SOLUTIONS - · PDF file13.09.2013 · COMPONENTS OF SOLUTION • SOLUTE –...

SOLUTIONS

Engr. Yvonne Ligaya F. Musico

SOLUTION

A homogeneous mixture of two or more

substances, the relative proportion of

which may vary within certain limits.


COMPONENTS OF SOLUTION

• SOLUTE – component which is in small

quantity

• SOLVENT – component that is in the

greatest abundance and typically determines

the physical state of the solution.


EXAMPLE

State of

Solution State of Solvent State of Solute Example

Gas

Liquid

Liquid

Liquid

Solid

Solid

Solid

Gas

Liquid

Liquid

Liquid

Solid

Solid

Solid

Gas

Gas

Liquid

Solid

Gas

Liquid

Solid

Air

Oxygen in water

Alcohol in water

Salt in water

Hydrogen in palladium

Mercury in silver

Silver in gold


SOLUBILITY – the amount of solute needed to form a saturated solution in a given quantity of solvent.

SATURATED – solution that can hold no more of the solute at a particular temperature.

UNSATURATED – solution which contains less amount of solute that is required to saturate it at that temperature.


FACTORS AFFECTING SOLUBILITY

• Solute-solvent interactions

• Pressure effect

• Temperature effect


Solute-Solvent Interaction

– The stronger the attraction between solute and solvent molecules, the greater the solubility.

– Polar liquids tend to dissolve readily in polar solvents.

– Non-polar liquids tend to be insoluble in polar liquids.

– Substance with similar intermolecular attractive forces tend to be soluble in one another

– “LIKE DISSOLVES LIKE”


MISCIBLE VS IMMISCIBLE

• MISCIBLE – liquids that mix in all

proportions.

• IMMISCIBLE – liquids that do not

dissolve significantly in one another.


Pressure Effect

The solubility of a gas in any solvent is increased

as the pressure of the gas over the solvent

increases.

By contrast, the solubility of solids and liquids are

not appreciably affected by pressure.

The solubility of the gas increases in direct

proportion to its partial pressure above the

solution.


Pressure Effect

The relationship between pressure and solubility of gas is expressed by HENRY’S LAW:

gg kPC

Where: Cg = concentration of the gas

k = proportionality constant or Henry’s

law constant

Pg = partial pressure of the gas over the

solution.


Sample Problem

Calculate the concentration of CO2 in a softdrink

that is bottled with a partial pressure of CO2 at

4.0 atm over the liquid at 25oC. The Henry’s

law constant for CO2 in water at this

temperature is 3.1 x 10-2 mol/L-atm.


Solution

= (3.1 x 10-2 mol/L-atm)(4.0

atm)

= 0.12 mol/L or 0.12 M

gg kPC


Practice Exercise

Calculate the concentration of CO2 in the

softdrink after the bottle is opened and

equilibrates at 25oC under CO2 partial pressure

of 3.0 x 10-4 atm.

Ans.: 9.3 x 10-6 M


Temperature Effect

– The solubility of most solid solutes in water

increase as the temperature of the solution.

– In contrast to gas solutes, the solubility of gases in

water decreases with increasing temperature.


Concentration

• It refers to the weight or volume of the solute

present in a specified amount of the solvent or

solution.


WAYS OF EXPRESSING

CONCENTRATIONS

• QUALITATIVELY

• QUANTITATIVELY


QUALITATIVELY

• DILUTE – a solution with a relatively small

concentration of solute.

• CONCENTRATED – a solution with a large

concentration.


QUANTITATIVELY

• Percent by weight

• Percent by volume

• Mole fraction and Mole Percent

• Molality (m)

• Molarity (M)

• Normality


1. Percent by Weight

100% xsolutionofweight

componentofweightweightby


Problem 1

A solution is made by dissolving 13.5 g of glucose, C6H12O6, in a 0.1 kg water. What is the weight percentage of solute in a solution?


