ChemistryChemistry

SolutionsSolutions

I. Introduction: I. Introduction: TerminologyTerminology

SoluteSolute: substance to be dissolved: substance to be dissolved

-- could be solid, liquid, or gas-- could be solid, liquid, or gas

SolventSolvent: substance doing the : substance doing the dissolvingdissolving

SolutionSolution: Solute + Solvent: Solute + Solvent

II. How Solutes Dissolve in II. How Solutes Dissolve in Liquid Solvents (Like Liquid Solvents (Like

dissolves like)dissolves like)Ionic Solutes:Ionic Solutes:

A. Non-polar solventsA. Non-polar solvents

-- do not dissolve-- do not dissolve

B. Polar solventsB. Polar solvents

-- Break into -- Break into ionsions in polar solvent in polar solvent

(-) ions attracted to (+) end of (-) ions attracted to (+) end of moleculemolecule

(+) ions attracted to (-) end of (+) ions attracted to (-) end of moleculemolecule

Molecular (covalent) solutes:Molecular (covalent) solutes:A. Non-polar solutesA. Non-polar solutes

-- only non-polar solvents-- only non-polar solventsB. Polar solutesB. Polar solutes

Dissolve in polar solvents (like HDissolve in polar solvents (like H22O)O)

e.g.. PHe.g.. PH33

(-) portion (P) is attracted to (-) portion (P) is attracted to (+) (+) H’s in waterH’s in water

(+) portion (H’s –in PH(+) portion (H’s –in PH33) are ) are attracted to (-) O’s in attracted to (-) O’s in

waterwater

PHPH33 in H in H22OO

How Solubility Works How Solubility Works (when the solvent is liquid)(when the solvent is liquid)

Solid solute – solvent must be able to get Solid solute – solvent must be able to get in between solute molecules (or ions) in between solute molecules (or ions) and pull them apartand pull them apart

Liquid solute – solvent just needs to get Liquid solute – solvent just needs to get in between molecules (or ions). Easier in between molecules (or ions). Easier because molecules are already further because molecules are already further apart apart

Gas solute – solvent has to pull gas Gas solute – solvent has to pull gas molecules molecules closer togethercloser together

III. III. SolubilitySolubility:: The maximum amount of solute that can The maximum amount of solute that can dissolve in a given amount of solvent dissolve in a given amount of solvent

-- Units = -- Units = grams of solutegrams of solute 100.0 grams of solvent100.0 grams of solvent

Saturated SolutionSaturated Solution: a solution in which : a solution in which the maximum amount of solute has been the maximum amount of solute has been dissolveddissolved

IV. Endothermic v. IV. Endothermic v. ExothermicExothermic

MostMost solid solutes solid solutes requirerequire NRG to dissolve NRG to dissolve

e.g., NaCl and He.g., NaCl and H22OO

--NRG is required for H--NRG is required for H22O molecules O molecules to get in to get in between the Nabetween the Na++ and Cl and Cl-- ions ions

--NRG is taken from H--NRG is taken from H22O molecules O molecules and and surroundings (outside)surroundings (outside)

--H--H22O molecules O molecules slow downslow down (lose (lose NRG) NRG) and the temperature of and the temperature of water decreaseswater decreases

EndothermicEndothermic: A process that absorbs : A process that absorbs heat (NRG)heat (NRG)

-- the (KE) NRG is -- the (KE) NRG is storedstored (PE) in the (PE) in the dissolving processdissolving process

-- the solution is -- the solution is coldercolder

Like…Like…

NaCl + HNaCl + H22O + NRG O + NRG →→ Na Na++ + Cl + Cl-- + + HH22OO

SomeSome solid solutes solid solutes releaserelease NRG when NRG when they are dissolvedthey are dissolved

Exothermic:Exothermic: Process that results in a Process that results in a release of heatrelease of heat

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NaOH + HNaOH + H22O O →→ Na Na++ + OH + OH-- + H + H22O + O + NRGNRG

(Remember: most solutes dissolve (Remember: most solutes dissolve endothermically)endothermically)

4 Rules for solubility4 Rules for solubility

Rule 1: Identity – Solubility of any Rule 1: Identity – Solubility of any solute depends solute depends bothboth on the identity on the identity of the solute and the identity of the of the solute and the identity of the solventsolvent

Rule 2: Temperature –Rule 2: Temperature –

-Increasing the temperature -Increasing the temperature usuallyusually (endothermic processes) increases solubility (endothermic processes) increases solubility for solidsfor solids

-Increasing temperature -Increasing temperature decreasesdecreases solubility solubility for gasesfor gases

When an endothermic solution is heated, then When an endothermic solution is heated, then cooled – solubility decreases and a cooled – solubility decreases and a precipitateprecipitate formsforms

