SOLUTIONS
are
HOMOGENEOUS MIXTURES
They consist of at least TWO (or more) COMPONENTS:
1.
SOLVENT which forms a
CONTINUUM MEDIUM
& is present in excess of the other component(s), which are
dispersed in it, down to individual particles of molecular size.
These components are:
2.
SOLUTE(S)
(one or more solutes in one & the same solvent).
SOLUTIONS are
SINGLE-PHASE SYSTEMS: ALL-GAS,
ALL-LIQUID, or ALL-SOLID.
PHYSICAL STATE of a SOLUTION is determined by
the PHYSICAL STATE of SOLVENT.
Pure solute (i.e. the substance to be dissolved in a solvent)
can be in any physical state: gas, liquid or solid.
GAS MIXTURE is a SOLUTION [Air]
SOLID SOLUTIONS of METALS are ALLOYS
[Brass: Cu & Zn; Steel: Fe & C & other metals, as Cr, Ni,
Mn, etc.]
No chemical reactions occur in the bulk of solids since their particles are
immobile (however, reactions at their surface are important).
GAS & LIQUID SOLUTIONS are the usual media for
chemical reactions.
Most abundant liquid solvent is WATER.
Other liquid solvents are acetone, ethanol, ether, etc.
Pure solute can be in any physical state:
Solid solute in liquid solvent: sugar in water, aqueous
solution of salt.
Liquid solute in liquid solvent: antifreeze - ethylene
glycol in water, vodka - ethanol in water, etc.
Gaseous solute in liquid solvent: soda -
CO2 in
water.
Often gases, liquid & solids serve as solutes all at once to form a solution:
soft drinks - sugar(s), CO2(g), alcohol(l).
DISSOLUTION PROCESS
Two aspects:
RATE of DISSOLUTION (time course of the process,
or its KINETICS. How fast does the dissolution occur?)
 SOLUBILITY
(EQUILIBRIUM does not depend on time:
How much can be dissolved, no matter how long it would take?)
To increase the Rate of dissolution:
1. Increase the surface area of the solute. [Powder dissolves
faster than a rock of the same material]
2. Agitate the solution [Magnetic stirrers]
3. Heat the solvent.
SOLUBILITY
is determined by
the
dissolution equilibrium,
i.e. the state of the system in which
rate of dissolution = rate of precipitation
(or any separation: crystallization of solids, gas escape, or separation of
another liquid).
SOLUTE + SOLVENT  SOLUTION
A solution which is, or can be, in equilibrium with pure solute is a
SATURATED SOLUTION
A solution that contains less solute
than a saturated solution is
UNSATURATED SOLUTION
A solution that contains more solute, than a saturated solution, is
SUPERSATURATED SOLUTION
To produce a supersaturated solution, a saturated
solution is first prepared at higher temperature, then
cooled.
Supersaturated solution is unstable.
The solute readily crystallizes from a supersaturated solution, when a
“seed” - a small crystal of the same substance - is dropped into it.
Saturated
at 75 oC
Supersaturated
at 20 oC,
unstable
cooling
Saturated
at 20 oC
adding seed,
precipitation
Solutions may be concentrated (over 10%) or diluted (under 1%)
That has nothing to do with them being saturated or unsaturated:
A saturated solution can be still diluted (at low solubility).
SOLUBILITY, S:
is the amount of solute required to form a saturated solution
with a specific amount of solvent at a specified temperature.
grams of solute per 100 g of solvent:
S(sugar, 20 C) = 204 g / 100 g water
o
SOLUBILITY depends on:
 nature of solvent & solute;
 temperature;
 pressure (for gases only)
GENERAL SOLUBILITY RULES
NATURE OF SOLVENT & SOLUTE:
“Like dissolves like”:
Ionic crystals & polar substances dissolve in water & other
polar solvents.
NaCl dissolves in water, ethanol in water, ethanol in acetone, etc.
But: NaCl, water do not dissolve in gasoline (nonpolar)
Nonpolar substances dissolve in nonpolar solvents.
Effect of TEMPERATURE:
Solubility changes with temperature.
In most cases, for solids & liquids, solubility increases
when temperature increases.
However, this is not a general
rule, & there are substances
whose solubility in water does
not essentially depend of
temperature, or even decreases
when temperature increases.
