SOLUTIONSA solution is a homogeneous mixture; particles are evenly distributed throughout the mixture.• Proportions may vary• Uniform ratio throughout the mixtureA liquid solution is clear. The particles are not visible, do not settle, and can not be filtered.
A solution differs from a suspension in that the particles of a suspension are visible, can be filtered, and settle.
A solution differs from a colloid in that the particles of a colloid exhibit Tyndall effect., yet do not settle.
(TYNDALL effect: the scattering of light by particles such as headlights in fog, flashlight through Jello or dilute milk)
A solution will not exhibit the Tyndall effect.
Cherry Kool-aidRed powder: flavor and colorWhite crystals: sugarClear liquid: water
One substance dissolved in another◦Solute: the substance being dissolved◦Solvent: the substance that dissolves
the solute Sugar is the SOLUTE (smaller quantity) Water is the SOLVENT (larger quantity)
Types of SolutionsGas (solvent is gas)
◦ Gas into Gas: air◦ Liquid into Gas: humidity◦ Solid into Gas: air pollution
Liquid (solvent is liquid)◦ Gas into Liquid: pop◦ Liquid into Liquid: vinegar◦ Solid into Liquid: sweet tea
Solid (solvent is solid)◦ Gas into Solid: absorbent charcoal◦ Liquid into Solid: dental fillings◦ Solid into Solid: alloys of metal
The Dissolving ProcessTwo factors affect the dissolving
process: dissolution◦The constant motion of the particles
(There’s that good old kinetic molecular theory again!)
◦The polarity of the solute and solvent (Recall that polarity is when a compound has partial charges because of uneven distribution of charges)
Steps of the Dissolving Process
1. Moving solvent particles cluster around solute molecules or particles at the surface of the solid.
2. Solvent molecules pull solute off of the solid surface and into solution.
3. Moving solvent particles continue to spread solute evenly throughout the solution,
The process repeats itself as fresh layers of the solute are exposed.
1. Solvent particles cluster around solute particles at the surface.
2. Solvent particles pull solute particles away from surface, into solution.
3. Moving solute particles continue to spread solute evenly through solution.
MOLECULAR
1. Solvent particles cluster around solute particles at the surface.
2. Solvent particles pull solute particles away from surface, into solution.
3. Moving solute particles continue to spread solute evenly through solution.
IONIC
IONIC COMPOUNDSWhen an ionic substance dissolves in
water, the forces of the solvent pulling on the ions is stronger than the forces holding the ions together.
The ions separate. This is called DISSOCIATION
Because charged ions are present in an ionic solution, ionic solutions conduct electricity and are called ELECTROLYTES.
EXAMPLE: NaCl
MOLECULAR COMPOUNDSCertain polar substances form
ions when they dissolve in water. This process is called IONIZATION.
Because ions are formed, the solution conducts electricity.
These substance are also ELECTROLYTES.
EXAMPLE: HCl, HC2H3O2
MOLECULAR COMPOUNDSOther polar substances do not
ionize in water.Because ions are not formed, the
solution does not conduct electricity.
These substances that do not ionize in water and do not conduct electricity are called NON-ELECTROLYTES.
EXAMPLE: sugar
ELECTROLYTES: substances that conduct electricity when dissolved in water
Ionic substances that separate into ions (dissociate) or polar molecular substances that form ions (ionization) when dissolved conduct electricity and are called ELECTROLYTES.
ReviewA substance whose water
solutions do not conduct electricity is a non-electrolyte.◦Many covalent compounds
A substance that separates into ions (dissociates) or forms ions (ionizes) in a water solution conducts electricity and is called an electrolyte◦All ionic and some covalent
compounds
COMPOUNDS THAT DISSOLVE IN WATER
ELECTROLYTESDO CONDUCT ELECTRICITY
IONIC COMPOU
NDS DISSOCIATE INTO
IONS
SOME POLAR
MOLECULAR
COMPOUNDS
IONIZE
NON-ELECTROLYTESDO NOT CONDUCT
ELECTRICITY
OTHER POLAR MOLECULAR COMPOUNDS
DISSOLVE WITH NO FORMATION OF IONS
Factors Solid in Liquid
Gas in Liquid
TEMPERATURE Temp > Rate >
Temp > Rate <
AGITATION Agitation > Rate >
Agitation > Rate <
SIZE OF PARTICLES(surface area)
Size < (surface area
>) Rate >
NA
PRESSURE NA Pressure > Rate >
WHY?(KMT)
The solute particles are
less energetic than solvent
The solute particles are
more energetic than
solvent
FACTORS THAT AFFECT THE RATE OF DISSOLUTION
Think BIGRecall the Pop and Mentos experiment!
The pop “explodes” because the dissolved gas rapidly leaves the solution because the
candy gives it surfaces to collect on (nucleation sites).
Particles in SolutionsSolvents with non-polar
molecules dissolve non-polar substances◦Oil, grease, dry cleaning fluid, paint,
turpentineSolvents with polar molecules
dissolve polar substances◦Water dissolves sugar, ionic
compounds
LIKE DISSOLVES LIKE
TermsSoluble: capable of being
dissolved in a particular solventInsoluble: incapable of being
dissolved in a particular solvent
Miscible: liquids that dissolve freely in any proportion
Immiscible: liquids that are not soluble in each other
Detergents and emulsifiersGrease is non-polarWater is polar
DETERGENT has ◦A non-polar end that dissolves the
grease◦A polar end that dissolves in the
water to rinse it away NONPOLAR DETERGENT
POLARGREASE WATER
SolubilityThere are limits to the amount of
solute that will dissolve in a given amount of solvent at a given temperature
There are some general terms:◦Unsaturated◦Saturated◦Super-saturated
Unsaturateda solution that can dissolve more
of a given solute at a certain temperature◦A crystal of solute added to an
unsaturated solution will dissolve When you add a second spoon of sugar
to your cup of tea, it dissolves. The tea was an unsaturated solution.
