ความเข้มข้นของสารละลาย ( Solution concentration)

: . สารละลาย ผู้�สอน อ ศราวุ�ทธ แสงอ�ไร

ความเข้�มข้�นข้องสารละลาย(Solution concentration)

วว3023130231 ปร�มาณส�มพั�นธ์� สถานะข้องสาร และปร�มาณส�มพั�นธ์� สถานะข้องสาร และเคม�ไฟฟ�าเคม�ไฟฟ�า

นายศราว�ทธ์ แสงอ�ไร นายศราว�ทธ์ แสงอ�ไร คร!ว�ชาการสาข้าเคม� โรงเร�ยนมหิ�ดลว�ทยาน�สรณ�คร!ว�ชาการสาข้าเคม� โรงเร�ยนมหิ�ดลว�ทยาน�สรณ�

ว�นท�' ว�นท�' 6 6 พัฤศจิ�กายน พัฤศจิ�กายน 25522552


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Concentration UnitsThe concentration of a solution is the amount of solute present in a given quantity of solvent or solution.Percent by Mass

% by mass = x 100%mass of solute

mass of solute + mass of solvent

= x 100%mass of solutemass of solution

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จิงค*านวณหิาความเข้�มข้�นเป+นร�อยละโดยมวลข้องสารละลายต่-อไปน�.โซเด�ยมคลอไรด� 500. g ในน*.า 2000. g

สารละลาย NaCl เข้�มข้�นร�อยละ 10 โดยมวล•จิงหิาเศษส-วนโมลข้อง NaCl•จิงหิาเศษส-วนโมลข้อง H2O

(Na = 23 , Cl = 35.5 , H = 1 , O = 16)


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Concentration Units

M =moles of solute

liters of solution

Molarity (M)

Molality (m)

m =moles of solute

mass of solvent (kg)

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Ex What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL?

m =moles of solute

mass of solvent (kg)M =

moles of solute

liters of solution

Assume 1 L of solution:5.86 moles ethanol = 270 g ethanol927 g of solution (1000 mL x 0.927 g/mL)

mass of solvent = mass of solution – mass of solute

= 927 g – 270 g = 657 g = 0.657 kg

m =moles of solute

mass of solvent (kg)=

5.86 moles C2H5OH

0.657 kg solvent= 8.92 m

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Solution Stoichiometry The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.

M = molarity =moles of solute

liters of solution

What mass of KI is required to make 500. mL ofa 2.80 M KI solution?

volume KI moles KI grams KIM KI M KI

500. mL = 232 g KI166 g KI

1 mol KIx

2.80 mol KI

1 L solnx

1 L

1000 mLx

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สารละลาย H2SO4 เข้�มข้�นร�อยละ 35 โดยมวล ความหินาแน-น 1.26 g/cm3 ท�' 20 C2.1 ในสารละลาย 100 cm3 ม� H2SO4 ละลายก�'กร�ม2.2 จิงหิาปร�มาต่รข้องสารละลายท�'ม� H2SO4 ละลายอย!- 500 กร�ม2.3 สารละลายน�.เข้�มข้�นเท-าใด ในหิน-วยโมลต่-อล!กบาศก�เดซ�เมต่ร2.4 สารละลายน�.เข้�มข้�นเท-าใด ในหิน-วยโมลต่-อก�โลกร�ม


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Ex สารละลาย HNO3 เข้�มข้�นร�อยละ 20 โดยมวล ม�ความหินาแน-น 1117. กร�มต่-อล!กบาศก�เซนต่�เมต่ร สารละลาย เข้�มข้�น ก�'โมลต่-อล�ต่ร


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Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution.

Dilution

Add Solvent

Moles of solutebefore dilution (i)

Moles of soluteafter dilution (f)=

MiVi MfVf=

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How would you prepare 60.0 mL of 0.2 MHNO3 from a stock solution of 4.00 M HNO3?

MiVi = MfVf

Mi = 4.00 Mf = 0.200 Vf = 0.06 L Vi = ? L

Vi =MfVf

Mi

=0.200 x 0.06

4.00= 0.003 L = 3 mL

3 mL of acid + 57 mL of water = 60 mL of solution

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ถ�าต่�องการเต่ร�ยมสารละลายแมกน�เซ�ยมซ�ลเฟต่ 01. mol/dm3

จิ*านวน 100 cm3 จิะต่�อง

ใช�แมกน�เซ�ยมซ�ลเฟต่ (MgSO4 . 7H2O) ก�'กร�ม (Mg = 24.3 , S = 32 , O = 16 , H = 1)


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จิงค*านวณหิาปร�มาต่รข้องสารละลายเร�'มต่�นท�'ต่�องน*ามาใช�เต่ร�ยมสารละลาย HCl 6.0 mol/dm3 จิ*านวน

