Solutions Chapter 14. Common Solutions Chemical solutions encountered in everyday life: aircoffee tap watergasoline shampoocough syrup orange sodaGatorade.

Solutions

Chapter 14

Common Solutions

Chemical solutions encountered in everyday life:Chemical solutions encountered in everyday life:

airair coffeecoffee

tap watertap water gasolinegasoline

shampooshampoo cough syrupcough syrup

orange sodaorange soda GatoradeGatorade

Solutions

. . . the components of a mixture . . . the components of a mixture are uniformly intermingled are uniformly intermingled (the mixture is (the mixture is homogeneoushomogeneous).).

A Solute

- dissolves in water (or other “solvent”)dissolves in water (or other “solvent”)

- changes phase changes phase (if different from the (if different from the solvent)solvent)

- is present in is present in lesserlesser amount (if the same amount (if the same phase as the solvent)phase as the solvent)

A Solvent

- retains its phaseretains its phase (if different from the (if different from the solute)solute)

- is present in is present in greatergreater amount (if the same amount (if the same phase as the solute)phase as the solute)

Figure 14.1: When solid sodium chloride dissolves, the ions are dispersed randomly throughout the solution

Aqueous Solutions

Aqueous solutions are solutions in which Aqueous solutions are solutions in which water is the solvent. water is the solvent.

Aqueous solutions are the most common type Aqueous solutions are the most common type of solution. of solution.

Some Properties of Water

Water is able to dissolve so many substances because:Water is able to dissolve so many substances because:

- Water is “bent” or Water is “bent” or V-shapedV-shaped..

- The O-H bonds are The O-H bonds are covalentcovalent..

- Water is a Water is a polarpolar molecule. molecule.

- HydrationHydration occurs when salts dissolve in water. occurs when salts dissolve in water.

04_40

H

O2

105

H

Water is a polar molecule because it is a bentmolecule. The hydrogen end is + while the oxygenend is -, Delta () is a partial charge--less than 1.

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+

+

++

+

+

+

+

++

+

OH

H

O HH

+

+

Cation

Anion

Polar water molecules interact with the positiveand negative ions of a salt, assisting in the dissolving process. This process is called hydration.

Solubility

The general rule for solubility is:The general rule for solubility is:

““Like dissolves like.”Like dissolves like.”

Polar water molecules can dissolve other Polar water molecules can dissolve other polar molecules such as alcohol and, also, polar molecules such as alcohol and, also, ionic substances such as NaCl.ionic substances such as NaCl.

Nonpolar molecules can dissolve other Nonpolar molecules can dissolve other nonpolar molecules but not polar or ionic nonpolar molecules but not polar or ionic substances. Gasoline can dissolve grease.substances. Gasoline can dissolve grease.

Miscibility

MiscibleMiscible -- two substances that will mix -- two substances that will mix together in any proportion to make a together in any proportion to make a solution. Alcohol and water are miscible solution. Alcohol and water are miscible because they are both polar and form because they are both polar and form hydrogen bonds.hydrogen bonds.

ImmiscibleImmiscible -- two substances that will not -- two substances that will not dissolve in each other. Oil and vinegar are dissolve in each other. Oil and vinegar are immiscible because oil is nonpolar and immiscible because oil is nonpolar and vinegar is polar.vinegar is polar.

Figure 14.6: An oil layer floating on water

Structure & Solubility

Like dissolves like.Like dissolves like.

Hydrophobic --water-fearing. Fat soluble Hydrophobic --water-fearing. Fat soluble vitamins such as A, D, E, & K.vitamins such as A, D, E, & K.

Hydrophilic --water-loving. Water soluble Hydrophilic --water-loving. Water soluble vitamins such as B & C.vitamins such as B & C.

Hypervitaminosis--excessive buildup of vitamins Hypervitaminosis--excessive buildup of vitamins A, D, E, & K in the body.A, D, E, & K in the body.

Solubility

How does the rule “Like dissolves like.” How does the rule “Like dissolves like.” apply to cleaning paint brushes used for apply to cleaning paint brushes used for latex paint as opposed to those used with latex paint as opposed to those used with oil-based paint? oil-based paint?

