MOLARITY
A measurement of the concentration of a solution
Molarity (M) is equal to the moles of solute (n) per liter of solution
M = n / V = mol / L
Calculate the molarity of a solution prepared by mixing 1.5 g of
NaCl in 500.0 mL of water.
First calculate the moles of solute:
1.5 g NaCl (1 mole NaCl) = 0.0257 moles of NaCl
58.45 g NaCl
Next convert mL to L:
0.500 L of solution
Last, plug the appropriate values into the correct
variables in the equation:
M = n / V = 0.0257 moles / 0.500 L = 0.051 mol/L
MOLARITY
M=n/V =
mol /
L
How many grams of LiOH is needed to prepare 250.0 mL of a
1.25 M solution?
First calculate the moles of solute needed:
M = n / V , now rearrange to solve for n:
n = MV
n = (1.25 mol / L) (0.2500 L) = 0.3125 moles of solute needed
Next calculate the molar mass of LiOH:
23.95 g/mol
Last, use diminsional analysis to solve for mass:
0.3125 moles (23.95 g LiOH / 1 mol LiOH) = 7.48 g of LiOH
MOLARITY
M=n/V =
mol /
L
What is the molarity of hydroiodic acid if the solution is
47.0% HI by mass and has a density of 1.50 g/mL?
First calculate the mass of solute in the 47.0% solution using the
density. The 1.50 g/mL is the density of the solution but only 47.0%
of the solution is the solute therefore:
47.0% of 1.50 g/mL = (0.470) (1.50 g/mL) = 0.705 g/mL
density of solute
Since molarity is given in moles per liter and not grams we must
convert the g/mL to mol/mL using the molar mass.
0.705 g/mL (1 mole/ 128 g) = 0.00551 mol/mL
Next convert mL to L:
0.00551 mol/mL (1000 mL/ 1L) = 5.51 mol/L = 5.51 M
MOLARITY & DILUTION
M 1 V1 = M 2 V2
The act of diluting a solution is to simply add more water
(the solvent) thus leaving the amount of solute unchanged.
Since the amount or moles of solute before dilution (nb)
and the moles of solute after the dilution (na) are the same:
nb = na
And the moles for any solution can be calculated by n=MV
A relationship can be established such that
MbVb = nb = na = MaVa
Or simply :
MbVb = MaVa
MOLARITY & DILUTION
Calculate the molarity of a solution prepared by diluting 25.0 mL
of 0.05 M potassium iodide with 50.0 mL of water (the densities
are similar).
M1 = 0.05 mol/L
M2 = ?
V1 = 25.0 mL
V2 = 50.0 + 25.0 = 75.0 mL
M 1 V1 = M 2 V 2
M1 V1 = M2 = (0.05 mol/L) (25.0 mL) = 0.0167 M of KI
V2
75.0 mL
MOLARITY & DILUTION
Given a 6.00 M HCl solution, how would you prepare
250.0 mL of 0.150 M HCl?
M1 = 6.00 mol/L
M2 = 0.150
V1 = ? mL
V2 = 250.0 mL
M 1 V1 = M 2 V 2
M2 V2 = V1 = (0.150 mol/L) (250.0 mL) = 6.25 mL of 6 M HCl
M1
6.00 mol/L
You would need 6.25 mL of the 6.00 M HCl reagent which would be
added to about 100 mL of DI water in a 250.0 mL graduated cylinder
then more water would be added to the mixture until the bottom of the
menicus is at 250.0 mL. Mix well.
MOLARITY & Stoichiometry
In the past chapters, you measured out a solid reactant
and needed to predict the mass of the product. You
learned to use the molar mass and mole ratio to achieve
that goal. The same reasoning applies with solution
chemistry but instead to using the molar mass you will use
the molarity to convert to moles.
Stoichiometry Flow chart
Use the mole ratio
from the balanced equation
use
Molarity
of A
moles of A
Volume of A
Use Molar
mass of A
mass of A
use
Molarity
of B
moles of B
Volume of B
Use Molar
mass of B
mass of B
MOLARITY & Stoichiometry
How much calcium carbonate will be precipitated by
adding 25.0 mL calcium chloride to 25.0 mL of 0.56 M
potassium carbonate?