Solution

But , weight of solution = weight of solute + weight of solvent = 13.5 g + 100 g = 113.5 g

100% 6126

6126 xsolutionofweight

OHCofweightOHCweightby

%9.111005.113

5.13% 6126 x

g

gOHCweightby


Problem 2

A commercial bleaching solution contain 3.62%

weight of sodium hypochlorite, NaOCl. What

is the weight of NaOCl in a bottle containing

2500 g of bleaching solution.


Solution

NaOClgg

NaOClweight

xg

NaOClweight

xsolutionofweight

NaOClweightNaOClweightby

5.90100

)2500)(62.3(

1002500

62.3

100%


2. Percent by Volume

100% xsolutionofvolume

soluteofvolumevolumeby


Sample Problem

What is % by volume of 12 ml of alcohol in a 100 ml of wine?


Solution

%12100100

12%

100%

xml

mlvolumeby

xsolutionofvolume

alcoholofvolumevolumeby


3. Mole Fraction and Mole Percent

100%

100%

xfractionmoleMole

or

xsolutionofmoles

componentofmolesMole

solutionofmoles

componentofmolescomponentoffractionMole


NOTE:

The sum of the mole fractions of solute and

solvent must be equal to one (1)


Sample Problem

Calculate the mole fractions and mole %

of ethyl alcohol, C2H5OH, and water,

H2O, in a solution by dissolving 13.8 g

of alcohol in 27 g of water.


Solution

molessolventofmolesoluteofmolesolutionofMoles

molemoleg

g

OHMW

OHweightOHMole

molemoleg

g

OHHCMW

OHHCweightOHHCMole

8.15.13.0

5.1/18

27

30.0/46

8.13

2

22

52

5252


Solution

%3.83)100)(833.0(100%

%7.16)100)(167.0(100%

833.0167.011

833.08.1

5.1

167.08.1

3.0

22

5252

522

2

52

52

52

xOHoffractionMoleOHMole

xOHHCoffractionMoleOHHCMole

OHHCoffractionMoleOHoffractionMole

or

moles

molesOHoffractionMole

moles

molesOHHCoffractionMole

solutionofmoles

OHHCofmolesOHHCoffractionMole


solventoframlogki

soluteofmolesmMolality )(

4. Molality (m)


Problem

Calculate the molality of a solution

made by dissolving 262 g of ethylene

glycol, C2H6O2, in 8000 g of water.


Solution

mkg

molem

molemoleg

g

OHCMW

OHCgOHCofmoles

OHoframlogki

OHCofmolesmMolality

5275.08

22.4

22.4/62

262

)(

262

262262

2

262


5. Molarity (M)

solutionofLiter

soluteofmolesMMolarity )(


Problem

Calculate the molarity of a solution made by dissolving 4.0 g of calcium bromide, CaBr2, in enough water to give 200 ml of solution.


Solution

ML

moleM

molemoleg

g

CaBrMW

CaBrofweightCaBrofmoles

solutionofLiter

CaBrofmolesMMolarity

1.02.0

02.0

02.0/200

0.4

)(

2

22

2


6. Normality (N)

soluteofweightequivalent

soluteofweightsoluteofequivalent

solutionofLiter

soluteofequivalentNNormality

)(


Equivalent Weight

• weight of substances that are equivalent to one

another in chemical reaction.

• The equivalent of an oxidizing agent or

reducing agent is the weight of the substance

required to gain or lose 1 mole of electron.


Equivalent Weight

Equivalent Weight = molecular weight (MW)

f (factor)


Reacting Capacity of Solute

ACID BASE SALT

f

Total number of H+

replaceable

Total number of replaceable

OH-

Total (+/-) charge


Sample Problem 1

Calculate the equivalents of:

(a) 20 g of H2SO4

(b) 50 g of Mg(OH)2

(c) 30 g of NaOH


Sample Problem 2

Calculate the normality of a Na2CO3 solution

containing 13.75 grams in 125 ml solution.


Solution

N = equivalent of solute

Liter of solution

equiv. of Na2CO3 = wt./equiv. wts.

= 13.75 g .

