CHEMISTRY 4
CHAPTER 16
SOLUTION
1 THE CHARACTERISTICS OF A SOLUTION
Solution is a homogenous mixture. A solution can exist in any of the three states: gas,
liquid, solid.
2 SOLUTION TERMINOLOGY
Concentrated and dilute.
A concentrated solution has a relatively large quantity of solute per unit amount of
solution, a dilute solution has a relatively small quantity of solute per unit amount of
solution.
Solubility. Saturated and unsaturated solutions.
A solution that can exist in equilibrium with undissolved solute is a saturated solution.
The concentration of the saturated solution is called the solubility of the solute.
A solution is unsaturated if its concentration is smaller than the solubility.
Supersaturated solutions
A solution is said to be supersaturated if its concentration is greater than the solubility.
Supersaturated solution is not stable, a slight physical disturbance can start
crystallization.
3 THE FORMATION OF A SOLUTION
The formation of a sodium chloride solution from sodium chloride crystals is made
possible by the interaction of water molecules with sodium and chloride ions. The
sodium and chloride ions in solution are surrounded by polar water molecules, and are
said to be hydrated.
NaCl (s) ↔ Na+ (aq) + Cl- (aq)
The dissolving process is reversible. When the rates of dissolving and crystallization are
the same the solution is saturated.
The rate of dissolving depends on
1 The surface area. A finely divided solid dissolves more rapidly.
2 The diffusion of the solute from the surface. Stirring increases the rate.
3 Temperature. The higher the temperature, the higher the rate.
4 FACTORS THAT DETERMINE SOLUBILITY
1 Intermolecular forces.
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Generally, if the forces between A molecules are about the same as the forces between B
molecules, A and B will probably dissolve in each other. The rule of solubility is like
dissolves like. A nonpolar solvent can dissolve a nonpolar solute. A polar solvent can
dissolve a polar or ionic solute.
2 Partial pressure of solute gas .
Pressure has little effect on the solubility of solids or liquids but has a pronounced effect
on the solubility of gases. For an ideal solution, the solubility of a gas in a liquid is
directly proportional to the partial pressure of the gas over the surface of the liquid.
3 Temperature.
The solubility of most solids increases with raising temperature. The solubility of gases
decreases with raising temperature.
5 SOLUTION CONCENTRATION: PERCENTAGE BY MASS
By definition, percentage by mass is grams of solute per 100 grams of solution.
% by mass = 100x g solute/100 g solution
Example : When 125 grams of a solution is evaporated to dryness, 42.3 grams of solute
was recovered. What was the percentage of the solute?
% by mass = 100 x 42.3 g/ 125 g = 33.8%
Example : You are to prepare 2.50x102 g of 7.00% Na2CO3 solution. How many grams
of sodium carbonate and how many milliliters of water do you use?
Grams Na2CO3= 2.50 x 102 g of solution x ( 7.00 g Na2CO3/ 100 g solution)
= 17.5 g Na2CO3
Gram of water = 2.50 x 102 g of solution - 17.5 g Na2CO3
= 233 g of water
Volume of water = 232 g of water x (1.00 mL water/ 1.00 g water) = 233 mL of water
6 SOLUTION CONCENTRATION:MOLARITY
The molarity , M, is the moles of solute per liter of solution
M = moles solute/liter solution
The unit of molarity is mol/L
Example Calculate the molarity of a solution made by dissolving 15 g of NaOH in water
and diluting to 100 mL
Moles of NaOH = 15 g NaOH x ( 1.00 mole NaOH/ 40.00 g NaOH) = 0.38 mol. NaOH
Molarity = 0.38 mol NaOH / 100 mL x ( 1000 mL/ 1.00 L ) = 3.8 mol/L = 3.8 M
Example How many grams of silver nitrate must be dissolved to prepare 5.00 x102mL of
0.150 M AgNO3
5.00 x102mL x ( 1.00 L/ 1000 mL)x(0.150 mol AgNO3/L)x(169.9 g AgNO3/mol AgNO3)
= 12.7 g AgNO3
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Example Find the volume of a 1.40 M solution that contains 0.287 mole of ammonia.
0.287 mole NH3 x ( 1L/ 1.40 mole NH3) = 0.205 L
7 SOLUTION CONCENTRATION: NORMALITY
Equivalent :
One equivalent of an acid is the quantity that yields one mole of hydrogen ions in a
chemical reaction.
One equivalent of a base is the quantity that reacts with one mole of hydrogen ions.
The number of equivalents in a mole of an acid depends on a specific reaction.
For hydrochloric acid, one mole of HCl is one equivalent.
