Concentrations of Solutions
•Behavior of solutions depend on compound
itself and on how much is present, i.e. on
the concentration.
•Two solutions can contain the same
compounds but behave quite different
because the proportions of those
compounds are different.
Concentrations of Solutions
• Concentration of a solution: the more
solute in a given volume of solvent, the
more concentrated
• 1 tsp salt (NaCl)/cup of water
vs
• 3 Tbsp salt/cup water
Molarity
Molarity is one way to measure the concentration of a
solution.
moles of solute
Molarity (M) =
volume of solution in liters
A 1.00 molar (1.00 M) solution contains 1.00 mol solute
in every 1 liter of solution.
Units of molarity are: mol/L = M
Preparing a 1.0 Molar Solution
One liter of a 1.00 M NaCl solution
• need 1.00 mol of NaCl
• weigh out 58.5 g NaCl (1.00 mole) and
• add water to make 1.00 liter (total volume) of
solution.
Copyright © 2005 by Pearson Education, Inc.
Publishing as Benjamin Cummings
Molarity Practice
• What is the molar NaCl concentration if
you have 0.5 mol NaCl in 1.00 L of
solution?
0.5 mol NaCl/1.00 L = 0.5 mol/L = 0.5 M
• What is the molar NaCl concentration if
you have 0.5 mol NaCl in 0.50 L of
solution?
0.5 mol NaCl/0.50 L = 0.5/0.50 mol/L = 1 mol/L = 1 M
Molarity Practice
What is the molar NaCl concentration if you have
10.0 g of NaCl in 1.00 L of solution?
Have grams not mols!
• Grams  mol
• Need molar mass
NaCl: 23.00 + 35.45= 58.45 g/mol
So: 10.0g x (1 mol/35.45g)=0.282 mol NaCl
Molar concentration: 0.282 mol/1.00 L = 0.282 M
Molarity – Moles - Volume
moles of solute
Molarity (M) =
volume of solution in liters
Molarity (M) =
•
•
•
•
mol
Volume (L)
Have mol and vol  molarity
Have molarity and vol  mol of solute
Have molarity and mol of solute  volume
AND: mol of solute  grams of solute
Practice
How many moles of HCl are present in 2.5 L of 0.10 M
HCl?
Given:
2.5 L of soln
0.10M HCl = 0.10 mol/1 L HCl
Find:
mol HCl
Use: mol = molarity x volume
mol HCl =0.10 M HCl x 2.5 L = 0.10 mol HCl x 2.5 L
1L
= 0.25 mol HCl
Practice
What volume of a 0.10 M NaOH solution is needed
to provide 0.50 mol of NaOH?
Given: 0.50 mol NaOH
0.10 M NaOH = 0.10 mol NaOH / 1L
Find: vol soln
Use: vol soln = mol solute / molarity
Vol soln = 0.50 mol NaOH = 0.50 mol NaOH
0.10 M NaOH
0.10 mol NaOH
1L
= 0.50 mol NaOH X 1L = 5 L
0.10 mol NaOH
More Practice
How many grams of CuSO4 are needed to
prepare 250.0 mL of 1.00 M CuSO4?
Given: 250.0 mL soln
1.00 M CuSO4
Find: g CuSO4
Use: mol CuSO4 = molarity x volume
Molarity = mol / 1L
Vol = 250.0 mL
Concentration of Solutions
Interconverting Molarity, Moles, and Volume
g CuSO4 = 250.0 mL soln x 1 L x 1.00 mol
1000 mL 1 L soln
x 159.6 g CuSO4
1 mol
= 39.9 g CuSO4
Steps involved in preparing
solutions from pure solids
Steps involved in preparing solutions
from pure solids
– Calculate the amount of solid required
– Weigh out the solid
– Place in an appropriate volumetric flask
– Fill flask about half full with water and mix.
– Fill to the mark with water and invert to mix.
You should be able to describe this process
(including calculating the mass of solid to use)
for any solution I specify.
Dilutions
• Many laboratory chemicals such as acids
are purchased as concentrated solutions
(stock solutions).
e.g. 12 M HCl
12 M H2SO4
• More dilute solutions are prepared by
taking a certain quantity of the stock
solution and diluting it with water.
Dilutions
• A given volume of a stock solution contains a specific
number of moles of solute.
e.g.: 25 mL of 6.0 M HCl contains 0.15 mol HCl
(How do you know this???)
25 mL HCl
25 mL H2O
0.15 mol
0.15 mol
+
50 mL
=
• If 25 mL of 6.0 M HCl is diluted with 25 mL of water, the
number of moles of HCl present does not change.