PrecipitationPrecipitation: the process by which a solute : the process by which a solute leaves the solution and becomes solid againleaves the solution and becomes solid again

Rule 3: Increased pressure increases Rule 3: Increased pressure increases solubility for gasessolubility for gases

Rule 4: When a solute dissolves in an Rule 4: When a solute dissolves in an exothermic manner, (sometimes) an exothermic manner, (sometimes) an increase in temperature increase in temperature decreasesdecreases solubilitysolubility

OYO’sOYO’s11.1 You must dissolve 50.0 g of KCrO11.1 You must dissolve 50.0 g of KCrO44

(a yellow-orange solid) into 250 mL of (a yellow-orange solid) into 250 mL of water. It’s not dissolving well. What water. It’s not dissolving well. What could you try?could you try?

11.2 A lab has a bottle of a solution made 11.2 A lab has a bottle of a solution made by dissolving CHby dissolving CH44 gas in octane. The lid gas in octane. The lid keeps popping off. What could you do keeps popping off. What could you do to prevent the lid from popping off?to prevent the lid from popping off?

11.3 You have to make a 500 mL 11.3 You have to make a 500 mL solution of 5.5 solution of 5.5 MM NaOH. It gets NaOH. It gets really hot – and it’s not dissolving really hot – and it’s not dissolving well. What should you do?well. What should you do?

V. Applying Stoichiometry V. Applying Stoichiometry to Solutionsto Solutions

Remember, Molarity (Remember, Molarity (MM) = moles/liter) = moles/liter

Therefore,Therefore,

L x L x MM (mol/L) = mol (mol/L) = mol

mol x 1/mol x 1/MM (L/mol) = L (L/mol) = L

(Just write (Just write MM as mol/L and you can as mol/L and you can figure it out. You could write mol/L or figure it out. You could write mol/L or L/mol and you don’t have to memorize L/mol and you don’t have to memorize the above)the above)

ExamplesExamples1. Sodium carbonate reacts with calcium 1. Sodium carbonate reacts with calcium

hydroxide to produce calcium carbonate and hydroxide to produce calcium carbonate and sodium hydroxide. How many grams of sodium hydroxide. How many grams of calcium carbonate will be made from 50.0 mL calcium carbonate will be made from 50.0 mL of 1.23 of 1.23 MM sodium carbonate and excess sodium carbonate and excess calcium hydroxide?calcium hydroxide?

2. Sodium chloride is reacted with excess 2. Sodium chloride is reacted with excess aluminum nitrate to make aluminum chloride aluminum nitrate to make aluminum chloride and sodium nitrate. How many L of a 0.245and sodium nitrate. How many L of a 0.245MM solution of NaCl will be necessary to make solution of NaCl will be necessary to make 100.0 g of AlCl100.0 g of AlCl33??

OYO’sOYO’s

11.4 The following reaction is carried 11.4 The following reaction is carried out in the laboratory:out in the laboratory:

5H5H22OO22(aq) + 2KMnO(aq) + 2KMnO44(aq) + 3H(aq) + 3H22SOSO44(aq) (aq) → 5O→ 5O22(g) + 2MnSO(g) + 2MnSO44(aq) + K(aq) + K22SOSO44(aq) (aq) + 8H+ 8H22OO

If 125 mL of 0.5 M HIf 125 mL of 0.5 M H22OO22 is reacted with is reacted with an excess of the other reactants, how an excess of the other reactants, how many grams of oxygen will be many grams of oxygen will be produced?produced?

11.511.5One way to purify a mixture of two metals is to add to the One way to purify a mixture of two metals is to add to the

mixture something that reacts with one of the metals mixture something that reacts with one of the metals and turns it into an aqueous compound. The mixture and turns it into an aqueous compound. The mixture can then be filtered, isolation the other metal. A can then be filtered, isolation the other metal. A mixture of magnesium and aluminum was treated in mixture of magnesium and aluminum was treated in such a way, to isolate the magnesium. Sodium such a way, to isolate the magnesium. Sodium hydroxide was added, which only reacts with the hydroxide was added, which only reacts with the aluminum:aluminum:

2Al(s) + 2NaOH (aq) + 6H2Al(s) + 2NaOH (aq) + 6H22O(l) → 2NaAl(OH)O(l) → 2NaAl(OH)44(aq) + (aq) + 3H3H22(g)(g)

The aluminum was converted to an aqueous salt, and the The aluminum was converted to an aqueous salt, and the magnesium stayed as a solid. The solution was then magnesium stayed as a solid. The solution was then filtered, isolating the pure magnesium. If a mixture of filtered, isolating the pure magnesium. If a mixture of these two metals contained 1.2 kg of Al, how many these two metals contained 1.2 kg of Al, how many liters of a 6.5 M NaOH solution would be needed to liters of a 6.5 M NaOH solution would be needed to make sure that all the aluminum was taken out of the make sure that all the aluminum was taken out of the mixture?mixture?