For phosphoric acid , depending on reaction conditions, the number of equivalents per
mole can be 1, 2 or 3.
NaOH + H3 PO4 
2 NaOH + H3 PO4 
3 NaOH + H3 PO4 
Na H2PO4
Na 2HPO4
Na3PO4
+
+
+
H2O
2 H2O
3 H2O
When an acid reacts with a base the number of equivalents of acid is equal to the number
of equivalents of the base.
Equivalent mass is the number of grams per equivalent. Equivalent mass can be
calculated by dividing molar mass by the number of equivalents in one mole.
Normality, N, is the number of equivalents per liter of solution
N = equivalents solute/liter solution
8 DILUTION OF CONCENTRATED SOLUTION
During dilution, only solvent is added, the amount of solute remains the same.
The number of moles of solute = Volume in liter x Molarity
The number of moles of solute remains the same :
Vc x Mc = :
Vd x Md
Example : How many milliliters of concentrated hydrochloric acid, which is 11.6 M,
should be used to prepare 5.50 liters of 0.500 M HCl ?
Vc x Mc = :
Vc x 11.6 M = :
Vd x Md
5.50 L x 0.500 M
Vc = 5.50 L x 0.500 M /11.6 M = 0.237 L = 237 mL
8 SOLUTION STOICHIOMETRY
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Example : How many grams of lead(II) iodide will precipitate when excess potassium
iodide solution is added to 50.0 mL of 0.811 M lead(II) nitrate.
Pb(NO3)2 (aq) + 2 KI(aq)  PbI2(s) + 2 KNO3(aq)
Given: 50.0 mL of 0.811 M lead(II) nitrate
Wanted: gram of lead(II) iodide.
Path:
mL of Pb(NO3)2 liters of Pb(NO3)2moles of Pb(NO3)2moles of PbI2gram of PbI2
Calculation :
50.0 mL of Pb(NO3)2 x (L of Pb(NO3)2/ 1000 mLof Pb(NO3)2)
x (0.811 mol of Pb(NO3)2/L Pb(NO3)2)
x (1 mol PbI2 /1 mol Pb(NO3)2)
x (461.0 g PbI2 /mol PbI2) = 18.7 g PbI2
9 TITRATION USING MOLARITY
Titration is the addition of a precise volume of one solution into another by means of a
burette. Usually an indicator is used to detect the moment the volume required to react
with a measured amount of another dissolved substance has been reached.
Titration can be used to standardized a solution. Sodium hydroxide cannot be weighed
accurately because it absorbs water from the air. Oxalic acid dihydrate which can be
weighed accurately may be used as a primary standard.
Example 1.18 g H2C2O4. 2H2O ( 126.07 g/mol) are dissolved in water and the solution is
titrated with a solution of NaOH of unknown concentration. 28.3 ml NaOH are required
to neutralize the acid. Calculate the molarily of the NaOH solution.
H2C2O4. + 2 NaOH = Na2C2O4. + 2 H2O.
Given 1.18 g H2C2O4. 2H2O, 28.3 mL (0.0283L) NaOH
Wanted mol/L NaOH
Path:
g H2C2O4. 2H2O mol H2C2O4. 2H2Omol H2C2O4mol NaOHmolarity NaOH
Factor
1 mol H2C2O4. 2H2O/ 126.07g H2C2O4. 2H2O,
2 mol NaOH/1 mol H2C2O4.
1.18 g H2C2O4. 2H2O x (1 mol H2C2O4. 2H2O/ 126.07g H2C2O4. 2H2O )
x (1 mol H2C2O4. / 1 mol H2C2O4. 2H2O)
x ( 2 mol NaOH/1 mol H2C2O4.) x ( 1/ 0.0283L NaOH)
= 0.661 mol/L
10 TITRATION USING NORMALITY
Normality is a convenient concentration in analytical chemistry. Since the number of
equivalents of acid and base in a reaction is the same, we have:
equivalents of acid = equivalents of base
or
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Na x Va = Nb x Vb
Example A 25.0 mL sample of an electroplating solution is analyzed for its sulfuric acid
concentration. It takes 46.8 mL of the 0.661 N NaOH to neutralize the sample. find the
normality of the acid.
Na = Nb x Vb /Va = (46.8 mL / 25.0 mL ) x 0.661 N = 1.24 N H2SO4
11 COLLIGATIVE PROPERTIES OF SOLUTIONS
Colligative properties are solution properties that are determined only by the number of
solute particles dissolved in a fixed quantity of solvent and not on the identity of the
solute particles.
Freezing point depression: Solution has lower freezing point.
Boiling point elevation: Solution has higher boiling point.
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