Saturateda solution that has dissolved all of the
solute that it can at a certain temperature◦ A crystal of solute added to a saturated
solution will drop to the bottom, un-dissolved. When you add three spoons of sugar to your tea,
some sugar drops to the bottom, undissolved. It is a saturated solution.
DYNAMIC EQUILIBRIUM exists: ◦ changing but balanced.◦ Some solid dissolves, but as some
dissolves, some re-crystallizes
Super-saturatedan unstable solution that contains
more solute than a saturated solution at a certain temperature◦ A crystal of solute added to a super-
saturated solution will cause crystallization. So will any disruption of the unstable solution.
◦ Make a saturated solution at an elevated temperature and cool it slowly. At the lower temperature, the solute will remain dissolved in an unstable situation. If disrupted, the solute crystallizes. Hot-packs and rock candy
Concentration: the amount of solute in a given amount of solvent or solutionDilute: a relatively small amount
of solute in a relatively large amount of solvent
Concentrated: a relatively large amount of solute in a relatively small amount of solvent
We can do better!
Percent by volumemL of solute /100 mL of solution
3% hydrogen peroxide3 mL H2O2 / 100 mL soln
70% isopropyl alcohol (what is different)
70 mL alcohol / 100 mL soln
Percent by massg of solute/100 mL of solutionIV saline.9 g NaCl/100 mL solution x
100% = .9%
IV glucose5 g glucose/100 mL soln x 100%
= 5%
PPM and PPBx/1,000,000x/1,000,000,000
MAWC Annual Report to Consumers
Serial dilutions are often used◦1x, 10x, 100x, 1000x, etc.◦ 1 mL “neat” stock to 10 mL total volume =
10x◦ 1 mL 10x to 10 mL total volume = 100x …
Mass per volumeg solute/1000 mL solution
9 g NaCl/1000mL = 9 g NaCl/LNot that useful to chemist…but
MOLES are!9 g NaCl | 1 mol NaCl = .15 mol
NaCl 1 L soln | 58.44 g NaCl L soln
This helps chemists relate solutions to amounts in chemical reactions, and there is a word for this…
MolarityM = mol solute/liter of solution
Chemists use this because it lets us work concentration into stoichiometry problems
SEE EXAMPLE PROBLEMS
Molarity example 1You have 3.50 L of solution that
contains 90.0 g of NaCl. What is the molarity?
M = mol/L
Molarity example 2You have .8 L of a .5 M HCl
solution. How many moles of HCl does this solution contain?
M = mol/L
To produce 40.0 g of silver chromate, you will need 23.4 g of potassium chromate in solution as a reactant. All you have on hand is 5 L of 6.0 M K2CrO4 solution. What volume of this solution is need to give you the 23.4 g K2CrO4 needed for this reaction?
M = mol/L
Molarity example 3
Molarity Practice
1. What is the molarity of a solution composed of 5.85 g of potassium iodide dissolved in enough water to make .125 L of solution?
2. How many moles of H2SO4 are present in .500L of a .150 M H2SO4 solution?
3. What volume of 3.00 M NaCl is needed for a reaction that requires 146.3 g NaCl?
Molalitym = mol solute/kg of solvent
Chemists sometimes use this because volume of liquids changes with temperature. Soon, we will be studying the effect of concentration as temperature changes.
SEE EXAMPLE PROBLEMS
Molality example 1A solution was made by
dissolving 17.1 g sucrose (molar mass 342.34 g/mole) in 125 g water. What is the molal concentration of this solution (the molality)?
m = mol solute/ kg solvent
Molality example 2A solution of I2 in carbon tetrachloride,
CCl4, is used for chemical tests for starch. How much iodine must be added to prepare a .480 m solution of iodine in CCl4 if 100.0 g of CCl4 is used?
m = mol solute/kg solvent
Practice Molality
1. What is the molality of acetone in a solution composed of 225 g of acetone, C3H6O, dissolved in 200. g water?
2. What quantity, in grams, of methanol CH3OH, is required to prepare a .244 m solution in 400. g of water?
Dilutions: a concentrated solution is diluted by adding more solvent to get the desired concentration.
M1V1 = M2V2
Moles before dilution = Moles after
dilution
◦M = molarity = mol/L◦V = volume = L
◦M x V = mol/L x L = # mol of solute
Dilutions: a concentrated solution is diluted by adding more solvent to get the desired concentration.
The # of moles before dilution is equal to the # of moles after dilution. The same # of moles is present in more solvent. The concentration (M, molarity) changes, the volume (V) changes, but not the # of moles of solute (M x V). ◦ Volume units may vary from L, but must be consistent
within problem.
So, M1V1 = M2V2
SEE EXAMPLE PROBLEMS
Dilution exampleI need 100. mL of 1.00 M HCl. I
have a stock solution of 12.0 M HCl. Describe how to make the solution I need.
M1V1 = M2V2
1.00 M (100. mL ) = 12.0 M (x mL)
X = 8.33 mL 12 M HCl diluted to a total volume of 100. mL
Dilution PracticeHow do you prepare 100 mL
of .20 M HCl from a stock solution of 2.0 M HCl?
M1V1 = M2V2
Making solutions1. “Mass” the solid2. Transfer solid to a small
beaker/flask3. Transfer all of solid by rinsing
weighing paper with water4. Swirl in small quantity of water5. Transfer to volumetric flask with
repeated rinses6. Mix in small quantity of water7. Fill carefully to the line8. Cover with parafilm and mix
thoroughly