SOLUTIONS

“A” students work (without solutions manual) ~10 problems/night.

solvent solute

Dr. Alanah Fitch Flanner Hall 402 508-3119 [email protected] Office Hours Th & F 2-3:30 pm Define some measurement scales and units

Module #13 Solution Properties Defining types of concentrations For mixtures

Method for Conversion from Molarity to molality 1. Assume 1 L volume 2. Calculate moles of solute ⎛ molessolute ⎞ ⎜ ⎟ (1L) = molessolute ⎝ Lsolution ⎠

3. Calculate g of solution from d, V

⎛ 10 3 mL ⎞ d solution ⎜ ⎟ (1Lsolution ) = g solution ⎝ L ⎠

1.

Molarity

2.

mole fraction

3.

molality

M=

G3: Science Is referential molessolute Lsolution

nA ≡ χA n A + nB . molessolute molality = m = kg solvent

Example Density of an aqueous solution of ammonium sulfate is 1.06g/mL and the molarity is 0.886. What is the molality? NH SO

(

)

4 2

1. Assume 1 L volume 2. Calculate moles of solute ⎛ molessolute ⎞ ⎜ ⎟ (1L) = molessolute = 0.886 ⎝ Lsolution ⎠ 3. Calculate g of solution from d, V

4

Atoms

amu

total

2N 8( H ) S 4( O) total

14.01 8(1008 . ) 32.07 4(16.00)

28.02 8.064 32.07 64 133.098

g ⎞ ⎛ 1000mL ⎞ ⎛ . ⎟ (L ⎟⎜ ⎜ 106 ) = 1060g solution ⎝ mL ⎠ ⎝ L ⎠ solution

4. Calculate g of solute from M

⎛ g ⎞ M (1Lsolution )⎜ solute ⎟ = g solute ⎝ molsolute ⎠

4. Calculate g of solute from M

g solution − g solute = g solvent

5. Subtract to get g of solvent

5. Subtract to get g of solvent

⎛ 133.098 g solute ⎞ mol ⎞ ⎛ ⎟ = 117.9248 g solute ⎜ 0.886 ⎟ (1Lsolution )⎜ ⎝ L ⎠ ⎝ molsolute ⎠

1060 g solution − 117.9248 g solute = 942.1g soluent

6. Calculate g solute/g solvent

molality = m =

molessolute kgsolvent

6. Calculate g solute/g solvent

molality = m =

molessolute 0.886molessolute = = 0.941 kg solvent 0.9421kg solvent

1

Method for Conversion from molality to Molarity

Example Density of an aqueous solution of KOH is 1.43g/mL and the molality is 14.2. What is the molarity?

1. Assume 1 kg solvent 2. Calculate moles of solute

1. Assume 1 kg solvent 2. Calculate moles of solute

⎛ molessolute ⎞ ⎜ ⎟ 1kg solvent = molessolute ⎝ kg solvent ⎠

⎡ 14.2molesKOH ⎤ ⎢ 1kgwater ⎥ [1kgwater ] = 14.2molesKOH ⎣ ⎦

3. Calculate g of solute from MM ⎛

(moles )⎜⎝ moles

g solute

solute

solute

⎞ ⎟ = g solute ⎠

3. Calculate g of solute from MM

. gKOH ⎞ ⎛ 39.10 + 16.00 + 108 ⎟ = 797.756 gKOH ⎝ ⎠ moleKOH

[14.2molesKOH ]⎜

4. Sum masses to get total mass of solution 1000g solvent + g solute = g solution

4. Sum masses to get total mass of solution

g solution = g H2 O + g KOH = 1000 g + 797.756 g = 1797.756 g solution

5. Calculate V of solution from density

⎛ L ⎞ solution ⎞ ⎛ ⎟⎜ ⎟ = ( Lsolution ) ( g solution )⎜⎝ mL g solution ⎠ ⎝ 10 3 mL ⎠

5. Calculate V of solution from density

(1797.756g

6. Calculate moles/V ⎛ molessolute ⎞ M=⎜ ⎟ ⎝ Lsolution ⎠

M=

45 40

12 35 10

30 25

8

20

6

15

weight percent

Molaity or (densityx10)

14.2 moles , KOH moles = = 11296 . Lsolution 1257 LKOHsolution .

“A” students work (without solutions manual) ~10 problems/night.

50

Density Molality Weight %

14

⎛ 1mLKOHsolution ⎞ ⎟ = 1,257.17mL KOHsolution ⎠

)⎜⎝ 143 . g

6. Calculate moles/V

Methods of measurement are related but not equivalent 16

solution

Dr. Alanah Fitch Flanner Hall 402 508-3119 [email protected] Office Hours Th & F 2-3:30 pm

4 10 2

5

0

Module #13 Solution Properties

0 0

2

4

6

8 Molarity

10

12

14

16

Water/Salt Solutions 1

2

SOLUTIONS Any two components A and B Mixed in any mole fraction

A mixture is usually plotted as variation in mole fractions

nB ≡ χB n A + nB .

χ A + χB =

nA nB nB + nB + = =1 n A + nB . n A + nB . n A + nB . nA ≡ χA n A + nB .