Common Terms of Solution Concentration

SaturatedSaturated - when a solution contains as much - when a solution contains as much solute as will dissolve at that temperature.solute as will dissolve at that temperature.

UnsaturatedUnsaturated - when a solution has not reached - when a solution has not reached its limit of solubility.its limit of solubility.

SupersaturatedSupersaturated - when a solution has more of - when a solution has more of the solute dissolved than it normally would --the solute dissolved than it normally would --very unstable.very unstable.

Common Terms of Solution Concentration

StockStock - routinely used solutions prepared in - routinely used solutions prepared in concentrated form.concentrated form.

ConcentratedConcentrated - - relativelyrelatively large ratio of solute large ratio of solute to solvent. (to solvent. (5.0 5.0 MM NaCl NaCl))

DiluteDilute - - relativelyrelatively small ratio of solute to small ratio of solute to solvent. (solvent. (0.01 0.01 MM NaCl NaCl))

Solution CompositionQualitative -- general Qualitative -- general and relative.and relative.

dilutedilute

concentratedconcentrated

Quantitative -- mathematically defined.Quantitative -- mathematically defined.

Mass % -- biologyMass % -- biology

Molarity -- most solutions in chemistryMolarity -- most solutions in chemistry

Normality -- chemical titrationsNormality -- chemical titrations

Molality -- molar mass, freezing point Molality -- molar mass, freezing point depressions, & boiling point elevationsdepressions, & boiling point elevations

Mass %

Mass (weight) percent =Mass (weight) percent =m ass o f so lu te

m ass o f so lu tio n1 0 0 % )(

Mass % Calculations

If 1.00 g of ethanol is added to 100.0 g of If 1.00 g of ethanol is added to 100.0 g of water, what is the mass % of the ethanol?water, what is the mass % of the ethanol?

OHHC

g

g

onmasssoluti

lmassethanoOHHC

52

52

%990.0

%1000.101

00.1

%100%

Mass % CalculationsCow’s milk typically contains 4.5 % by mass of the sugar Cow’s milk typically contains 4.5 % by mass of the sugar

lactose, Clactose, C1212HH2222OO1111. Calculate the mass of lactose present in . Calculate the mass of lactose present in

175 g of milk.175 g of milk.

Mass % =Mass % =

mass of solute = (mass %)(mass of solution)/(100%)mass of solute = (mass %)(mass of solution)/(100%)

mass of solute = (4.5 % lactose)(175 g)/(100%)mass of solute = (4.5 % lactose)(175 g)/(100%)

mass of solute = 7.9 g lactosemass of solute = 7.9 g lactose

m ass o f so lu te

m ass o f so lu tio n1 0 0 % )(

Molarity

Molarity (Molarity (MM) = moles of solute per volume of ) = moles of solute per volume of solution in liters:solution in liters:

M

M

molaritymoles of soluteliters of solution

HClmoles of HCl

liters of solution3

62

Molarity Calculations

If 1.00 g of ethanol is dissolved in enough If 1.00 g of ethanol is dissolved in enough water to make 101 mL of solution, what is water to make 101 mL of solution, what is the molarity of the solution?the molarity of the solution?

Mg

mol

L

mL

mL

gethanol215.0

07.46

1

1

1000

101

00.1

Molarity Calculations

Calculate the molarity of a solution prepared by Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution.water to make 1.50 L of solution.

(11.5g NaOH/1.50L)(1 mol NaOH/40.00g NaOH)(11.5g NaOH/1.50L)(1 mol NaOH/40.00g NaOH)

= 0.192 M NaOH= 0.192 M NaOH

Molarity Calculations

Calculate the molarity of a solution prepared by Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl into enough dissolving 1.56 g of gaseous HCl into enough water to make 26.8 mL of solution.water to make 26.8 mL of solution.

(1.56g HCl/26.8mL)(1mol/36.5g)(1000mL/1L) = (1.56g HCl/26.8mL)(1mol/36.5g)(1000mL/1L) = 1.59 M HCl solution1.59 M HCl solution

Molarity Calculations

How many moles of nitrate ions are present in How many moles of nitrate ions are present in 25.00 mL of a 0.75 M Co(NO25.00 mL of a 0.75 M Co(NO33))22 solution? solution?

