1. SOLUTIONS
Solutions : Homogeneous mixture of two or more substances. Consist of a solute and a solvent.
Properties of a solution
Solutions have variable composition
Solutions are always clear ( transparent to light)
Solutions are homogeneous
Solutions do not settle out
Solutions can be separated by physical means
Solutions pass through filter paper

2. SOLUBILITY (FORMATION) OF SOLUTIONS
Ionic substances (NaCl) will dissolve in polar solvents (H2O)
The positive hydrogen atoms of water attract the negative Cl- ions.
The negative oxygen atoms of water attract the positive Na+ ions.
This results in hydration (solvation) of the Na+ and Cl- ions.

3. NaCl dissolves in water because it is ionic (polar)
Nonpolar compounds like Cl2 and oil do not dissolve in polar solvents like water.
Like dissolves like.

4. Factors affecting solubilities.
* Temperature.
Solubility of solid solutes generally increases with increase in temperature. However there are exceptions,
e.g. Ce(SO4)3 solubility decreases.
NaCl little change.
Gases become less soluble with incr. in temperature, e.g. SO2.
* Pressure.
Solids and liquids are unaffected by change in pressure.
For gases solubility in liquid increases with increase in partial pressure above the liquid. (Henry’s Law)
e.g. Soda Can.

5. IONIZATION.
Some solutes ionize in solution.
Ionization results in the formation of ions or electrolytes. (Biologically referred to as minerals)
Electrolytes are substances whose water solution conducts electricity. They dissociate into ions when placed in water (ionisation). Ions are charged and the net charge is zero.
Nonelectrolytes do not conduct electricity.

6. Conductivity.
When electrolytes are placed in solution they dissociate into ions. The ions are attracted to oppositely charged electrodes, i.e. cations attracted to cathode (- electrode) and anions attracted to anode (+ electrode). This can result flow of electricity.
Electrolytes
Strong electrolytes: ionize completely in solution
e.g. NaCl → Na Cl -
Weak electrolytes: ionize partially in solution.
e.g. Acetic acid (vinegar)
CH3COOH H CH3COO-

7. Equivalents
UNIT usED to EXPRESS the amount of each ion in solution.
Amount of ion in solution (e.g. body fluids) determined by equivalent (Eq).
This is the amount of each ion (positive or negative ion) in 1 mole of solution.
Relates the amount of an ion to its charge.
e.g. NaCl → Na Cl-
1 mole of Na+ ions has 1 equivalent (Eq) of positive charges
1 mole of Cl- ions has 1 equivalent of negative charges.
MgO → Mg O2-
1 mole of Mg2+ ions has 2 equivalent (Eq) of positive charges
1 mole of O2- ions has 2 equivalent of negative charges
NB: Number of positive ions always equal number negative ions in neutral solution.

8. How many equivalents (Eq) are present in each of the following?
0.5 mol K+
0.75 mol Ca2+
0.5 mol PO43-

9. Milliequivalents (mEq) 1 Eq = 1000 mEq Used to determine concentration of ions in the body. e.g. How many mEq of the Calcium ion are there in 0.4 g of CaBr2. e.g. The blood Magnessium level for a patient was found to be 16.0 mEq/L. How many moles of magnessium ion are in L of blood?

10. Concentration Units Weight/Volume Percent Concentration
The concentration of a solution tells how much
solute is dissolved in a given amount of solution.
Weight/volume percent concentration, (w/v)%, is the number of grams of solute dissolved in 100 mL of solution.
Weight/volume
percent concentration
mass of solute (g)
(w/v)% = x
100%
volume of solution (mL)

11. Concentration Units Weight/Volume Percent Concentration
For example, vinegar contains 5 g of acetic acid in
100 mL of solution, so the (w/v)% concentration is
5 g acetic acid
100 mL vinegar solution
x 100% = 5% (w/v)

12. Concentration Units Volume/Volume Percent Concentration
volume of solute (mL)
(v/v)% =
x 100%
volume of solution (mL)
For example, if a bottle of rubbing alcohol contains
70 mL of 2-propanol in 100 mL of solution, then the
(v/v)% concentration is
70 mL 2-propanol
100 mL rubbing alcohol
x 100% = 70% (v/v)

13. Concentration Units Using a Percent Concentration as a Conversion Factor
Sample Problem 7.6
A saline solution used in intravenous drips contains
0.92% (w/v) NaCl in water. How many grams of NaCl
are contained in 250 mL of this solution?
Step [1]
Identify the known quantities and the
desired quantity.
0.92% (w/v) NaCl solution
250 mL
known quantities
? g NaCl
desired quantity

14. Concentration Units Using a Percent Concentration as a Conversion Factor
Step [2]
Write out the conversion factors.
100 mL solution
0.92 g NaCl
0.92 g NaCl
100 mL solution or
Choose this one to cancel mL.
Step [3]
Solve the problem.
Answer
0.92 g NaCl
100 mL solution
250 mL x =
2.3 g NaCl
Milliliters cancel.

