Ions in Aqueous solution & Colligative Properties Chpt. 14
I. Compounds in aqueous solution
A. Dissociation -separation of an ionic compound that occurs when it
dissolves in water
1. always assume the ions have 100% dissociation
2. indicate ions produced & # of moles produced
NaCl(s)  Na+(aq) + Cl–(aq)
FeCl3
(s)
Fe+3
(aq)
+ 3Cl-1 (aq)
3.Not all ionic compounds are soluble in water.
4. Precipitation Reactions occur when one or more ionic comounds are
insoluble. To determine which it is use a solubility chart .
a. precipitate – a solid produced from a chem reaction in a solution– the
solid settles out of the solution
5. Net ionic equation = an equation that includes only those cmpds & ions that
undergo a chem change in a reaction in an aq solution
a. Spectator ions = ions that don’t take part in a chem reaction & are
found in solution both BEFORE & AFTER the reaction – these ions are
NOT written in a net ionic equation
b. LOOK FOR A PRECIPITATE! These ions are the ones involved in the
reaction!
c. How to write net ionic equations:
1) Start with a balanced equation (predict products if needed)
(NH4)2S
(s)
+ Cd(NO3)2(s)
water
2NH4NO3 +CdS
2) Write all strong electrolytes as ions (write them dissociated) :
electrolytes conduct electricity in solution
Use your Ion Chart!!!
(NH4)2S
(s)
a) Strong Acids: HCl, HBr, HI, and any oxygen containing acid
with 2 or more oxygens (all organic acids --contain carbon-- are
weak electrolytes)
b) Strong Bases: hydroxides of group I metals and barium and
strontium are strong electrolytes
c) Soluble salts: NaCl
→ NH4+(aq) + S-2(aq)
Cd(NO3)2(s) → Cd+2 (aq) + 2NO3-1(aq)
NH4+(aq) + S-2(aq) + Cd+2 (aq) + 2NO3-1(aq)
soooooo…
double replacement reaction
Hints: Oxides, gases,and water are always written as molecular or undissociated
3) Check your solubility chart to see which new cmpd is insoluble.
NH4+(aq) + S-2(aq) + Cd+2 (aq) + 2NO3-1(aq)
NH4NO3(aq) + CdS (s)
4) Cross out all spectator ions (ions that are the same on the
reactant and product side of the equation –if they stay dissolved
then they are considered the same so, look for a precipitate….
something insoluble in water!)
NH4+(aq) + S-2(aq) + Cd+2 (aq) + 2NO3-1(aq)
NH4+(aq) +
NO3-1(aq) + CdS (s)
5) Write the Net Ionic equation using the substance NOT crossed
out!
Cd+2 (aq) + S-2(aq)
CdS (s)
B. Ionization - breaking apart of some polar molecular compounds into
aqueous ions (it’s the creation of ions where none previously
existed!)
1. acids are polar molecules that ionize in water
HNO3(aq) + H2O(l)  H3O+(aq) + NO3–(aq)
(H3O+ is the hydronium ion we’ll talk about it with acids)
C. Molecular Solvation the molecules dissolve in water but they stay intact
C6H12O6(s)  C6H12O6(aq)
these substances are nonelectrolytes!
II. Colligative Property- property that depends on the concentration of solute
particles, not their identity---- Changes in the behavior of the solutions that
are due primarily to the concentration of solute particles rather than the
typical properties of those substances are known as colligative properties.
A. Types
1. Freezing Point Depression (tf)
a. f.p. of a solution is lower than f.p. of the pure solvent
t:
kf:
m:
change in temperature (°C)
constant based on the solvent (°C·kg/mol) page p.445 in the book
molality (m)
n:
# of (ion) particles in solution (1 for non-electrolytes & you must write net
ionic equations to find # for electrolytes)
2. Boiling Point Elevation (tb)
a. b.p. of a solution is higher than b.p. of the pure solvent
t:
kb:
m:
n:
change in temperature (°C)
constant based on the solvent (°C·kg/mol) page p.445 in the book
molality (m)
# of (ion) particles in solution (1 for non-electrolytes & you must write net
ionic equations to find # for electrolytes)
3. Osmotic pressure – external pressure that must be applied to STOP
osmosis
a. osmosis – movement of water through a semipermeable membrane
from the side of lower [SOLUTE] to the side of higher [SOLUTE]
OR
the diffusion of water through a semipermeable membrane
b. eventually the concentrations of the 2 solutions become equal
(equilibrium)
c. increasing the number of dissolved particles increases the osmotic
pressure
d. Osmotic Pressure of the Blood
1) Cell membranes are semipermeable
2) The osmotic pressure of blood cells cannot change or damage
occurs.
3) The flow of water between a red blood cell and its surrounding
environment must be equal (isotonic solution)
a) Medically 5% glucose and 0.9% NaCl are used (their solute
concentrations provide an osmotic pressure equal to that of red
blood cells)
b) Hypotonic Solutions
Lower osmotic pressure than red blood cells
Lower concentration of particles outside than inside RBCs
In a hypotonic solution, water flows in to the RBC
The RBC undergoes hemolysis; it swells and may burst.
*Greater concentration of water outside cell than inside so
water moves into the cell.
c) Hypertonic Solutions
Has higher osmotic pressure than RBC
Has a higher particle concentration
In hypertonic solutions, water flows out of the RBC
The RBC shrinks in size (plasmolysis)
*Greater concentration of H2O inside cell than outside so
water moves out
B. Applications
1. salting icy roads – lowers the f.p. of water so it must be colder outside in
order to form ice
2. making ice cream-lowers the f.p. of ice & causes it to melt – when
something melts it is an endothermic process!! It pulls E (heat) from the
milk mixture making it colder
3. antifreeze – lowers f.p. of water & increases b.p. of water in radiator
(-64°C to 136°C)
4. Dialysis
a. Occurs when solvent (water) and small solute particles pass through
a semipermeable membrane in kidney
b. Large particles retained inside blood small stuff moves through to
kidney & then to bladder
c. Hemodialysis is used medically (artificial kidney) to remove waste
particles such as urea from blood when kidney no longer functions
C. Calculations
1.Find the freezing point of a saturated solution of NaCl containing 28 g NaCl
in 100. mL water. NaCl
Na+ + Clt:
?
kf:
1.86 (°C·kg/mol)
m: 4.8 m 28g convert to moles = .48mol then… .48mol/.100kg = 4.8 m
n:
2 ions
t = kf x n x
m
Δt = 1.86(°C·kg) x 2 x 4.8mol = 17.856 °C you’re not finished yet!!
mol
kg
normal f.p. of water is 0°C so 0°C – 17.856 = -17.856°C