VI. Molality (m)VI. Molality (m)

Molarity (Molarity (MM) = moles (solute) /liters ) = moles (solute) /liters (solution)(solution)

- liter - liter includesincludes the solute + solvent the solute + solvent

Molality (m) = moles (solute)/ kg of Molality (m) = moles (solute)/ kg of solventsolvent = moles (solute)/mass = moles (solute)/mass of solvent of solvent onlyonly

ExamplesExamples

A. A chemist mixes 3.4 moles NaCl A. A chemist mixes 3.4 moles NaCl with 4.5 kg Hwith 4.5 kg H22O. What is the O. What is the molality?molality?

B. A solution is made with 45.0 g B. A solution is made with 45.0 g ZnClZnCl22 and 150.0 g H and 150.0 g H22O. What is its O. What is its molality?molality?

OYO’sOYO’s11.6 11.6 Calculate the molality of the Calculate the molality of the

following solutions:following solutions:

a. 2.3 moles of NaOH dissolved in 1.2 kg a. 2.3 moles of NaOH dissolved in 1.2 kg of waterof water

b. 250.0 g KNOb. 250.0 g KNO33 dissolved in 2.43 kg of dissolved in 2.43 kg of waterwater

c. 24.5 g NaBr dissolved in 578 g of c. 24.5 g NaBr dissolved in 578 g of waterwater

VII. Freezing Point VII. Freezing Point DepressionDepression

A. Why things freezeA. Why things freeze

- Things freeze when molecules slow - Things freeze when molecules slow downdown

- For this to happen, NRG must be - For this to happen, NRG must be removed from the liquidremoved from the liquid

C. We can predict by how much the temperature C. We can predict by how much the temperature will go downwill go down

∆∆T = -i KT = -i Kff m m

∆∆T = change in the temperatureT = change in the temperature

-i = the number of molecules (or ions) the -i = the number of molecules (or ions) the solute solute splits intosplits into

KKff = the freezing point depression constant = the freezing point depression constant (it is (it is different for every solvent). Units = different for every solvent). Units = ooC/mC/m

m = molality (moles of solute/kg of solvent)m = molality (moles of solute/kg of solvent)

The negative sign before i means that The negative sign before i means that you must you must subtract this amount from the original subtract this amount from the original freezing point freezing point (use 0.00 (use 0.00 ooC for the freezing C for the freezing point of water .point of water .

KKff H2O H2O = 1.86 = 1.86 ooC/m C/m

KKff acetic acidacetic acid = 3.59 = 3.59 ooC/mC/m

Example 11.3Example 11.3

A. When 15.0 grams of NaCl are added to A. When 15.0 grams of NaCl are added to 100.0 g H100.0 g H22O, what is the change in the O, what is the change in the water’s freezing point?water’s freezing point?

B. What is the freezing point of a solution B. What is the freezing point of a solution that is made by mixing 2.5 g Al(NOthat is made by mixing 2.5 g Al(NO33))33 in in 150.0 g of water?150.0 g of water?

C. What is the freezing point of a C. What is the freezing point of a solution that is made by dissolving solution that is made by dissolving 100.0 grams of table sugar 100.0 grams of table sugar (C(C1212HH2222OO1111) into 950.0 grams of ) into 950.0 grams of water? water?

KKff H2OH2O = 1.86 = 1.86 ooC/mC/m

OYO’sOYO’s11.7 In order to prevent a car’s water from 11.7 In order to prevent a car’s water from

freezing, people all “antifreeze” to the water. freezing, people all “antifreeze” to the water. The antifreeze is typically ethylene glycol The antifreeze is typically ethylene glycol (C(C22HH66OO22). What is the freezing point of a ). What is the freezing point of a radiator solution made from 4.00 kg water and radiator solution made from 4.00 kg water and 1.00 kg ethylene glycol? K1.00 kg ethylene glycol? Kf H2Of H2O = 1.86 = 1.86 ooC/mC/m

11.8 11.8 If a solution is made by dissolving 2.30 g If a solution is made by dissolving 2.30 g Mg(OH)Mg(OH)22 in to 25.0 grams of acetic acid in to 25.0 grams of acetic acid (C(C22HH44OO22), what is the freezing point? (K), what is the freezing point? (Kff acetic acetic

acidacid = 3.59 = 3.59 ooC/m, freezing point of pure acetic C/m, freezing point of pure acetic acid is 16.6 acid is 16.6 ooC)C)