(25.00mL)(1L/1000mL)(0.75mol Co(NO(25.00mL)(1L/1000mL)(0.75mol Co(NO33))22/1L)/1L)

(2 mol NO(2 mol NO33--/1 mol Co(NO/1 mol Co(NO33))22) = 3.8 x 10) = 3.8 x 10-2 -2 mol NOmol NO33

--

Molarity Calculations

Typical blood serum is about 0.14M NaCl. What Typical blood serum is about 0.14M NaCl. What volume of blood contains 1.0 mg of NaCl?volume of blood contains 1.0 mg of NaCl?

(1.0mg NaCl)(1g/1000mg)(1mol/58.45g)(1L/0.14mol) (1.0mg NaCl)(1g/1000mg)(1mol/58.45g)(1L/0.14mol) = 1.2 x 10= 1.2 x 10-4-4 L blood L blood

serumserum

Molarity Calculations

A chemist needs 1.00 L of an aqueous 0.200 M A chemist needs 1.00 L of an aqueous 0.200 M KK22CrCr22OO77 solution. How much solid potassium solution. How much solid potassium

dichromate must be massed to make this dichromate must be massed to make this solution?solution?

(1.00L)(0.200mol K(1.00L)(0.200mol K22CrCr22OO77/L)(294.2g/1mol) = /L)(294.2g/1mol) =

58.8 g K58.8 g K22CrCr22OO77

Standard Solution

A standard solution is a solution whose A standard solution is a solution whose concentration is accurately known. concentration is accurately known.

A volumetric flask is used to make a standard A volumetric flask is used to make a standard solution.solution.

04_44

Volume marker(calibration mark)

Weighedamountof solute

Wash Bottle

(a) (b) (c) (d)

Steps involved in making a standard solution.

Dilution of Stock Solutions

When diluting stock solutions, the When diluting stock solutions, the moles of solute after dilution must equal the moles of solute before dilution.

moles of solute before dilution = moles of solute after dilutionmoles of solute before dilution = moles of solute after dilution

Stock solutions are diluted using either a Stock solutions are diluted using either a measuring or a delivery pipet and a volumetric measuring or a delivery pipet and a volumetric flask.flask.

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(a) (b) (c)

Rubber bulb

500 mL

Steps to dilute a stock solution.

Dilution Calculations

What volume of 16 M sulfuric acid must be used to What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M Hprepare 1.5 L of a 0.10 M H22SOSO44 solution? solution?

(0.10mol H(0.10mol H22SOSO44/1L)(1.5L)(1L/16mol)(1000mL/1L)/1L)(1.5L)(1L/16mol)(1000mL/1L)

= 9.4 mL conc = 9.4 mL conc HH22SOSO44

Dilution Calculations

What volume of 12 M HCl must be taken to What volume of 12 M HCl must be taken to prepare 0.75 L of 0.25 M HCl?prepare 0.75 L of 0.25 M HCl?

(0.75L)(0.25mol HCl/L)(1L/12mol)(1000mL/1L) (0.75L)(0.25mol HCl/L)(1L/12mol)(1000mL/1L) = 16 mL conc. = 16 mL conc.

HClHCl

Steps For Solving Solution Stoichiometry Problems

1.1. Write the balanced reaction.Write the balanced reaction.

2.2. Calculate moles of reactants.Calculate moles of reactants.

3.3. Determine limiting reactant.Determine limiting reactant.

4.4. Calculate moles of required reactant/product.Calculate moles of required reactant/product.

5.5. Convert to grams or volume, as required.Convert to grams or volume, as required.

Precipitation CalculationsWhen aqueous solutions of NaWhen aqueous solutions of Na22SOSO44 & Pb(NO & Pb(NO33))22 are mixed, are mixed,

PbSOPbSO4 4 precipitates. Calculate the mass of PbSOprecipitates. Calculate the mass of PbSO44 formed formed

when 1.25 L of 0.0500 M Pb(NOwhen 1.25 L of 0.0500 M Pb(NO33))22 & 2.00 L of 0.0250 M & 2.00 L of 0.0250 M

NaNa22SOSO44 are mixed. are mixed.