500 cm3 จิากสารละลาย HCl 12 mol/dm3

(H = 1 , Cl = 35.5)


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ในการเต่ร�ยมสารละลายซ�ลเวอร�ไนเต่รต่

1) ถ�าต่�องการเต่ร�ยมสารละลายเข้�มข้�น 01.mol/dm3 จิ*านวน 250 cm3 จิะต่�องใช�ซ�ลเวอร�ไนเต่รต่(AgNO

3 ) ก�'กร�ม

2) ถ�าต่�องการเต่ร�ยมสารละลายซ�ลเวอร�ไนเต่รต่เข้�มข้�น 0025. mol/dm3 จิ*านวน 500

cm3 จิากสารละลายในข้�อ 1 ) จิะต่�องใช�สารละลายก�' cm3


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2) ถ�าต่�องการเต่ร�ยมสารละลายซ�ลเวอร�ไนเต่รต่เข้�มข้�น 0025. mol/dm3 จิ*านวน 500

cm3 จิากสารละลายในข้�อ 1 ) จิะต่�องใช�สารละลายก�' cm3


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Ex ถ�าต่�องการเต่�รยมสารละลาย HCl 0.5 โมลต่-อล�ต่ร จิ*านวน 500 ล!กบาศก�เซนต่�เมต่ร จิะต่�องน*าสารละลาย HCl 0.2 โมลต่-อล�ต่ร และสารละลาย HCl 1.0 โมลต่-อล�ต่รมาผสมก�นอย-างละก�'ล!กบาศก�เซนต่�เมต่ร


Temperature and Solubility

Solid solubility and temperature

solubility increases with increasing temperature

solubility decreases with increasing temperature

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Fractional crystallization is the separation of a mixture of substances into pure components on the basis of their differing solubilities.

Suppose you have 90 g KNO3 contaminated with 10 g NaCl.

Fractional crystallization:

1. Dissolve sample in 100 mL of water at 600C

2. Cool solution to 00C

3. All NaCl will stay in solution (s = 34.2g/100g)

4. 78 g of PURE KNO3 will precipitate (s = 12 g/100g). 90 g – 12 g = 78 g

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Temperature and Solubility

Gas solubility and temperature

solubility usually decreases with

increasing temperature

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Pressure and Solubility of Gases

The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law).

c = kP

c is the concentration (M) of the dissolved gas

P is the pressure of the gas over the solution

k is a constant (mol/L•atm) that depends onlyon temperature

low P

low c

high P

high c

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Chemistry In Action: The Killer Lake

Lake Nyos, West Africa

8/21/86CO2 Cloud Released

1700 Casualties

Trigger?

• earthquake

• landslide

• strong Winds

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How would you prepare 60.0 mL of 0.2 MHNO3 from a stock solution of 4.00 M HNO3?

MiVi = MfVf

Mi = 4.00 Mf = 0.200 Vf = 0.06 L Vi = ? L

Vi =MfVf

Mi

=0.200 x 0.06

4.00= 0.003 L = 3 mL

3 mL of acid + 57 mL of water = 60 mL of solution

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Gravimetric Analysis1. Dissolve unknown substance in water

2. React unknown with known substance to form a precipitate

3. Filter and dry precipitate

4. Weigh precipitate

5. Use chemical formula and mass of precipitate to determine amount of unknown ion

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TitrationsIn a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.

Equivalence point – the point at which the reaction is complete

Indicator – substance that changes color at (or near) the equivalence point

Slowly add baseto unknown acid

UNTIL

the indicatorchanges color

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What volume of a 1.420 M NaOH solution isRequired to titrate 25.00 mL of a 4.50 M H2SO4 solution?

WRITE THE CHEMICAL EQUATION!

volume acid moles acid moles base volume base

H2SO4 + 2NaOH 2H2O + Na2SO4

4.50 mol H2SO4

1000 mL solnx

2 mol NaOH

1 mol H2SO4

x1000 ml soln

1.420 mol NaOHx25.00 mL = 158 mL

M

acid

rx

coef.

M

base

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Chemistry in Action: Metals from the Sea

CaCO3 (s) CaO (s) + CO2 (g)

Mg(OH)2 (s) + 2HCl (aq) MgCl2 (aq) + 2H2O (l)

CaO (s) + H2O (l) Ca2+ (aq) + 2OH (aq)-

Mg2+ (aq) + 2OH (aq) Mg(OH)2 (s)-

Mg2+ + 2e- Mg

2Cl- Cl2 + 2e-

MgCl2 (l) Mg (l) + Cl2 (g)

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Colligative Properties of Nonelectrolyte Solutions

Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.