15. Concentration Units Parts Per Million
When a solution contains a very small concentration
of solute, it is often expressed in parts per million.
Parts per million
mass of solute (g)
ppm = x
mass of solution (g) or
volume of solute (mL)
volume of solution (mL)
ppm = x

16. Concentration Units Molarity
Molarity is the number of moles of solute per liter of
solution, abbreviated as M.
moles of solute (mol)
liter of solution (L)
Molarity = M =

17. Concentration Units Molarity
HOW TO Calculate Molarity from a Given Number of
Grams of Solute
Calculate the molarity of a solution made
from 20.0 g of NaOH in 250 mL of solution.
Example
Identify the known quantities and the desired
quantity.
Step [1]
20.0 g NaOH
250 mL solution
known quantities
? M (mol/L)
desired quantity

18. Concentration Units Molarity
HOW TO Calculate Molarity from a Given Number of
Grams of Solute
Convert the number of grams of solute to
the number of moles. Convert the volume
to liters if necessary.
Step [2]
Use the molar mass to convert grams of NaOH to
moles of NaOH (molar mass 40.0 g/mol).
1 mol
40.00 g NaOH
20.0 g NaOH x =
0.500 mol NaOH
Grams cancel.

19. Concentration Units Molarity
HOW TO Calculate Molarity from a Given Number of
Grams of Solute
Convert milliliters of solution to liters of solution
using a mL–L conversion factor.
1 L
1000 mL
250 mL solution x =
0.25 L solution
Milliliters cancel.

20. Concentration Units Molarity
HOW TO Calculate Molarity from a Given Number of
Grams of Solute
Divide the number of moles of solute by the
number of liters of solution to obtain the
molarity.
Step [3]
moles of solute (mol) mol NaOH
V (L) L solution
M = =
= 2.0 M
Answer

21. Dilution Dilution is the addition of solvent to decrease the
concentration of solute. The solution volume changes,
but the amount of solute is constant.
moles of solute (mol) = molarity (M) x volume (V)
M1V = M2V2
initial values
final values

22. initial values final values
Dilution
Sample Problem 7.13
How many milliliters of a 4.0% (w/v) solution must be used to prepare 250 mL of a 0.080% (w/v) solution?
Identify the known quantities and the desired
quantity.
Step [1]
C1V1 = C2V2
initial values final values
C1 = 4.0% (w/v) C2 = 0.080% (w/v)
V2 = 250 mL
known quantities
? V(L)
desired quantity 22 22

23. Dilution
Write the equation and rearrange it to isolate the desired quantity, V1, on one side.
Step [2]
Solve for V1 by dividing both sides by C1.
C1V1 = C2V2
V1 = C2V2 C1
Step [3]
Solve the problem.
V1 = (0.080%)(250 mL)
4.0% =
5.0 mL dopamine solution
Answer 23 23

24. Osmosis The membrane that surrounds living cells is a
semipermeable membrane.
Semipermeable membranes allow water and small
molecules to pass across, but ions and large molecules cannot.
Osmosis is the passage of water across a semipermeable membrane from a solution of low solute concentration to a solution of higher solute concentration.
Osmotic pressure is the pressure that prevents
the flow of additional solvent into a solution on
one side of a semipermeable membrane.

25. Causes hemolysis ( bursting of red blood cells) Hypertonic solutions.
Osmosis and Biological Membranes
Isotonic solutions.
Two solutions with the same solute concentration/osmotic pressure are said to be isotonic.
A 0.92% m/v sodium chloride solution is called a physiologic saline solution. It is isotonic with blood, i.e. has the same salt concentration as blood.
Hypotonic solutions.
A solution that contains a lower solute concentration compared to another is said to be hypotonic.
Causes hemolysis ( bursting of red blood cells)
Hypertonic solutions.
Hypertonic solutions contain a higher solute concentration compared to another.
Causes plasmolysis/crenation (shrinking of the red blood cells)

26. isotonic
solution
hypotonic
solution
hypertonic
solution
Hemolysis
(burst)
Crenation
(shrink)