Still contains 0.15 mol HCl
Dilutions
moles solute
before dilution
=
moles solute
after dilution
• Although the number of moles of solute does
not change, the volume of solution does
change.
• The concentration of the solution will change
since
Molarity = moles solute
Volume of solution
Dilution Calculation
• When a solution is diluted, the concentration of
the new solution can be found using:
Mc x Vc = Md x Vd
where Mc= initial concentration (mol/L)= more
concentrated
Vc = initial volume of more conc. solution
Md =final concentration (mol/L) in dilution
Vd = final volume of diluted solution
Dilution Calculation
What is the concentration of a solution prepared by
diluting 25.0 mL of 6.00 M HCl to a total volume of
50.0 mL?
Given: Vc = 25.0 mL
Note: Vcand Vd do not have to
be in liters, but they
Mc = 6.00 M
must be in the same
Vd = 50.0 mL
units.
Find: Md
Use Vcx Mc= Vdx Md
Solve for Md
Dilution
• Make a diluted solution once you know Vc and Vd
– Use a pipet to deliver a volume of the concentrated solution
to a new volumetric flask.
– Add solvent to the line on the neck of the new flask.
– Mix well.
Practice
• How many mL of 5.0 M K2Cr2O7 solution must
be diluted to prepare 250 mL of 0.10 M solution?
Vc = ?
Mc = 5.0M
Vd = 250 mL
Md = 0.10M
• If 10.0 mL of a 10.0 M stock solution of NaOH is
diluted to 250 mL, what is the concentration of
the resulting solution?
Md = ?
Vc = 10.0 mL
Mc = 10.0M Vd = 250 mL
Solution Stoichiometry
• Remember: reactions occur on a mole to mole basis.
– For pure reactants, we measure reactants using
mass
– For reactants that are added to a reaction as aqueous
solutions, we measure the reactants using volume of
solution.
Solution Stoichiometry
grams
A
Molar
mass
moles
A
Molarity
A
Vol
Soln A
Molar
ratio
grams
B
Molar
mass
moles
B
Molarity
B
Vol
Soln B
Solution Stoichiometry Practice
If 25.0 mL of 2.5 M NaOH are needed to
neutralize (i.e. react completely with) a solution of
H3PO4, how many moles of H3PO4 were present
in the solution?
3NaOH (aq) + H3PO4 (aq)  Na3PO4 (aq) + 3H2O(l)
Given: 25.0 mL 2.5 M NaOH
balanced eqn: 3 mol NaOH/1 mol H3PO4
Find: moles of H3PO4
Approach
Molarity
moles
NaOH
NaOH
2.5 M (=mol/L)
Molar
ratio
3 mol NaOH/1 mol H3PO4
0.025 L NaOH soln
moles
H3PO4
Mol NaOH = 25.0 mL x
Vol
NaOH
Soln
1L
2.5 mol
x
1000 mL
1L
25.0 mL NaOH soln
= 0.0625 mol NaOH
More practice
What mass of aluminum hydroxide is
needed to neutralize 12.5 mL of 0.50 M
sulfuric acid?
Solution Stoichiometry
• Solution stoichiometry can be used to
determine the concentration of aqueous
solutions used in reactions.
• Concentration of an acid can be determined
using a process called titration.
reacting a known volume of the acid with a
known volume of a standard base solution
(i.e. a base whose concentration is known)
Titration
Practice
If 35.50 mL of 2.5 M NaOH are needed to
neutralize 50.0 mL of an H3PO4 solution, what is
the concentration (molarity) of the H3PO4
solution?
3NaOH (aq) + H3PO4 (aq)  Na3PO4 (aq) + 3H2O(l)
Given: 35.50 mL 2.5 M NaOH
50.0 mL of H3PO4 sol’n
Find: molarity (mol/L) H3PO4
Strategy: M = moles
L
• To find the concentration of H3PO4 soln, we need
both # moles and volume of H3PO4.
• Since volume is given, we can simply find moles
and plug into the equation for M.
Plan
moles
NaOH
Molar
ratio
moles
H3PO4
Molarity
NaOH
Vol
NaOH
Soln
Mol H3PO4 = 35.5 mL x
1L
1000 mL
x 2.50 mol NaOH x 1 mol H3PO4
1L
3 mol NaOH
=
0.0296 mol H3PO4
We’re not done….we need
molarity.
Molarity of H3PO4
Molarity = moles
L
= 0.0296 mol H3PO4 x 1000 mL
50.0 mL
L
= 0.592 M H3PO4