Vapor-Pressure Lowering

Raoult’s law

If the solution contains only one solute:

X1 = 1 – X2

P 10 - P1 = P = X2 P 1

0

P 10 = vapor pressure of pure solvent

X1 = mole fraction of the solvent

X2 = mole fraction of the solute

P1 = X1 P 10

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PA = XA P A0

PB = XB P B0

PT = PA + PB

PT = XA P A0 + XB P B

0

Ideal Solution

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PT is greater thanpredicted by Raoults’s law

PT is less thanpredicted by Raoults’s law

ForceA-B

ForceA-A

ForceB-B< &

ForceA-B

ForceA-A

ForceB-B> &

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Fractional Distillation Apparatus 40


Boiling-Point Elevation

Tb = Tb – T b0

Tb > T b0 Tb > 0

T b is the boiling point of the pure solvent

0

T b is the boiling point of the solution

Tb = Kb m

m is the molality of the solution

Kb is the molal boiling-point elevation constant (0C/m)

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Freezing-Point Depression

Tf = T f – Tf0

T f > Tf0 Tf > 0

T f is the freezing point of the pure solvent

0

T f is the freezing point of the solution

Tf = Kf m

m is the molality of the solution

Kf is the molal freezing-point depression constant (0C/m)

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What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g.

Tf = Kf m

m =moles of solute

mass of solvent (kg)= 2.41 m=

3.202 kg solvent

478 g x 1 mol62.01 g

Kf water = 1.86 0C/m

Tf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C

Tf = T f – Tf0

Tf = T f – Tf0 = 0.00 0C – 4.48 0C = -4.48 0C

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Colligative Properties of Nonelectrolyte Solutions


Vapor-Pressure Lowering P1 = X1 P 10

Boiling-Point Elevation Tb = Kb m

Freezing-Point Depression Tf = Kf m

Osmotic Pressure () = MRT

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Colligative Properties of Electrolyte Solutions

0.1 m NaCl solution 0.1 m Na+ ions & 0.1 m Cl- ions


0.1 m NaCl solution 0.2 m ions in solution

van’t Hoff factor (i) = actual number of particles in soln after dissociation

number of formula units initially dissolved in soln

nonelectrolytes

NaCl

CaCl2

i should be

1

2

3

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Which would you use for the streets of Bloomington to lower the freezing point of ice and why? Would the temperature make any difference in your decision?

a) sand, SiO2

b) Rock salt, NaCl

c) Ice Melt, CaCl2

Change in Freezing Point Change in Freezing Point

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Boiling-Point Elevation Tb = i Kb m

Freezing-Point Depression Tf = i Kf m

Osmotic Pressure () = iMRT

Colligative Properties of Electrolyte Solutions 48


Change in Freezing Point Change in Freezing Point Common Applications of Common Applications of Freezing Point DepressionFreezing Point Depression

Propylene glycol

Ethylene glycol – deadly to small animals

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Change in Boiling Point Change in Boiling Point

Common Applications of Common Applications of Boiling Point ElevationBoiling Point Elevation

04/20/23


At what temperature will a 5.4 molal solution of At what temperature will a 5.4 molal solution of NaCl freeze?NaCl freeze?

SolutionSolution

∆∆TTFPFP = K = Kff • m • i • m • i

∆ ∆TTFPFP = (1.86 = (1.86 ooC/molal) • 5.4 m • 2C/molal) • 5.4 m • 2

∆ ∆TTFP FP = 20.1= 20.1 ooCC

FP = 0 – 20.1 = -20.1FP = 0 – 20.1 = -20.1 ooCC

Freezing Point DepressionFreezing Point Depression 51


The Cleansing Action of Soap 52


Osmotic Pressure ()

Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one.

A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules.

Osmotic pressure () is the pressure required to stop osmosis.

dilutemore

concentrated

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HighP

LowP

Osmotic Pressure ()

= MRT

M is the molarity of the solution

R is the gas constant

T is the temperature (in K)

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A cell in an:

isotonicsolution

hypotonicsolution

hypertonicsolution

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Chemistry In Action: Desalination

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A colloid is a dispersion of particles of one substance throughout a dispersing medium of another substance.

Colloid versus solution

• collodial particles are much larger than solute molecules

• collodial suspension is not as homogeneous as a solution

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Colloids

• Tyndall Effect• Brownian motion

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แหิล-งอ�างอ�ง• Martin S. Silberberg, Chemistry: The Molecular Nature

of Matter and Change, McGraw-Hill Higher Education, 2004

• Raymond Chang, Chemistry, Williams College, McGraw-Hill Higher Education, 2002

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ความเข้มข้นของสารละลาย ( Solution concentration)