VIII. Boiling Point VIII. Boiling Point ElevationElevation

A. About boiling point when solutes are presentA. About boiling point when solutes are present- The boiling point is higher when a solute is - The boiling point is higher when a solute is dissolved in a solventdissolved in a solvent- The bond of attraction is even stronger than - The bond of attraction is even stronger than the Hthe H22O– HO– H22O bondO bond- You need to add even more NRG to break - You need to add even more NRG to break bondsbonds

- When something boils, the gas H- When something boils, the gas H22O molecules O molecules must escape. The more ions that are present, must escape. The more ions that are present, the more NRG is needed to break the the more NRG is needed to break the attraction between them and the Hattraction between them and the H22O O molecules molecules

B. The formula for boiling point B. The formula for boiling point elevationelevation

∆∆T = i KT = i Kbb m m

Note:Note:

-There is not a negative sign in front -There is not a negative sign in front of “i” – the boiling point goes of “i” – the boiling point goes upup when when solutes are added to solventssolutes are added to solvents

- K- Kbb is the “Boiling point elevation is the “Boiling point elevation constant,” and is constant,” and is differentdifferent from the K from the Kff

For Example:For Example:

In cooking pasta, a chef added 20.0 In cooking pasta, a chef added 20.0 grams of table salt (NaCl) to 1.45 kg grams of table salt (NaCl) to 1.45 kg of water and brought the water to a of water and brought the water to a boil. What is the temperature of this boil. What is the temperature of this boiling solution? Kboiling solution? Kbb H2OH2O =0.521 =0.521 ooC/m C/m (Use 100.0 (Use 100.0 ooC as the boiling point for C as the boiling point for water)water)

OYO’sOYO’s11.9 11.9 Acetic acid has a KAcetic acid has a Kbb of 3.08 of 3.08 ooC/m and a C/m and a

normal boiling point of 118.5 normal boiling point of 118.5 ooC. What C. What would be the boiling point of a solution would be the boiling point of a solution made by dissolving 50.0 g of CaCOmade by dissolving 50.0 g of CaCO33 in in 500.0g of acetic acid?500.0g of acetic acid?

11.10 11.10 If you wanted to boil a solution made If you wanted to boil a solution made by dissolving 250.0 grams of table sugar by dissolving 250.0 grams of table sugar (C(C1212HH2222OO1111) into 1.3 kg of water, to what ) into 1.3 kg of water, to what temperature would you have to raise it? temperature would you have to raise it? (K(Kb H2Ob H2O = 0.521 = 0.521 ooC/m)C/m)

IX. Solutions, Enthalpy & IX. Solutions, Enthalpy & EntropyEntropy

Enthalpy (Enthalpy (HH) = used to describe heat ) = used to describe heat changes that take place during a changes that take place during a process or reaction process or reaction

∆∆HHrxnrxn = ∆ = ∆HHproductsproducts - ∆ - ∆HHreactantsreactants

- ∆- ∆HH is is negativenegative for exothermic for exothermic

- ∆- ∆H H is is positivepositive for endothermic for endothermic

A reaction is A reaction is favorablefavorable if if exothermicexothermic

∆Hsoln = ∆H1 + ∆H2 + ∆H3

But, if most things dissolve But, if most things dissolve endoendothermically, and thermically, and exoexothermic thermic processes are favorable – why do things processes are favorable – why do things dissolve at all?dissolve at all?

There are 2 factors that govern all There are 2 factors that govern all chemical and physical processes chemical and physical processes (determine whether a reaction will occur (determine whether a reaction will occur naturally – spontaneously):naturally – spontaneously):

1. NRG – exothermic or endothermic1. NRG – exothermic or endothermic

2. 2. Disorder – Disorder – everything tends to everything tends to disorder disorder (think of your bedroom!)(think of your bedroom!)

Entropy Entropy (S): a direct measure of disorder(S): a direct measure of disorder

22ndnd Law of Thermodynamics: Law of Thermodynamics: The universe The universe naturally tends to disorder (entropy naturally tends to disorder (entropy increases)increases)

The solute-solvent mixture is more The solute-solvent mixture is more disordereddisordered than the solute-solute or the than the solute-solute or the solvent-solventsolvent-solvent

-- -- EntropyEntropy increases increases

OYO’sOYO’s

11.11 Draw the diagram of the solution process. 11.11 Draw the diagram of the solution process. Label the steps (1, 2, & 3) and ∆Label the steps (1, 2, & 3) and ∆HH11, ∆, ∆HH22, ∆, ∆HH33..Put a negative or plus sign under each ∆Put a negative or plus sign under each ∆HH to to describe if the step was endo- or exo- thermic.describe if the step was endo- or exo- thermic.

11.12 What 2 factors govern all chemical and 11.12 What 2 factors govern all chemical and physical reactions?physical reactions?

11.13 In terms of those 2 factors, what type of 11.13 In terms of those 2 factors, what type of reaction will reaction will alwaysalways be spontaneous? be spontaneous?