CaCl2 + K2CO3  CaCO3 + 2 KCl
V = 25.0 mL
M = 0.56 mol/L
m=?
V = 25.0 mL
First convert volume of A to moles of A:
0.025 L K2CO3 (0.56 mol/L) = 0.014 moles of K2CO3
Now convert moles of A to moles of B:
0.014 mol K2CO3(1 mol CaCO3/1 mol K2CO3) =0.014 mol CaCO3
Next convert moles of B to grams of B:
0.014 mol CaCO3 (100 g/mol) = 1.40 g of CaCO3
MOLARITY & Stoichiometry
What would be the molarity of the potassium chloride
solution from the last problem?
CaCl2 + K2CO3  CaCO3 + 2 KCl
V = 25.0 mL
M = 0.56 mol/L
M=?
V = 25.0 mL
First convert volume of A to moles of A:
0.025 L K2CO3 (0.56 mol/L) = 0.014 moles of K2CO3
Now convert moles of A to moles of B:
0.014 mol K2CO3(2 mol KCl /1 mol K2CO3) =0.028 mol KCl
Next convert moles of B to molarity of B:
0.028 mol KCl / 0.050 L = 0.56 M of KCl
PRACTICE PROBLEMS #24a
2.4 M
_________1.
What is the concentration of 250.0 mL of 0.60
moles of HCl?
3.00 L
_________ 2. What volume of 0.7690 M LiOH will contain 55.3
g of LiOH?
1.80 L
_________ 3. How many liters of water must be added to
100.0 mL of 4.50 M HBr to make a solution that is 0.250 M
HCl?
39.6 g
_________ 4. How many grams of barium sulfate that will
precipitate when 500.0 mL of 0.340 M BaCl2 and 300.0 mL of
1.70 M Na2SO4 are mixed?
5. How would you prepare 850.0 mL of a 0.020 M ferric chloride solution if
Weigh
g of of
theFeCl
hydrated
salt to a graduated cylinder with
.
you
start out
with4.59
crystals
3 6H2O?
800.0 mL then add enough DI water to make 850 mL exactly.
PRACTICE PROBLEM #24b
1. What is the concentration of 35.0 mL of 0. 0556 moles of
KCl?
1.59 M
2. How many grams of KCl is needed to prepare 50.0 mL of a
0.10 M solution?
0.38 g
3. How many milliliters of water must be added to 30.0 mL of
9.0 M KCl to make a solution that is 0.50 M KCl? 510 mL
4. How many grams of calcium carbonate will precipitate
when 500.0 mL of 0.340 M CaCl2 and 300.0 mL of 1.70 M
Na2CO3 are mixed?
17.0 g
5. What is the concentration of the product solution
(assuming the volumes are additive) when 500.0 mL of 0.340
M CaCl2 and 300.0 mL of 1.70 M Na2CO3 are mixed? 0.425 M
GROUP STUDY PROBLEM #24a
______1. How many grams of K3PO4 is needed to prepare 500.0 mL
of a 0.00567 M solution?
______2. What volume of 0.7690 M LiOH will contain 55.3 g of LiOH?
______3. How many milliliters of water must be added to 600.0 mL
of 3.40 M HCl to make a solution that is 0.500 M HCl?
______ 4. What is the concentration of the product solution
(assuming the volumes are additive) when 24.9 mL of 0.70 M BaCl2
and 45.0 mL of 0.15 M Na3PO4 are mixed?
GROUP STUDY PROBLEM #24b
______1. What is the concentration of 250.0 mL of 0.60 moles of NaOH?
______2. What mass of MgCl2 will be required to prepare 500 mL of 0.150
M solution?
______3. How many milliliters of water must be added to 50.0 mL of 1.97
M NaOH to make a solution that is 0.025 M NaOH?
______ 4. How many grams of barium sulfate that will precipitate when
500.0 mL of 0.340 M BaCl2 and 300.0 mL of 1.70 M Na2SO4 are mixed?
______ 5. What is the concentration of the product solution (assuming
the volumes are additive) when 500.0 mL of 0.340 M BaCl2 and 300.0 mL of
1.70 M Na2SO4 are mixed?