Chapter 16 Solution

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Solution Formation

The compositions of the solvent and the solute determine whether a sub will dissolve.

Polar

 substances (e.g. ionic cpds, acids, bases, salts  usually dissolve in polar solvents (e.g. water)

Nonpolar

 substances e.g. oil, wax,  usually dissolve in nonpolar solvents (e.g. gasoline) 2

Factors Affecting Speed of dissolving Factors

that determine how fast (not the amt) a sub dissolves are • stirring (agitation) • Temperature--heat • the particle size (surface area of the dissolving particles)—fine particles 3

Solution Formation

A cube of sugar in cold tea dissolves slowly.

4

Solution Formation

Granulated sugar dissolves in cold water more quickly than a sugar cube, especially with stirring.

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Solution Formation

Granulated sugar dissolves very quickly in hot tea.

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Solution Formation Stirring and Soln Formation

 speeds up the dissolving process because  fresh solvent (water) is continually brought into contact with the surface of the solute (sugar).

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Solution Formation Temperature and Solution Formation

 higher temp → heat → higher KE of water molecules → move faster.  …more collision with the solvent molecules 8

Solution Formation Particle Size and Soln Formation

 granules dissolves more quickly than a sugar cube because  smaller particles expose a much

greater surface area

to the colliding water molecules.

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Solubility

A

saturated solution

• contains the max amt of solute for a given qty of solvent at a given temp and pressure. • An

unsaturated solution

contains less solute than a saturated soln at a given temp and pressure.

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Solubility

In a saturated soln, • the rate of dissolving = the rate of crystallization, • the total amt of dissolved solute remains constant.

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Solubility

The

solubility

of a sub • the amt of solute that dissolves in a given qty of a solvent at a specified temp and (pressure) to produce a saturated soln.

• Solubility is often expressed in grams of solute per 100 g of solvent. e.g. 10 g of sugar per 100 g of water at 25°C.

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Solubility

2 (

l

) are

miscible

 dissolve in each other in all proportions. (e.g. water and alcohol).

2 (

l

) are

immiscible

 insoluble in each other. (e.g. water and oil) 13

Solubility

• Oil and water are

immiscible

.

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Solubility

• Vinegar and oil are immiscible.

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Factors Affecting Solubility Factors Affecting Solubility (the amt)

 temperature (affects both speed and amt)  solids …  gases …  pressure (affect gases only) 16

Factors Affecting Solubility Temperature

 High temp increases solubilities of ( solutes

s

) and (

l

)  High temp decreases solubilities of gaseous solutes

Pressure

(

gaseous

solutes only)  gases… (directly proportional) e.g. bottled soda … Higher pressure increases the solubility of 17

Factors Affecting Solubility Temperature

 solubility of most (

s)

sub increases as the temp of the solvent increases.  solubilities of most

gases

are greater in cold water than in hot .

 e.g. open the lid of a can of soda, put it under the sun for 30 minutes ….

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Factors Affecting Solubility

Crystals begin to form in the soln immediately after the addition of a seed crystal.

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Factors Affecting Solubility

A supersaturated soln is clear before a seed crystal is added.

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Factors Affecting Solubility

Excess solute crystallizes rapidly.

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Factors Affecting Solubility Pressure

 (

g

) solubility

increases

as the partial pressure of the (

g

) above the soln increases.

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Temperature in °C KCl 23

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Molarity Concn units

can vary greatly. •

Molarity

defined as the # of moles of solute per L of soln.

# mol solute Molarity

=

Volume of solution (L)

1 L solution ≠ 1 L solvent 26

Molarity

volume →

total soln

vol that results (not just the volume of solvent alone). • Suppose you need 1.0 L of the salt soln. 27

Molarity

• In order to be at the same [ ] as the salt in the patient’s blood, it needs to have a concn of 0.15 moles of NaCl per L of soln.

• molarity of soln = 0.15M. 28

Molarity

• 0.15

M

NaCl, where the

M

stands for “moles/L” and represents the word

molar

.

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Molarity

Need to know 3 things when working quantitatively: (1) the concn (M) , (2) the amt of solute (g), and (3) the total vol of soln (L) needed.

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Preparing 1 L of 0.15 M NaCl Solution (1)

How to prepare 1.0 L of a 0.15

M

NaCl soln? 31

Preparing 1 L of 0.15 M NaCl Solution (2)

1

mole

 58 .

