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SOLUTIONS Chapter 6 A solution is a homogeneous mixture of two or more substances;
the dissolved substance is called the solute and the
substance in which the solute is dissolved is called the solvent.
The relative amounts of the substances in the solution determine which is
solvent and solute; the one in the greater amount is solvent and solute; the
one in the greater amount is considered the solvent. Concentration of a solution refers to the amount of solute in a
given amount of solution. Methods of expressing concentration are: Dilute
and Concentrated A
dilute solution contains a relatively small proportion of solute, while a
concentrated solution contains a relatively large proportion of solute.
Concentrated soltuions are only possible when the solute is very soluble. Unsaturated, Saturated
and Supersaturated An
unsaturated solution is one that can still dissolve more solute at that
temperature. A saturated solution is one in which the solute in solution is in
equilibrium with the pure undissolved solute; in other words, the solution
contains all of the solute it can dissolve at that particular temperature. A supersaturated solution contains more
solute in solution than it would ordinarily hold at that temperature. Example:
If in hot coffee one dissolves all the sugar it can possibly hold and carefully
cools it, the coffee still holds all the sugar that was dissolved and a
supersaturated solution exists. If a small crystal of sugar is introduced into
this cooled coffee, some sugar will crystallize out and a saturated solution
will remain. Types of Solutions p 143 gas
in liquid carbonated
beverages liquid
in liquid alcohol in water solid
in liquid salt in water gas
in solid hydrogen in
platinum liquid
in solid mercury in silver
(dental fillings) solid
in solid copper in silver
(sterling silver) PERCENT COMPOSITION - Think in terms of 100 p 147 PERCENT = PART
x 100 WHOLE 1. Percent - Mass in a total
volume (m/v) Most frequently used form of %. A 5% solution means 5 grams in 100 mL volume. Or 2.5 grams in 50 mL volume. A 5 mg% solution means 5 mg in 100 mL volume. Same as 5 mg/dL. (dL = 100 mL) 2. Percent- Mass of solute
and mass of solution (m/m)
This refers to the amount of solute in grams in the total
solution also in grams. Example: A 10% aqueous NaCl solution contains 10 g of NaCl in 90
g of water (or 10 g NaCl per 100
g of solution). 3. Percent - Volume of solute
in volume of solution. (v/v) This refers to relative amounts of two liquids in solution.
Example: An alcoholic beverage that
is 7% alcohol contains 7 ml of alcohol per 100 ml of solution. Ratios of individual liquids is generally given rather than
%. Example: A chromatography solvent was chloroform: methanol: water (6:3:1).
Problems 1. How many grams of sugar
are needed to make 100 ml of a 20% solution? 2. What is the concentration
in mg % of 100 mg of NaCl dissolved in 500 ml of solution? 3.
How many grams of sugar are needed to make 150 ml of a 30% solution: 4.
I have 250 ml of a .9% NaCl solution (saline). How many grams of salt do I have
? 5.
How many grams of sugar are need to make 50 grams of a 3% solution? MOLAR SOLUTIONS p 149 Molar solutions are used more frequently and are more
precise than the percent solutions. Molarity = moles liters 1. How would I make up 500 mL
of a 1 M solution of NaCl? moles =
grams Molarity
= moles MW L 1 =
x moles 0.5 L x =
0.5 moles 0.5 = x grams 58.5 x =
29.25 grams Put
29.25 g NaCl in flask, stir, add H2O and bring up to 500 ml 2. How would we make up 250 mL of a 3 M solution of NaCl? 3.
How much NaCl would I need to make up 10 ml of a 6 M solution? 4. How many grams of Ca(CO3)
would I need for 100 ml of a 2 M solution? 5. I have 29.25 g NaCl in 500
ml. What is its molarity? COLLIGATIVE PROPERTIES p160 Any property of a solution
that depends only on the number of solute particles dissolved in a given
amount of solvent and not on the nature of these particles is called a
colligative property. Lowering of Freezing Point One mole of any particle dissolved in 1000g of water lowers the
freezing point by 1.86EC. Note:
1000 g of water is 1 liter. We do
not have a final volume of 1 liter.
