solutions. solutions a solution is a homogeneous mixture. a solution is composed of a solute...
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SolutionsSolutions
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Solutions• A solution is a homogeneous mixture.• A solution is composed of a solute dissolved in a
solvent.• Solutions exist in all three physical states:
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Gases in Solution• Temperature effects the solubility of gases.
• The higher the temperature, the lower the solubility of a gas in solution.
• An example is carbon dioxide in soda:– Less CO2 escapes when you open a cold soda than
when you open the soda warm
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Polar Molecules• When two liquids make a solution, the solute is
the lesser quantity, and the solvent is the greater quantity.
• Recall, that a net dipole is present in a polar molecule.
• Water is a polar molecule.
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Polar & Nonpolar Solvents• A liquid composed of polar molecules is a polar
solvent. Water and ethanol are polar solvents.
• A liquid composed of nonpolar molecules is a nonpolar solvent. Hexane is a nonpolar solvent.
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Like Dissolves Like• Polar solvents dissolve in one another.
• Nonpolar solvents dissolve in one another.
• This it the like dissolves like rule.
• Methanol dissolves in water but hexane does not dissolve in water.
• Hexane dissolves in toluene, but water does not dissolve in toluene.
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Miscible & Immiscible• Two liquids that completely
dissolve in each other are miscible liquids.
• Two liquids that are not miscible in each other are immiscible liquids.
• Polar water and nonpolar oil are immiscible liquids and do not mix to form a solution.
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Solids in Solution• When a solid substance dissolves in a liquid, the
solute particles are attracted to the solvent particles.
• When a solution forms, the solute particles are more strongly attracted to the solvent particles than other solute particles.
• We can also predict whether a solid will dissolve in a liquid by applying the like dissolves like rule.
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Like Dissolves Like for Solids• Ionic compounds, like sodium chloride, are
soluble in polar solvents and insoluble in nonpolar solvents.
• Polar compounds, like table sugar (C12H22O11), are soluble in polar solvents and insoluble in nonpolar solvents.
• Nonpolar compounds, like naphthalene (C10H8), are soluble in nonpolar solvents and insoluble in polar solvents.
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The Dissolving Process• When a soluble crystal is placed into a solvent, it
begins to dissolve.
• When a sugar crystal is placed in water, the water molecules attack the crystal and begin pulling part of it away and into solution.
• The sugar molecules are held within a cluster of water molecules called a solvent cage.
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Dissolving of Ionic Compounds• When a sodium chloride crystal is place in water,
the water molecules attack the edge of the crystal.• In an ionic compound, the water
molecules pull individual ions off of the crystal.
• The anions are surrounded by the positively charged hydrogens on water.
• The cations are surrounded by the negatively charged oxygen on water.
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Rate of Dissolving• There are three ways we can speed up the rate of
dissolving for a solid compound.
• Heating the solution:– This increases the kinetic energy of the solvent and the
solute is attacked faster by the solvent molecules.
• Stirring the solution:– This increases the interaction between solvent and
solute molecules.
• Grinding the solid solute:– There is more surface area for the solvent to attack.
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Solubility and Temperature• The solubility of a compound is the maximum
amount of solute that can dissolve in 100 g of water at a given temperature.
• In general, a compound becomes more soluble as the temperature increases.
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Saturated Solutions• A solution containing exactly the maximum
amount of solute at a given temperature is a saturated solution.
• A solution that contains less than the maximum amount of solute is an unsaturated solution.
• Under certain conditions, it is possible to exceed the maximum solubility of a compound. A solution with greater than the maximum amount of solute is a supersaturated solution.
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Supersaturated Solutions• At 55C, the solubility of NaC2H3O2 is 100 g per
100 g water.
• If a saturated solution at 55C is cooled to 20C, the solution is supersaturated.
• Supersaturated solutions are unstable. The excess solute can readily be precipitated.
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Supersaturation• A single crystal of sodium acetate added to a
supersaturated solution of sodium acetate in water causes the excess solute to rapidly crystallize from the solution.
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Concentration of Solutions• The concentration of a solution tells us how much
solute is dissolved in a given quantity of solution.
• We often hear imprecise terms such as a “dilute solution” or a “concentrated solution”.
• There are two precise ways to express the concentration of a solution:– mass/mass percent
– molarity
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Mass Percent Concentration• Mass percent concentration compares the mass of
solute to the mass of solvent.
• The mass/mass percent (m/m %) concentration is the mass of solute dissolved in 100 g of solution.
mass of solutemass of solution
× 100% = m/m %
g soluteg solute + g solvent
× 100% = m/m %
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Calculating Mass/Mass Percent• A student prepares a solution from 5.00 g NaCl
dissolved in 97.0 g of water. What is the concentration in m/m %?
