0
)150.0750.0()150.0)(36.0(
Pg. 102 Exercises
(78,80,82,83,85,87,89)
1. It is important to be able to find the
concentration of ions in a solution once the
ionic compound has dissociated because…
1. It is the ions in the solution that are
responsible for any conductivity or
chemical reactions that
may occur.
2. Often the [ions] will
determine whether a
reaction will occur or not.
Polyatomic ions (e.g , )
etc… remain intact during dissociation.
2
4SO3
4PO
3NO
)()()( aqaqs ClNaNaCl
)(2
4)()(42 2 aqaqsSONaSONa
)(3
4)(2
)(243 23)( aqaqsPOMgPOMg
Calculating ion concentrations!!e.g. What is the concentration of all ions in a 1.0 M
NaCl solution?
(1) Write out the dissociation equation for the ionic
compound.
(2) Write in the concentration before dissociation
occurs.
(3) Write in the concentration after dissociation
occurs.
ClNaNaCl
ClNaNaCl
1.0M 0.0M 0.0M
ClNaNaCl1.0M 1.0M0.0M
ClNiNiCl 33
3
Before: 2.0M 0.0M 0.0M
After: 0.0M 2.0M 3(2.0)M = 6.0M
2. The final molarity(M) of the ions is in the same ratio as the moles in the dissociation equation.
E.g. Find the ion concentrations in a 2.0M NiCl3 solution.
1. Sometimes two ionic
solutions are mixed
together that have a
common ion
present.
2. Until you learn
differently in
Chemistry 12,
assume that no
reaction occurs
when two solutions
are mixed.
1. Dilution method formula (if needed)
2. Dissociation equation
3. Calculation [ions] in solution
solution
solute
V
nM
ffii VMVM
ClMgMgCl 22
2
1) Don’t need a dilution equation as we are not
adding/diluting it
2) Write out the dissociation equation for each ionic
compound.
3) Find the concentration of each ion after dissociation.
AlCl3 Al 3+ + 3Cl-
Before: 0.25M 0.0M 0.0M
After: 0.0M 0.025M 3(0.025M) = 0.75M
Cl Ions
(1) Find the new concentration of each after
mixing using the dilution formula.
M
LDivideby
LM
LMLM
VMVM
M NaCl0.1
0.2
)0.2(0.2
)0.2)(()0.1)(0.2(
2211
(1) Find the new concentration of each after
mixing using the dilution formula.
M
LDivideby
LM
LMLM
VMVM
M NiCl0.2
0.2
)0.2(0.4
)0.2)(()0.1)(0.4(
3
2211
2. Write out the dissociation equation for each ionic compound.
3. Find the concentration of each ion after dissociation.
ClNaNaCl ClNiNiCl 33
3
Before: 1.0M 0.0M 0.0M
After: 0.0M 1.0M 1.0M
2.0M 0.0M 0.0M
0.0M 2.0M 6.0M
(4) Look for ions common to both equations
in order to find the total ion concentration by
adding them together.
MNa 0.1][
M
MMCl
0.7
)0.60.1(][
MNi 0.2][ 3
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