1 liter of water in addition to the solute. Problem: We add 275 g of ethylene glycol (C2H6O2)
per 1000 g of water in a car radiator, what will the freesing point of the
solution be? One mole of ethylene glycol
gives 1 mole of particles. How many moles of ethylene
glycol are present? If 1 mole lowers the freezing
point by 1.86EC, how much will the ethylene glycol lower the
freezing point? OSMOLARITY p
161-163 Osmosis
is the passage of a solvent through a semi-permeable membrane from a less
concentration to a more concentrated concentration. (Think of water trying to
dilute the concentrated solution) |
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-------- | | -------- | __________________ _________________ membrane membrane The
water will pass over until gravity becomes too great, and there is no further
increase. If we wanted to make the two
sides equal, we could put pressure on the higher side until the solvent is
forced back. This is called osmotic
pressure. THE NUMBER OF PARTICLES
IS IMPORTANT IN OSMOLARITY! If
we had 1 mole of NaCl, we would have 2 moles of particles. Osmolarity =
(molarity)(number of particles) (osmol) A
1 M solution of NaCl would have 2 osmol in particles A
1 M solution of Na3(PO4)
Problem: Physiological saline is 0.89% NaCl. a) What is the molarity of
the solution? b)
What is the osmolarity of the solution? ISOTONIC,
HYPOTONIC, AND HYPERTONIC SOLUTIONS Isotonic
means the same osmotic strength on both sides of the semipermeable
membrane. Physiological saline (0.89%
NaCl) is isotonic with blood. The cell retains is=s shape and function. Hypotonic means the
solution has a lower concentration of particles than the cell. If a red blood cell were placed in distilled
water, there would be more water moving into the cell and eventually burst it. Hemolysis (If you soak limp lettuce in water it
will perk up. Why/) Hypertonic means the
solution has a higher concentration of particle than the cell. If a red blood cell were place in brine
(concentrated NaCl), the cell would shrivel up. This method is used for curing meats like country ham. The bacteria can=t survive in the high salt concentration. (What is on the outside of country ham?) DIALYSIS In osmosis the membrane allows only the solvent to
pass through the semipermeable membrane.
The pores of the membrane can be increased selectively to allow passage
of different molecular weight compounds.
Quite often it is used to remove salt or other ions from a
solution. Hemodialysis uses this principle to filter out the
buildup of toxic compounds in the body.
The function of the kidneys is to remove these waste products through
dialysis. The kidneys can also reabsorb
ions to keep a balance. There is a threshold for reabsorption for glucose. When the blood concentration is 150 mg%, the
sugar is seen in the urine.
In the dialysis machine the blood is circulated through a long
tube of cellophane in an isotonic solution and then returned to the patient=s vein. The
ions can pass out and the large proteins remain. When protein is found in
the urine, that means there is some type of damage to the kidney. The isotonic solution has 0.6% NaCl, 0.04%
KCl, 0.2% NaHCO3 , and 0.72% glucose (w/v). Thus no sodium ions or glucose is lost from
the blood. The isotonic solution is
changed every 2 hours and the patient stays on the dialysis machine for 4 to 7
hours. Rates of Dissolution Rate a solute dissolves
depends on 1. surface area 2. agitation 3. temperature Factors Affecting
Solubility 1. pressure - gas in soda 2. temperature 3. chemical structure Problem Set 1. Calculate the molar concentration
of the following solutions a. 36.5 g HCl in 1 liter of solution ________________________ b. 36.5 g HCl in 5 liters of solution _______________________ c. 36.5 g HCl in 0.25 L of solution _________________________ d. 73.0 g HCl in 1 liter of solution ________________________ e. 73.0 g HCl in 5 L of solution ____________________________ f. 73.0 g of HCl in .25 L of solution________________________ 2.
Calculate the weight of solute contained in each of the following solutions a.
250 mL of a 1 M solution of H2SO4
_______________________ b. 3000 mL of 0.5 M solution of HC2H3O2
_________________ c. 2 liters of 1.5 M solution of H2CO3
_______________________ d. 100 mL of 2 M solution of KOH
__________________________ 3.
Calculate the volume of solution (1.5 M) that would contain a. 40 g of NaOH _______________ b. 80 g of Ca(OH)2
______________ c. 20 g of HCl _______________ d. 40 g H2CO3
________________ Problems at end of Chapter
6 -
6.30; 6.31; 6.34;
6.35; 6.36; 6.42;
6.65; 6.69 |
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Wednesday, March 21, 2007 01:28:01 PM |