Chemistry: Atoms First
Julia Burdge & Jason Overby
Chapter 9 .5
Homework: 61, 63, 65, 67,
71, 73, 75, 77, 79 and 81
Kent L. McCorkle
Cosumnes River College
Sacramento, CA
Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
A solution is a homogenous mixture of 2 or more
substances
The solute is(are) the substance(s) present in the
smaller amount(s)
The solvent is the substance present in the larger
amount
Solution
Solvent
Solute
Soft drink (l)
H2O
Sugar, CO2
Air (g)
N2
O2, Ar, CH4
Soft Solder (s)
Pb
Sn
4.1
Solutions
Solvent
solute
When the solvent is water the solution is said to be aqueous
9.1
General Properties of Aqueous Solutions
A solution is a homogenous mixture of two or more substances.
The substance present in the largest amount (moles) is referred to as
the solvent.
The other substances present are called the solutes.
A substance that dissolves in a particular solvent is said to be soluble
in that solvent.
9.5
Concentration of Solutions
Molarity (M), or molar concentration, is defined as the number of
moles of solute per liter of solution.
molarity =
moles solute
liters solution
Other common
rearrangements:
L=
mol
M
mol = M  L
Worked Example 9.8
For an aqueous solution of glucose (C6H12O6), determine (a) the molarity of
2.00 L of a solution that contains 50.0 g of glucose, (b) the volume of this solution
that would contain 0.250 mole of glucose, and (c) the number of moles of glucose
in 0.500 L of this solution.
Strategy Convert the mass of glucose given to moles, and use the equations for
Think About
Check
see that
magnitude
of your answers
interconversions
of M,Itliters,
andtomoles
to the
calculate
the answers.
are logical. For example, the mass given in the problem corresponds
50.0 g
to 0.277 mole
of solute.
If you
(b),mol
for the
moles
of glucose
= are asked, as in=part
0.277
180.2
g/mol
volume that contains a number of
moles
smaller than 0.277, make
sure your answer is smaller than the original volume.
Solution
0.277 mol C6H12O6
(a) molarity =
= 0.139 M
2.00 L solution
0.250 mol C6H12O6
(b) volume =
= 1.80 L
0.139 M solution
(c) moles of C6H12O6 in 0.500 L = 0.500 L×0.139 M = 0.0695 mol
4.5
Dilution is the procedure for preparing a less concentrated
solution from a more concentrated solution.
Dilution
Add Solvent
Moles of solute
before dilution (i)
=
Moles of solute
after dilution (f)
MiVi
=
MfVf
4.5
How would you prepare 60.0 mL of 0.2 M
HNO3 from a stock solution of 4.00 M HNO3?
MiVi = MfVf
Mi = 4.00
Vi =
Mf = 0.200
MfVf
Mi
Vf = 0.06 L
Vi = ? L
0.200 x 0.06
=
= 0.003 L = 3 mL
4.00
3 mL of acid + 57 mL of water = 60 mL of solution
Another Dilution Problem
If 32 mL stock solution of 6.5 M H2SO4 is diluted to a volume of 500 mL
What would be the resulting concentration?
M1*V1 = M2*V2
(6.5M) * (32 mL) = M2 * (500.0 mL)
M2 =
6.5 M * 32 mL
500 mL
M2 =
0.42 M
Concentration of Solutions
Dilution is the process of preparing a less concentrated solution from
a more concentrated one.
moles of solute before dilution = moles of solute after dilution
Concentration of Solutions
In an experiment, a student needs 250.0 mL of a 0.100 M CuCl2
solution. A stock solution of 2.00 M CuCl2 is available.
How much of the stock solution is needed?
Solution: Use the relationship that moles of solute before dilution =
moles of solute after dilution.
Mc × Lc = Md × Ld
(2.00 M CuCl2)(Lc) = (0.100 M CuCl2)(0.2500 L)
Lc = 0.0125 L or 12.5 mL
To make the solution:
1) Pipet 12.5 mL of stock solution into a 250.0 mL volumetric flask.
2) Carefully dilute to the calibration mark.
Concentration of Solutions
Because most volumes measured in the laboratory are in milliliters
rather than liters, it is worth pointing out that the equation can be
written as
Mc × mLc = Md × mLd
Worked Example 9.9
What volume of 12.0 M HCl, a common laboratory stock solution, must be used
to prepare 150.0 mL of 0.125 M HCl?
Strategy Mc = 12.0 M, Md = 0.125 M, mLd = 250.0 mL
Solution
12.0 M × mLc = 0.125 M × 250.0 mL
mLc =
0.125 M × 250.0 mL
= 2.60 mL
12.0 M
Think About It Plug the answer into Equation 9.4, and make sure that the
product of concentration and volume on both sides of the equation give the same
result.
Worked Example 9.10
Starting with a 2.0 M stock solution of hydrochloric acid, four standard solutions
(1 to 4) are prepared by sequential diluting 10.00 mL of each solution to
250.00 mL. Determine (a) the concentrations of all four standard solutions and
(b) the number of moles of HCl in each solution.
Strategy (a) Md =
Mc× mLc
-1 L
;
(b)
mol
=
M×L,
250.00
mL
=
2.500×10
mLd
Solution
2.00 M × 10.00 mL
-2 M
(a) Md1 =
=
8.00×10
250.00 mL
8.00×10-2 M × 10.00 mL
= 3.20×10-3 M
Md2 =
250.00 mL
3.20×10-3 M × 10.00 mL
Md3 =
= 1.28×10-4 M
250.00 mL
1.28×10-4 M × 10.00 mL
= 5.12×10-6 M
Md4 =
250.00 mL
Worked Example 9.10 (cont.)
Solution
(b) mol1 = 8.00×10-2 M × 2.500×10-1 L = 2.00×10-2 mol
mol2 = 3.20×10-3 M × 2.500×10-1 L = 8.00×10-4 mol
mol3 = 1.28×10-4 M × 2.500×10-1 L = 3.20×10-5 mol
mol4 = 5.12×10-6 M × 2.500×10-1 L = 1.28×10-6 mol
Think About It Serial dilution is one of the fundamental practices of
homeopathy. Some remedies undergo so many serial dilutions that very few (if
any) molecules of the original substance still exist in the final preparation.
Worked Example 9.11
Using square-bracket notation, express the concentration of (a) chloride ion in a
solution that is 1.02 M in AlCl3, (b) nitrate ion in a solution that is 0.451 M in
Ca(NO3)2, and (c) Na2CO3 in a solution in which [Na+] = 0.124 M.
Strategy Use the concentration given in each case and the stoichiometry
indicated in the corresponding chemical formula to determine the concentration of
the specified ion or compound.
Solution (a) There are 3 moles of Cl- ion for every 1 mole of AlCl3,
AlCl3(s) → Al3+(aq) + 3Cl-(aq)
so the concentration of Cl- will be three times the concentration of AlCl3.
[Cl-]
3 mol Cl= [AlCl3] ×
1 mol AlCl3
1.02 mol AlCl3
3 mol Cl=
×
L
1 mol AlCl3
3.06 mol Cl=
= 3.06 M
L