Ch15 Solutions Notes

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Chapter 15 - KEY
Solutions
Solution - homogeneous mixture containing two or more substances called the solute and
the solvent
A solution may exist as a gas (air), liquid (sugar and water), or solid (braces – titanium and nickel).
Most solutions are liquid.
Aqueous solutions - reactants and products that are mixed in water.
Water is the most common solvent among liquid solutions. It is known as the "universal solvent."
soluble - a substance that dissolves in a substance.
Ex. -sugar is soluble in water.
insoluble - a substance that does not dissolve in a solvent.
Ex. - sand is insoluble in water.
immiscible - 2 liquids that do not dissolve each other Ex. - oil and vinegar
miscible - 2 liquids that are soluble in each other Ex. - food coloring and water
solvation - the process of surrounding solute particles with solvent particles to form a
solution.
If the attractive forces between a solute and a solvent are greater than the attractive forces holding the solute
together, then the solvent particles pull the solute particles apart and surround them.
General rule is that "like dissolves like" (polar dissolves polar, nonpolar dissolves nonpolar, etc.)
Ex. - water and NaCl
Factors that affect rate of solvation 1. agitation - moves the solute particles around and thereby allows for more collisions
2. Increase the surface area of the solute -this also allows more collisions to occur
(tablespoon of sugar dissolves faster than a cube).
3. raising the temperature of the solvent -this increases the kinetic energy of its .particles,
resulting in more frequent collisions.
Heat of Solution - the overall energy change that occurs during the solvation process.
Energy is required to overcome the attractive forces within the solute and solvent (endothermic)
When solute and solvent mix, the particles attract each other and release energy (exothermic)
Solubility - maximum amount of solute that will dissolve in a given amount of
solvent at a specified temp. and pressure.
Saturated solution - contains the maximum amount of dissolved solute for a given amount of solvent at a specific
temp. and pressure
Unsaturated Solution - contains less dissolved solute for a given temp. and pressure than a saturated solution
Supersaturated Solution - contains more dissolved solute than a saturated solution at the same temperature. (it is
formed at a high temp. and then cooled slowly, allowing the solute to remain dissolved)
Ex. - rock candy
Pressure and solubility- the solubility of a gas in any solvent increases as its external pressure increases
(carbonated beverages work this way)
Henry’s Law- at a given temperature, the solubility (S) of a gas in a liquid is directly
proportional to the pressure (P) of the gas above the liquid
Example Problem:
If 0.85 g of a gas at 4.0 atm of pressure dissolves in 1.0 L of water at 25" C, how much will dissolve in 1.0 L of
water at 1.0 atm of pressure and the same temperature?
S1 = 0.85 g/L
P1 = 4.0 atm
P2 = 1.0 atm
S2 = ? g/L
𝑆2 = 𝑆1 π‘₯
𝑃2
𝑃1
= 0.85 g⁄L π‘₯
1.0 atm
4.0 π‘Žπ‘‘π‘š
= 0.21 𝑔⁄𝐿
More Example Problems1. If 0.55 g of a gas dissolves in 1.0 L of water at 20.0 kPa of pressure, how much will dissolve at 110.0 kPa of
pressure?
S1 = 0.55 g/L
P1 = 20.0 kPa
P2 = 110.0 kPa
S2 = ? g/L
𝑆2 = 𝑆1 π‘₯
𝑃2
𝑃1
= 0.55 g⁄L π‘₯
110.0 π‘˜π‘ƒπ‘Ž
20.0 π‘˜π‘ƒπ‘Ž
= 3.025 𝑔⁄𝐿
2. A gas has a solubility of 0.66 g/L at 10.0 atm of pressure. What is the pressure o n a 1.0 L sample that
contains 1.5 g of gas?
𝑆2
𝑃2 = 𝑃1 π‘₯
𝑆1
S1 = 0.66 g/L
P1 = 10.0 atm
S2 = 1.5 g/L
P2 = ? atm
= 10.0 atm π‘₯
= 22.73 π‘Žπ‘‘π‘š
1.5 𝑔/𝐿
0.66 𝑔/𝐿
Solution concentration - a measure of how much solute is dissolved in a specific amount of solvent or
solution.
Concentrations can be described qualitatively using the words concentrated or dilute, but they are more often
described quantitatively. Some quantitat ive descriptions are percent by mass or volume, molarity, and
molality .
1. Percent by Mass - usually used when a solid is dissolved in a liquid solution
% 𝑏𝑦 π‘šπ‘Žπ‘ π‘  =
π‘šπ‘Žπ‘ π‘  π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’
π‘šπ‘Žπ‘ π‘  π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
π‘₯ 100
(mass of solution = grams of solute + grams of solvent)
Example problem:
In order to maintain a sodium chloride concentration similar to ocean water, an aquarium must contain 3.6 g
NaCl per 100.0 g of water. What is the percent by mass of NaCl in the solution?
mass of solution = 3.6 g + 100.0 g
= 103.6 g
% 𝑏𝑦 π‘šπ‘Žπ‘ π‘  =
π‘šπ‘Žπ‘ π‘  π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’
π‘₯ 100
π‘šπ‘Žπ‘ π‘  π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
3.6 𝑔
=
103.69 𝑔
π‘₯ 100 = 3.5%
2. Percent by Volume - usually describes solutions that both the solute and solvent are liquids
% 𝑏𝑦 π‘£π‘œπ‘™π‘’π‘šπ‘’ =
π‘£π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’
π‘£π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
π‘₯ 100
Example Problem What is the percent by volume of ethanol in a solution that contains 35 mL of ethanol dissolved in 115 mL
of water?