Solubility of Gases always decreases when temperature
increases. Liquids can be degassed by heating.
For GASES only:
PRESSURE EFFECT
SOLUBILITY increases when PRESSURE of
the gas INCREASES
EXAMPLE: determine in which points the solution is saturated,
unsaturated, supersaturated.
If 100 g of a solution saturated at 60oC was cooled down to 14oC, how
many grams of precipitate were formed?
EXAMPLE: one of the substances in the graph below is a gas.
Identify it.
There is only one solute, A, for which solibility decreases while
temperature increases. This behavior is typical for gases.
Reasons for dissolution:
1. The common reason for dissolution of solids & liquids is
the general trend of the nature to a disorder, chaos,
which is provided by heat, i.e. random motion of
particles:
When a crystal dissolves, its constituent particles (atoms, ions or
molecules) initially ordered in a crystalline lattice, pass to a solution
with no ordering.
When two liquids mix, the extent of ordering is reduced, too.
2. Energy relationships. When a crystal dissolves, the attraction
between its particles is replaced by the attraction between these
particles & solvent molecules. The strength of these interactions is
different & may be stronger than that of particles in the solid itself.
The physical meaning of the solubility rule:
“Like dissolves like”:
The attraction forces between particles in pure solute, pure solvent,
& solute-solvent are similar. Then the trend to chaos prevails (two
polar, or two non-polar substances).
Dissolved ions or molecules are solvated
(hydrated for aqueous solutions).
Water properties are very special.
H – O – H There are 2 highly polar covalent bonds & 2 lone pairs
on oxygen. Water is a strong dipole:
O
H
H
Dipoles, unlike ions, do not travel in electric field, but orient in it:
Battery
No electric
field
Electric
field
Unlike other polar substances, water has a special ability to form two additional
bonds by attracting hydrogens (protons), from neighbor water molecule, to its
oxygen lone electron pairs, so that each OXYGEN is surrounded by 4, rather than 2
hydrogens, forming tetrahedron. This is hydrogen bond: -O…H – O – H…
0.28 nm
ICE: each O is a center of tetrahedron, with 4 O-H bonds. Overall, it is a an
infinite network – a crystalline lattice.
LIQUID WATER: similar, but partially disordered, 3.4 bonds for each O on
average. Some water molecules occupy empty spaces in the lattice, so that
density of liquid water is higher than that of ice.
Due to two-fold H-bonding, water
behave like it were higher molecular
mass, than H2O
(18 amu), e.g. it boils much higher
than its analogs:
H2S, H2Se, H2Te & H2Po
H-bonds form not only between water molecules, but also by other compounds with
three most electronegative elements: O, N, F.
H–F…H–F…H–F… is also a liquid at room temp., while formally heavier HCl,
HBr & HI are all gases.
Non-polar substances, as gasoline, do not dissolve in water because their
dissolution would require disruption of H-bonds in water.
Substances that repel water are called hydrophobic.
The ultimate reason of hydrophobicity is H- bonding of water.
Polar substances, especially those that can either form H-bonds
themselves (if liquid, they are called protonic solvents), or at least
have electronegative atoms with lone electron pairs, to accept protons
from water, dissolve well in water:
H
|
aprotonic solvent:
Ethanol: C2H5 – O – H…O – C2H5
Acetone: CH3 – C – CH3
Ethanol-water
H
||
|
O – H…O
C2H5 – O – H…O – H
|
H
Hydrocarbons, such as methane, CH4, or gasoline, C8H18, have no electronegative atoms, so
that they can’t form H bonds, & are therefore water-insoluble.
H bonds are important in biochemistry, maintaining 3D structure of proteins & nucleic acids.
The ultimate reason of hydrophobicity is H-bonding of water.
Ionic compounds dissolve in water by electrolytic dissociation:
ions separate & are hydrated: NaCl  Na+(aq) + Cl-(aq)
Dissolution equation
Ions, due to their electrical charges, orient water molecules around
themselves:
+
+
+
-
-
+
-
-
-
K+
+
-
+
+
+
-
+
-
Cl+
+
-
+
+
-
+
+
-
+
-
Solubility of ionic compounds depends on the relative strength of
interactions between ions in the crystalline lattice, & interaction of ions with
water molecules.