1

mole fraction

0.8

0.6

0.4

0.2

0

B

0

A Solute is usually Considered the “dissolved”

Solvent is usually Considered the “dissolver”

0.2

Pure B

0.4

0.6

0.8

moles of A

1

Pure A

mixture

SOLUTIONS solvent solute

Solvation Diagrammed as an Example of Hess’s Law 2+ 2 MgSO4 ,s ⎯⎯⎯ ⎯→ Mg aq + SO42,−aq H O

∆ H = − 912 .

kJ mol

Any two components A and B Mixed in any mole fraction

-1284.9kJ mole -1376.1kJ mole

water 2+ 2 MgSO4 ,s ⎯⎯⎯ ⎯→ Mg aq + SO42,−aq H O

∆ H = − 912 .

kJ mol

3

NH 4 NO3( s ) → NH 4+( aq ) + NO3−( aq )

∆ H = + 281 . kJ

NH 4 NO3( s ) ⎯⎯⎯⎯→ NH 4+( aq ) + NO3−( aq )

∆ H = + 281 . kJ

water

http://www.qtp.ufl.edu/~roitberg/pdf/2002_04.pdf http://www.lsbu.ac.uk/water/magic.html

NH4 NO3( s) + heat → NH4+( aq ) + NO3−( aq )

For most salts, heat will increase solubility

∆ H = + 281 . kJ

Aquated Ammonium

Aquated nitrate

SOLUTIONS

“A” students work (without solutions manual) ~10 problems/night.

solvent

solute Dr. Alanah Fitch Flanner Hall 402 508-3119 [email protected] Office Hours Th & F 2-3:30 pm

Module #13 Solution Properties water

Organic/Solute Solutions

2+ 2 MgSO4 ,s ⎯⎯⎯ ⎯→ Mg aq + SO42,−aq H O

organic ∆ H = − 912 .

kJ mol

4

Like Dissolves Like What trends do you see here?

s=

g solute g solvent

similar intermolecular forces between solute-solute solvent-solvent imply solute-solvent interaction will be decent

What trends do you see here?

Soluble in water?

Soluble in water?

Context Slide

Context Slide Solute-Solvent interactions: Vitamin C and Lead Vitamin C is more soluble in the aqueous Phase of the body (urine) than other Vitamins which are stored in fat.

Vitamin C is water soluble so it is excreted from the body

the longest you can last without serious disease without consuming fresh vitamin C is 6 months

Sir Franklin

5

Context Slide

Context Slide

John Hartnell

Context Slide “A” students work (without solutions manual) ~10 problems/night.

Dr. Alanah Fitch Flanner Hall 402 508-3119 [email protected] Office Hours Th & F 2-3:30 pm

Module #13 Solution Properties Gas/Solvent Solutions Raoult’s Law, Freezing Point, Boiling Pt changes Osmotic Pressure

6

All ⎯⎯⎯⎯⎯⎯→ Ag

nA ≡ χA =1 n A + nB .

rate escapes

SOLUTIONS

Number of escapes will depend on 1. Number of molecules with energy>> intermolecular forces 2. Fraction of molecules with that energy at some the given temperature 3. The Surface area available for escape 4. Escapes independent of what gas is doing

solvent solute

Agg ⎯⎯⎯⎯⎯⎯→ Al rate sticking

Liquid Liquid Gas Solution Component B

Predict: 1. T8 6more gas vapor 2. T8 6less sticking of gases more gas vapor

Consider first case where A Is a pure substance

Component A

nA ≡ χA =1 n A + nB .

Intermolecular forces

nA ≡ χA =1 n A + nB .

Prediction 1 1. T8, more escapes, more gas phase A 2. T8, less sticking (fewer returns)

1 2 mv 2

µ

⎡ − ∆ Hvaporization ⎤ ⎢ ⎥ RT ⎦

urms =

Exponential represents fraction of molecules at temperature T with sufficient energy to break intermolecular forces ()Hvaporization)

Ppure liquid = e

Liquid Phase In a container

From earlier chapter we learned of kinetic energy: Maxwell’s Distribution

Ek =

Ppure liquid = e ⎣

1 ⎡ − ∆ Hvaporization ⎤ ⎢ ⎥ T⎣ R ⎦

Number of returns will depend on 1. Intermolecular forces in gas phase greater than collisional energy 2. Fraction of molecules at with less energy than intermolecular forces at that temperature 3. Sticking followed by dropping does not depend upon the surface area

3kT

1 E k = mass of 2 Liquid Phase In a container

−−11

TT↑↑;;eeTT ↑↑; PAo ↑ Math for the fraction of molecules with sufficient energy Is on next slide, if desired

⎛1 E k = ⎜ mass of ⎝2

RT

⎡ ⎤ 3kT ⎥ molecule ⎢ mass of molecule ⎣ ⎦ molecule

⎛ 3kT ⎞ Ek = ⎜ ⎟ ⎝ 2 ⎠ Review

σ

g molecule

⎞ 3kT ⎞⎛ ⎟ ⎟⎜ ⎠ ⎝ mass of molecule ⎠

f ( x) = e

1⎛ x− µ ⎞ 2 − ⎜ ⎟ 2⎝ σ ⎠

αEK

2

However, only a fraction of molecules at a given emperature that average speed, and therefore, Average energy If you assume a normal distribution for the “bell curve” you can calculate the fraction of molecules with an energy above the average value

f ( x) = e

−

E molecule kT

=e

−

E mole RT

7

All ⎯⎯⎯⎯⎯⎯→ Ag rate escapes

SOLUTIONS

Number of escapes will depend on 1. Number of molecules with energy>> intermolecular forces 2. Fraction of molecules with that energy at some the given temperature 3. The Surface area available for escape 4. Escapes independent of what gas is doing

solvent solute

Agg ⎯⎯⎯⎯⎯⎯→ Al rate sticking

Number of returns will depend on 1. Intermolecular forces in gas phase greater than collisional energy 2. Fraction of molecules at with less energy than intermolecular forces at that temperature 3. Sticking followed by dropping does not depend upon the surface area

Liquid Phase In a container Prediction 2 1. Surface Area 9, less escapes, less gas A 2. Surface Area 9, no impact on rate sticking

Liquid Liquid Gas Solution Component B

One way to diminish Surface area is to add solute to Pure substance A

Component A

nA ≡ χA