(106 g/mol)/(2 equiv/mol)

= 0.259 equiv/mol

N = (0.259)

(0.125)

N = 2.075 N


RELATIONSHIP BETWEEN NORMALITY AND

MOLARITY

N = n solute x f

liter of solution

N = molarity (M) x f


Sample Problem 5

Find:

(a) The molarity (M) of 0.12 N H3PO4

(b) The normality (N) of 0.25 M NaOH

(c) The normality (N) of 2.50 M HCl


Dilution

• It refers to the reduction of concentration of a

solution.

• The most important thing to remember

concerning dilution is that you are only adding

solvent.

• You are not adding solute when you dilute.


Dilution

Therefore:

moles of solute before dilution = moles of solute after dilution

or

Number of millimoles of solute before = millimoles of solute after dilution

or

Number of grams of solute before dilution = number of grams of solute after dilution

Since the definition for Molarity is:

Molarity = moles of solute / volume of solution in liters

Solving for moles of solute gives:

moles of solute = M x V of soln in liters

or

millimoles of solute = M x V of solution in ml Engr. Yvonne Ligaya F. Musico

Dilution

If

Moles of solute before dilution = moles of solute after dilution

Then

M x V in liters before dilution = M x V in liters after dilution

or

M1V1 = M2V2

where

M1 = Molarity before dilution

V1 = volume of solution before dilution

M2 = Molarity of solution after dilution

V2 = Volume of solution after dilution Engr. Yvonne Ligaya F. Musico

Dilution

Actually one can use Molarity or mass % as the

concentration term so you could have the

following alternative where mass % is used:

mass % x grams solution before dilution = mass % x grams of solution after dilution


Sample Problem 1

How would you prepare 500 ml of 3 M HCl

using 6 M HCl from the stockroom.


Solution

HClMofmlV

M

mlMV

mlMVM

VMVM

6250

6

5003

50036

1

1

1

2211

1. Determine the volume of 6M of HCl

to use applying the dilution equation


Solution

2. Determine the amount of water to be added to 6M

HCl

Since the total volume after dilution is 500 ml and the

volume of 6M to use is 250 ml then:

Volume of water = 500 - 250 = 250 ml of water


Sample Problem 2

A chemist starts with 50.0 mL of a 0.40 M NaCl

solution and dilutes it to 1000. mL. What is the

concentration of NaCl in the new solution?


Solution

MM

mlMM

mlMmlM

VMVM

02.0

1000

504.

1000504.0

2

2

2

2211


Sample Problem 3

How would you prepare 800 grams of a 3%

Hydrogen Peroxide solution using 10% H2O2

solution


Solution

1. Determine the mass of 10% H2O2 to be used using the dilution equation mass of H2O2 before dilution = mass of H2O2 after dilution

mass % x mass of solution before dilution = mass % x mass of solution after dilution

10 ( mass of solution before) = (3) (800)

mass of solution before = (3) (800)

10

mass of solution before = 240 grams


Solution

2. Determine the mass of water to be added

Total mass of 3% = mass of 10% + mass of water added

mass of water to be added = 800 - 240

mass of water to be added = 560 g


Sample Problem 4

A chemist wants to make 500. mL of 0.050 M

HCl by diluting a 6.0 M HCl solution. How

much of that solution should be used?


Solution

HClMofmlV

M

mlMV

mlMVM

VMVM

0.617.4

0.6

50005.0

500050.00.6

1

1

1

2211


Practice Exercise

1. How much 2.0 M NaCl solution would you need to make 250 mL of 0.15 M NaCl solution?

2. How would you prepare 1000 ml of a 5% glucose solution using a 20% glucose solution. How much 20% glucose and how much water to be mixed?

3. What would be the concentration of a solution made by diluting 45.0 mL of 4.2 M KOH to 250 mL?

4. What would be the concentration of a solution made by adding 250 mL of water to 45.0 mL of 4.2 M KOH?

5. How much 0.20 M glucose solution can be made from 50. mL of 0.50 M glucose solution?


Thank You for Listening


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