Solubility rules:
(each previous rule is senior to all subsequent)
Generally Soluble:
All salts of alkaline metals & NH4+
No common exceptions
____________________________
All nitrates, NO3-, nitrites NO2-,
Chlorates, ClO3-, Perchlorates, ClO4Permanganates, MnO4Acetates, CH3COO-,
Hydrogen-containing anions, such as HCO3No common exceptions
____________________________
Chlorides, ClExceptions: Ag+, Hg+, Pb2+
Bromides, Br-, & iodides, IExceptions: Heavy metals, AgBr, AgI, PbI2
____________________________
Sulfates, SO42Exceptions: Ba2+, Pb2+, Hg2+
Generally Insoluble:
Sulfides, S2Exceptions: Mg2+, Ca2+
_________________________________________________
Carbonates, CO32-, Phosphates, PO43-, Arsenates, AsO43-
CONCENTRATION of SOLUTIONS
MASS PERCENT (%):
grams of solute per 100g of solution
EXAMPLE:
To prepare 200 g of 5% KOH, we need 5 g KOH per each 95 g of water, i.e. to dissolve 10 g of KOH in
190 g of water.
The answer will not change for whatever other substance: to prepare
200g of 5% H2SO4, we still mix 10g of sulfuric acid with 190g of
water. Molar mass is not involved. Mostly used in technology, pharmacy, other
applications.
MOLAR CONCENTRATION,
or
MOLARITY (M, i.e. mol/L)
moles of solute per 1L of solution.
EXAMPLE:
To prepare 1L of 0.05M NaOH we dissolve 0.05mol NaOH, i.e.
(Na-23, O-16. H-1)
0.05mol  40g/mol = 2g NaOH, in water, adjusting the overall volume
of the solution to 1.00L.
Thus, concentrations may be based on mass or moles of solute, per mass or volume of solution.
Stoichiometry of Solutions
Typical Problems
1. Determine the amount of substance, in moles, or grams, in a given
volume of a solution of known molarity:
Example:
How many moles of NaCl are there in 200. mL of 0.0300 M NaCl?
Solution: Molarity = number of moles in a unit of volume, liter:
M = n/V
n = MV
NaCl(mol) = 0.03 mol/L  0.200L = 0.006 mol = 6 mmol
Example:
How many grams of NaCl are needed to prepare 200 mL of
0.03M NaCl?
Solution: same as above, but one extra step to convert moles into grams using molar
mass of the solute:
NaCl(gram) = 0.03mol/L0.200L58.5g/mol = 0.035gNaCl
23+35.5=58.5g/mol
How solutions of a given molarity are technically prepared starting
with a pure (solid) solute:
1. Take a mass of solute using balance.
2. Place it in a graduated flask of a desired volume.
3. Add water to the upper mark.
2. Prepare a solution of a given molarity by dilution of a more
concentrated solution.
Example: How many milliliters of a 0.50M KOH must be taken to
prepare 3.0L of 0.025M KOH?
Solution: At dilution, total number of moles does not change. Since
the number of moles is: n = M1V1 = 0.025mol/L3.0L = 0.075mol
V2= n/M2 = 0.075mol / 0.50mol/L = 0.15L = 150mL
Or:
Example: How many milliliters of water must be added to 200.mL of
1.5M HNO3 in order to prepare 0.50M HNO3?
Solution: Again, the number of moles of HNO3 is:
n = M1V1 = 1.5 mol/L  0.200L = 0.30 mol.
the volume of the solution to prepare will be:
V2 = n/M2 = 0.30 mol/0.50mol/L = 0.60 L
the volume of water to add: 0.60 – 0.20 = 0.40L = 400mL.
(here again, we assumed the densities of solutions same as water)
3. Determine the molarity of a solution prepared by mixing two
solutions of different molarities.
Example: What will be the molarity of a solution prepared by mixing
0.25L of 0.020M & 1.75L of 0.010M solutions of sulfuric acid?
Solution: M = n/V
Here n = n1 + n2; V = V1 + V2; n1 = M1V1 n2 = M2V2
M = (M1V1 + M2V2) / (V1+V2) =
(0.25 0.02 + 1.75 0.01) / (0.25+1.75) = 1.12 mol/L
(assuming the volumes additive, i.e. densities of both solutions are the same, &, for diluted solutions, 1.0g/mL)