Solutions are homogeneous mixtures. The substance in the solution that is present
in the largest quantity is called the solvent. The substance dissolved in the solvent is the
solute.
The amount of solute
dissolved in a specific amount of solvent is referred to as the concentration
of the solution.
Percent by mass
mass of solute x
100
mass of solution
Ex:
8.1 g of sugar dissolved in 45.0 g of water.
mass of solute = 8.1 g
mass of
solution = 8.1 g + 45.0 g = 53.1 g
Mass
percent = (8.1 g/53.1 g) x 100
= 15% by mass
(15
grams solute for every 100 grams solution)
Calculate the percent by
mass of the following solutions.
17.25 g KCl in 435 mL of
water
0.25 g phenolphthalein in
275 mL of ethanol
How much solute and
solvent would be required to prepare the following solutions?
100.0
g of 2.5 % NaCl solution
250.0
g of 2.5% NaCl solution
745.0
g of 1.8% NaHCO3 solution
Molarity
-- expression of concentration using number of solute particles (moles) present
rather than mass of solute present
Molarity
(M) = moles
of solute
liters of solution
Example: 5.85 g of NaCl in 500.0 mL of solution
5.85
grams NaCl x 1 mol NaCl = 0.100 mol NaCl
58.44 g NaCl
Molarity
= 0.100 mol = 0.200
M
0.5000 L
Question: How many grams of NaCl would be needed to
make 200.0 mL of 0.300 Molar NaCl solution?
0.300 mol x 0.2000
L = 0.0600 mol
1 L
0.0600
mol x 58.44
g = 3.51 g NaCl
1 mole
Calculate
the molarity of the following solutions.
170.0
g of BaCl2 in 250.0 mL of solution
11.5
g of NaOH in 1.50 L of solution
1.56
g HCl in 26.8 mL of solution
Calculate
the number of moles of solute present in the following solutions.
25
mL of 0.75 M AgNO3
1.75
L of 1.0 x 10-3 M ZnCl2
2.5
L of 12.3 M formaldehyde (CH2O)
Calculate
the mass of solute needed to prepare the following solutions.
1.0
L of 0.200 M K2Cr2O7
3.75
L of 0.200 M NaIO3
525
mL of 0.050 M CuSO4
Note:
To prepare solutions of a known concentration (standard solutions), a volumetric flask is used to accurately measure the volume
of the final solution. (See Fig 4.10)
What
mass of solute is contained in 1.5 L of 0.10 M H2SO4?
What
volume of 0.14 M NaCl would contain 1.0 mg NaCl?
Dilution
of Solutions: Solutions of high
concentration can be diluted by adding more solvent
Diluting
only adds solvent, so the number of moles of solute remains the same.
Initial
Molarity(M1) x Initial Volume(V1) = moles of solute
Final
Molarity(M2) x Final Volume(V2) = moles of solute
Dilution
Equation: M1(V1) = M2(V2)
100.0
mL of a 2.0 M solution of AgNO3 is diluted by adding 900.0 mL of
water. What is the concentration of the
new solution?
M1
= 2.0 M V1
= 100.0 mL =
0.1000 L
M2
= ? V2
= 0.1000 L +
0.9000 L = 1.000 L
2.0
M x 0.1000 L = M2
x 1.000 L
M2
= 2.0
(0.1000) = 0.20 M
1.000
How
many milliliters of water must be added to 50.0 mL of a 0.40 M NaCl solution to
obtain a 0.050 M NaCl solution?
M1 = 0.40 M V1
= 0.0500 L
M2 = 0.050 M V2
= ?
0.40
(0.0500) = 0.050(V2) V2 =
0.40 L
0.40
L – 0.0500 mL = 0.35 L of water must be added
Solution Stoichiometry
Reactions which take place
in solution still involve use of mole ratios.
Problem
1: Sodium bicarbonate can be added to spilled sulfuric
acid to neutralize it as shown:
2NaHCO3 + H2SO4
---> Na2SO4 + 2H2O + 2CO2
If 30.0 mL of 6.0 M sulfuric acid are spilled, what is the
minimum mass of sodium bicarbonate which must be used to neutralize it?
·
An indicator is
used which changes color when exactly enough of Reactant 1 has been added to
Reactant 2 to completely use up all of Reactant 2.
·
When the color
change is noted, the addition of Reactant 1 is stopped and the volume of
Reactant 1 added is measured. This
volume can be used along with the concentration of Reactant 1 to determine the
number of moles of Reactant 1 added.
·
Stoichiometry can
be used to determine the number of moles of Reactant 2 that must have been present
to react with Reactant 1.
·
The number of
moles and the volume of Reactant 2 can be used to determine the molarity of
Reactant 2.
·
An acid-base
titration involves the addition of an acid of known concentration to a base of
unknown concentration, or vice-versa.
Example: 25.00 mL of an unknown concentration NaOH
solution could be titrated with a 1.25 M H2SO4
solution. A few drops of phenolphthalein
indicator added to the NaOH solution before the titration begins would make the
NaOH solution turn pink. When exactly
enough acid is added to completely neutralize all of the
base, the solution will turn colorless.
If the following data was
obtained, calculate the concentration of the NaOH solution.
Volume
of NaOH used: 25.00 mL
Volume
of H2SO4 used: 37.62
mL
Concentration
of H2SO4: 1.25
M
What volume of 0.0496 M HClO4
is needed to neutralize 25.0 mL of 0.505M KOH?
HClO4
(aq) + KOH(aq) à KClO4 (aq) + H2O(l)
What is the concentration of
a NaOH solution if
36.2 mL of this solution is needed to neutralize 25.0 mL of
0.0513 M HNO3?
HNO3
(aq) + NaOH(aq) à NaNO3 (aq) +
H2O(l)
What mass of calcium
carbonate is needed to neutralize 26.7 mL of 0.0887 M HCl?
2
HCl (aq)+
CaCO3 (s) à CaCl2 (aq)+H2O(l)+
CO2(g)
What is the concentration of
a magnesium hydroxide solution if 10.00 mL of this solution is needed to
neutralize 27.25 mL of 0.174 M H3PO4?
2
H3PO4 (aq)+3
Mg(OH)2 (aq) à Mg3(PO4)2
(aq)+6 H2O(l)
What is the concentration of
a phosphoric acid solution if 15.00 mL of this solution is needed to neutralize
24.69 mL of 2.65 M Mg(OH)2?
2
H3PO4 (aq)+3
Mg(OH)2 (aq) à Mg3(PO4)2
(aq)+6 H2O(l)