5

g x

 8 .

8

g

0 .

15

mole x

• thus measure 8.8 g of NaCl 32

Preparing 1 L of 0.15 M NaCl Solution (3)

(2) Add some H 2 O to dissolve it, and then (3) Add enough additional H 2 O to bring the total vol of the soln to 1.0 L.

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Preparing a Different Vol of a Glucose Soln

How would you prepare 5.0L of a 1.5

M

glucose, C 6 H 12 O 6 ?

soln of (1) Determine the # of g of glucose to add to a 5.0-L container. 34

Preparing a Different Vol of a Glucose Soln

• The 1.5

M

soln must contain 1.5 mol of glucose per L of soln. 1

mole

 180

g

1 .

5

mole x x

 270

g

• We need 270 g of glucose for 1 L of solution.

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Preparing a Different Vol of a Glucose Soln

• Add the measured amt of glucose (1350 g) to a 5.0-L container, add enough H 2 O to dissolve the glucose, and fill with H 2 O to the 5.0-L mark. 36

Calculating Molarity Problem

: You add 32.0 g of KCl to a container and add enough H 2 O to bring the total soln volume to 955 mL. What is the molarity of this soln? 37

Calculating Molarity

Molar mass of KCl = 40g + 35g = 75g Mass Cl # moles of KCl = Molar mass KCl  32

g

75

g

 0 .

427

mole

KCl Molarity of KCl = 0 .

427

mole

0 .

955

L

 0 .

447

M

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Example 1

Intravenous saline solutions are often administered to patients in the hospital. One saline solution contains 0.90 g NaCl in exactly 100 mL of solution. What is the molarity of the solution?

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answer

0.15M

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Example 2

Household laundry bleach is a dilute aq soln of sodium hypochlorite (NaClO). How many moles of solute are present in 1.5 L of 0.70M NaClO? How many g of NaClO was used?

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Answer

1.05 moles

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Dilution problems Dilution

• Add water • Volume of solution increases • Lower concentration → molarity decreases • #mol of solute remains constant → molarity x volume = constant • M 1 V 1 = M 2 V 2 43

Example 3

How many mL of aq 2.00M MgSO 4 soln must be diluted with water to prepare 100.0 mL of aq 0.400M MgSO 4 ?

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% solutions

Volume of solute

% by volume (v/v) =

Volume of solution X 100%

e.g. in one L of 30% rubbing alcohol contains 300 mL of pure alcohol and 700mL of water

Mass of solute

% by mass (m/m) =

Mass of solution X 100%

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ppm Parts per million (parts)

• [ ] represented by # of parts of solute in one million parts of soln • e.g. 56 parts of Cu in 1,000,000 parts of ocean water --- 56 ppm • For solutes present in very small amts 46

Example 4

What is the % by vol of ethanol in the final soln when 85 mL of ethanol is diluted to a vol of 250 mL with water?

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Example 5

If 10 mL of acetone is diluted with water to a total soln volume of 200 mL, what is the % by volume of acetone in the soln?

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Example 6

How many g of K 2 SO 4 would you need to prepare 1500 g of 5.0% K 2 SO 4 (m/m) soln?

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CST problem 1

How many moles of HNO 3 are needed to prepare 5.0 L of a 2.0 M soln of HNO 3 ?

A 2.5

B 5 C 10 D 20 50

CST problem 2

The Dead Sea is the saltiest sea in the world. It contains 332 g of salt per 1000 g of water. What is the [ ] in parts per million (ppm)?

A 0.332 ppm B 332 ppm C 33,200 ppm D 332,000 ppm 51

CST problem 3

If the solubility of NaCl at 25 ° C is 36.2 g/100 g H 2 O, what mass of NaCl can be dissolved in 50.0 g of H 2 O?

A 18.1 g B 36.2 g C 72.4 g D 86.2 g 52

CST problem 4 Substance

Magnesium chloride Ammonia Ethanol Benzoic acid

Solubility of substances in water @20 ° C Formula/State

MgCl 2 / solid NH 3 / gas CH 3 CH 2 OH / liquid C 6 H 5 COOH / solid

Solubility (g/100g H 2 O)

54.6

34 infinite 0.29

Which of the substances in the table can act as either the solute or the solvent when mixed with 100 g of water at 20°C?

A B C D NH 3 C 6 H 5 COOH MgCl 2 CH 3 CH 2 OH 53

The End

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