5.50 g NaCl5.00 g NaCl + 97.0 g H2O
× 100% = m/m %
5.00 g NaCl102 g solution
× 100% = 4.90 %
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Mass Percent Unit Factors• We can write several unit factors based on the
concentration 4.90 m/m% NaCl:
4.90 g NaCl100 g solution 4.90 g NaCl
100 g solution
4.90 g NaCl95.1 g water 4.90 g NaCl
95.1 g water
95.1 g water100 g solution 95.1 g water
100 g solution
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Mass Percent Calculation• What mass of a 5.00 m/m% solution of sucrose
contains 25.0 grams of sucrose?
• We want grams solution, we have grams sucrose.
100 g solution5.00 g sucrose
= 500 g solution25.0 g sucrose ×
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Molar Concentration• The molar concentration, or molarity (M), is the
number of moles of solute per liter of solution, is expressed as moles/liter.
• Molarity is the most commonly used unit of concentration.
moles of soluteliters of solution
= M
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Calculating Molarity• What is the molarity of a solution containing
18.0 g of NaOH in 0.100 L of solution?
• We also need to convert grams NaOH to moles NaOH (MM = 40.00 g/mol).
= 4.50 M NaOH×18.0 g NaOH
0.100 L solution1 mol NaOH
40.00 g NaOH
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Molarity Unit Factors• We can write several unit factors based on the
concentration 4.50 M NaOH:
4.50 mol NaOH1 L solution 4.50 mol NaOH
1 L solution
4.50 mol NaOH1000 mL solution 4.50 mol NaOH
1000 mL solution
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Molar Concentration Problem• How many grams of K2Cr2O7 are in 250.0 mL of
0.100 M K2Cr2O7?
• We want mass K2Cr2O7, we have mL solution.
= 7.36 g K2Cr2O7
0.100 mol K2Cr2O7 1000 mL solution250.0 mL solution × ×
294.2 g K2Cr2O7
1 mol K2Cr2O7
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Molar Concentration Problem• What volume of 12.0 M HCl contains 7.30 g of
HCl solute (MM = 36.46 g/mol)?
• We want volume, we have grams HCl.
= 16.7 mL solution
1 mol HCl 36.46 g HCl
7.30 g HCl × ×1000 mL solution
12.0 mol HCl
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Dilution of a Solution• Rather than prepare a solution by dissolving a
solid in water, we can prepare a solution by diluting a more concentrated solution.
• When performing a dilution, the amount of solute does not change, only the amount of solvent.
• The equation we use is: M1 × V1 = M2 × V2
– M1 and V1 are the initial molarity and volume and M2 and V2 are the new molarity and volume
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Dilution Problem• What volume of 6.0 M NaOH needs to be diluted
to prepare 5.00 L if 0.10 M NaOH?
• We want final volume and we have our final volume and concentration.
M1 × V1 = M2 × V2
(6.0 M) × V1 = (0.10 M) × (5.00 L)
V1 = = 0.083 L(0.10 M) × (5.00 L)
6.0 M
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Solution Stoichiometry• In Chapter 10, we performed mole calculations
involving chemical equations, stoichiometry problems.
• We can also apply stoichiometry calculations to solutions.
molarity known moles known moles unknown mass unknown
solutionconcentration
balancedequation
molar mass
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Solution Stoichiometry Problem• What mass of silver bromide is produced from the
reaction of 37.5 mL of 0.100 M aluminum bromide with excess silver nitrate solution?
AlBr3(aq) + 3 AgNO3(aq) → 3 AgBr(s) + Al(NO3)3(aq)
• We want g AgBr, we have volume of AlBr3
= 2.11 g AgBr
37.5 mL soln ×3 mol AgBr
1 mol AlBr3
0.100 mol AlBr3
1000 mL soln×
1 mol AgBr
187.77 g AgBr×
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Conclusions
• Gas solubility decreases as the temperature increases.
• Gas solubility increases as the pressure increases.
• When determining whether a substance will be soluble in a given solvent, apply the like dissolves like rule.
– Polar molecules dissolve in polar solvents.
– Nonpolar molecules dissolved in nonpolar solvents.
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Conclusions Continued• Three factors can increase the rate of dissolving
for a solute:– Heating the solution
– Stirring the solution
– Grinding the solid solute
• In general, the solubility of a solid solute increases as the temperature increases.
• A saturated solution contains the maximum amount of solute at a given temperature.
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Conclusions Continued• The mass/mass percent concentration is the mass
of solute per 100 grams of solution:
• The molarity of a solution is the moles of solute per liter of solution.
moles of soluteliters of solution
= M
mass of solutemass of solution
× 100% = m/m %
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Conclusions Continued
• You can make a solution by diluting a more concentrated solution:
M1 × V1 = M2 × V2
• We can apply stoichiometry to reactions involving solutions using the molarity as a unit factor to convert between moles and volume.