1. Pb(NO1. Pb(NO33))2(aq)2(aq) + Na + Na22SOSO4(aq)4(aq) ----> PbSO ----> PbSO4(s) 4(s) + 2NaNO+ 2NaNO3(aq)3(aq)

2. (1.25L)(0.0500mol Pb(NO2. (1.25L)(0.0500mol Pb(NO33))22/L) = 0.0625 mol Pb(NO/L) = 0.0625 mol Pb(NO33))22

(2.00L)(0.0250mol Na(2.00L)(0.0250mol Na22SOSO44/L) = 0.0500 mol Na/L) = 0.0500 mol Na22SOSO44

Precipitation CalculationsContinued

3.3. (0.0625mol Pb(NO(0.0625mol Pb(NO33))22)(1mol Na)(1mol Na22SOSO44/1mol Pb(NO/1mol Pb(NO33))2 2 ==

0.0625 mol Na0.0625 mol Na22SOSO44

NaNa22SOSO44 is the limiting reactant.is the limiting reactant.

4.4. (0.0500mol Na(0.0500mol Na22SOSO44)(1mol PbSO)(1mol PbSO44/1mol Na/1mol Na22SOSO44))

(303.3g/1mol) = 15.2 g PbSO(303.3g/1mol) = 15.2 g PbSO44

Neutralization Reactions

Acids and bases react to neutralize each other Acids and bases react to neutralize each other and form a salt and water. This type of and form a salt and water. This type of reaction is called a neutralization reaction.reaction is called a neutralization reaction.

HClHCl(aq)(aq) + + NaOHNaOH(aq)(aq) ----> NaCl ----> NaCl(aq) (aq) + HOH+ HOH(l)(l)

Acid-Base Calculations

What volume of a 0.100 M HCl solution is needed to What volume of a 0.100 M HCl solution is needed to neutralize 25.0 mL of 0.350 M NaOH?neutralize 25.0 mL of 0.350 M NaOH?

HClHCl(aq)(aq) + NaOH + NaOH(aq)(aq) ----> HOH ----> HOH(l) (l) + NaCl+ NaCl(aq)(aq)

(25.0mL)(0.350mol NaOH/1L)(1mol HCl/1mol (25.0mL)(0.350mol NaOH/1L)(1mol HCl/1mol NaOH)(1L/0.100mol) = 87.5 mL HCl solutionNaOH)(1L/0.100mol) = 87.5 mL HCl solution

Normality

Acid-Base Equivalents = (moles) (total (+) Acid-Base Equivalents = (moles) (total (+) charge)charge)

N = M ( total (+) charge)N = M ( total (+) charge)

ionLitersolut

ssoluteequivalentNNormality )(

Normality Calculations

.250 M H3PO4 =______N

N = M(total(+) charge)

N = (0.250)(3)

N = 0.750 N H3PO4

Normality Calculations

A solution of sulfuric acid contains 86 g of HA solution of sulfuric acid contains 86 g of H22SOSO44

per liter of solution. What is its normality?per liter of solution. What is its normality?

(86g H(86g H22SOSO44/1L)(1mol/98.0g) = 0.88 M H/1L)(1mol/98.0g) = 0.88 M H22SOSO44

N = M(total (+) charge)

N = (0.88M)(2)

N = 1.8 N H2SO4

Normality Stoichiometry Calculations

What volume of a 0.075 N NaOH solution is required What volume of a 0.075 N NaOH solution is required to react exactly with 0.135 L of 0.45 N Hto react exactly with 0.135 L of 0.45 N H33POPO44??

NNaa = 0.45 N = 0.45 N

VVaa = 0.135 L = 0.135 L

NNbb = 0.075 N = 0.075 N

VVbb = ? = ?

NaVa = NbVb

Vb = (NaVa)/Nb

Vb = (0.45N)(0.135L)/(0.075N)

Vb = 0.81 L NaOH