volume of solute = 35 mL
volume of solution = 150 mL
% 𝑏𝑦 π‘£π‘œπ‘™π‘’π‘šπ‘’ =
=
3. π‘€π‘œπ‘™π‘Žπ‘Ÿπ‘–π‘‘π‘¦ (𝑀) =
π‘£π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’
volume π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
35 π‘šπΏ
π‘₯ 100
150 π‘šπΏ
π‘₯ 100
=
23%
π‘šπ‘œπ‘™π‘’π‘  π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’
π‘™π‘–π‘‘π‘’π‘Ÿπ‘  π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›
Example Problem:
An IV solution contains 5.10 g of glucose (C6H1206) in 110.5 mL of water. What is the molarity of
this solution? The molar mass of glucose is 180 g/mol.
mass of solute = 5.10 g C6H12O6
5. 10 𝑔 𝐢6 𝐻12 𝑂6 π‘₯
molar mass of C6H12O6 = 180 g/mol
volume of solvent = 100.5 mL
110.5 mL x
1𝐿
1000 π‘šπΏ
1 π‘šπ‘œπ‘™π‘’πΆ6 𝐻12 𝑂6
180 𝑔 𝐢6 𝐻12 𝑂6
= 0.028 π‘šπ‘œπ‘™ 𝐢6 𝐻12 𝑂6
= 0.1105 𝐿 𝐻2 𝑂
so
0.028 π‘šπ‘œπ‘™ 𝐢6 𝐻12 𝑂6
0.1105 𝐿
= 0.25 M
Diluting Solutions- M1V1 = M2V2 (M =molarity)
Example Problem:
What volume in milliliters of 2.00 M calcium chloride stock solution would you use to make 0.50 L of 0.300 M
calcium chloride solution?
𝑉1 =
M1 = 2.00 M CaCl2
M2 = 0.300 M
V2 = 0.50 L
V1 = ? mL
=
𝑀2 𝑉2
𝑀1
0.300 𝑀 π‘₯ 0.50 𝐿
2.00 𝑀
Then to change L to mL…..
4. π‘€π‘œπ‘™π‘Žπ‘™π‘–π‘‘π‘¦ (π‘š) =
0.075 𝐿 π‘₯
=
0.075 𝐿
1000 π‘šπΏ
1𝐿
= 75 mL
π‘šπ‘œπ‘™π‘’π‘  π‘œπ‘“ π‘ π‘œπ‘™π‘’π‘‘π‘’
π‘˜π‘–π‘™π‘œπ‘”π‘Ÿπ‘Žπ‘š π‘œπ‘“ π‘ π‘œπ‘™π‘£π‘’π‘›π‘‘
Example Problem :
In the lab, a student adds 4.5 g of sodium chloride to 100.0 g of water. Calculate the molality of the solution.
1 π‘šπ‘œπ‘™ π‘π‘ŽπΆπ‘™
58.55 𝑔 π‘π‘ŽπΆπ‘™
g of H2O = 100.0 g
#1.
4.5 g NaCl x
g of NaCl = 4.5 g
m = ? mol/kg
#2.
100 𝑔 𝐻2 𝑂 π‘₯
#3.
m=
1 π‘˜π‘” 𝐻2 𝑂
1000 𝑔 𝐻2 𝑂
0.077 π‘šπ‘œπ‘™ π‘π‘ŽπΆπ‘™
0.10 π‘˜π‘” 𝐻2 𝑂
= 0.077 π‘šπ‘œπ‘™ π‘π‘ŽπΆπ‘™
= 0.10 kg H2O
= 0.77 π‘šπ‘œπ‘™/π‘˜π‘”
Colligative Properties - physical properties of solutions that are affected by the number of particles but not the
identity of dissolved solute particles.
Ex. -vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure
1. Vapor Pressure Lowering
Vapor pressure is the pressure exerted in a closed container by liquid particles that have escaped the liquid 's
surface and entered the gaseous state.
The greater the number of solute particles in a solvent, the lower the resulting vapor pressure.
2. Boiling Point Elevation
Liquid will boil when its vapor pressure equals atmospheric pressure.
Definition - The temperature difference between a solution's boiling point and a pure solvent 's boiling point.
The grater the .number .of solute particles in the solution; the greater the boiling point elevation.
3. Freezing Point Depression
The freezing point of a solution is always lower than that of a pure solvent.
Freezing point depression - the difference in temperature between its freezing point and the freezing point of its pure
solvent.
Examples: putting salt in water to boil adding salt to ice on roadways in the winter, adding salt to water when
making ice cream.
4. Osmosis and Osmotic Pressure
Osmosis – the diffusion of solvent particles across a semi permeable membrane from an area
of higher solvent concentration to an area of lower solvent concentration.
* the membranes of all living cells are semi permeable
Osmotic pressure - the additional pressure needed to reverse osmosis
Ex. -sugar water:
The water will cross the membrane, but the sugar will not. Therefore, the pure water will cross over to the
solution side since more water molecules are contacting the membrane. The additional water on the solution
side creates a pressure that will push the water molecules back across the membrane.
Not all mixtures are solutions (homogeneous). Some are heterogeneous.
1. Suspensions – a mixture containing particles that settles out if left undisturbed (large particles)
Ex.- cornstarch and water, muddy water
2. Colloids - a heterogeneous mixture of intermediate sized particles (between the size of solution and
suspension particles)
Ex. -milk, butter, cheese, blood
***Many mixtures are solutions, but not all mixtures are solutions!!
Brownian Motion -the erratic m ovement of colloid particles
These collisions prevent the particles from settling out. The particles have polar or charged atomic groups
which cause an attraction and the collisions.
You can interfere with these attractions if you add an electrolyte or heat a colloid.
Tyndall effect – the scattering of light by colloidal particles.
Ex. - mix milk in water and shine a laser pointer through it.
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