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                                       Molarity

Molarity video

Concentrations of Solutions 

The concentration of a solution is the "strength" of a solution.

A solution typically refers to the dissolving of some solid substance in a liquid, such as dissolving salt in water.

It is also often necessary to figure out how much water to add to a solution to change it to a specific concentration.

The concentration of a solution is typically given in molarity.

Molarity is defined as the number of moles of solute (what is actually dissolved in the solution) divided by the volume in liters of solution (the total volume of what is dissolved and what it has been dissolved in).

Molarity =
moles of solute

liters of solution

Molarity is probably the most commonly used term because measuring a volume of liquid is a fairly easy thing to do.

Example: If 5.00 g of NaOH are dissolved in 5000 mL of water, what is the molarity of the solution?


One of our first steps is to convert the amount of NaOH given in grams into moles:

5.00g NaOH

1
x
1 mole

(22.9 + 16.00 + 1.008)g
= 0.125 moles

Now we simply use the definition of molarity: moles/liters to get the answer

Molarity =
0.125 moles

5.00 L of soln
= 0.025 mol/L

So the molarity (M) of the solution is 0.025 mol/L.

Molarity:  What it is and how to calculate it
Molarity is equal to the moles of solute divided by the liters of   solution.                       
 
To put it in the form of an equation:

molarity =

moles of solute
liter of solution

Example:  What's the molarity of a solution that contains 5.5 moles of sodium chloride in 10.5 liters of solution?

                  Answer:  M = moles / liters, or (5.5 moles) / (10.5 liters) = 0.52 M.

In this case, the unit is M. 

M stands for "molarity". 

A 0.52 M solution is referred to as being a "0.52 molar" solution.     

Example:  If I have 3.50 grams of sodium chloride in 1250 mL of a solution, what's the molarity?

Solution: 

To find molarity, we need to convert grams to moles and milliliters to liters. 

To convert grams to moles, we first need to divide the number of grams by the molar mass of sodium chloride.  (3.5/58.5) = 0.060 moles. 

To convert milliliters to liters, multiply by 0.001.  (1250 x 0.001) = 1.25 liters. 

In the final step, divide the number of moles by the number of liters to get the molarity. 

Since 0.060 / 1.25 = 0.048, the molarity of the solution is 0.048 M.

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                                      Solutions

Explanation of what a solution is

http://www.northland.cc.mn.us/biology/Biology1111/animations/dissolve.html

Whenever you dissolve something in a liquid, you've made a solution. 

 Another way of thinking about it is that if you have a liquid and it's not a pure substance, it's a solution. 

The liquid part is the "solvent" and the stuff you dissolved (usually a solid) is the "solute".  An example: 

If you dissolve salt in water, the salt is the solute and the water is the solvent. 

A handy rule of thumb: 

If somebody asks you to tell you what the solvent in a solution is and you have no idea, say "water". 

Water is by far the most common solvent for solutions that you're likely to run into.

When we're talking about how much of the solute is dissolved in the liquid, we're talking about the concentration. 

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There are four terms we can use to describe the concentration of a solution:

unsaturated: 

This means that if you were to add more solute to the liquid, it would keep dissolving. 

 For example:

if you take one teaspoon of salt and put it in a bucket of water, you've made an unsaturated solution. 

(In other words, if you added another teaspoon of salt, it would dissolve, too).

saturated: 

This means that the liquid has dissolved all of the solute that is possible. 

If you add one teaspoon of sugar to iced tea, you've got an unsaturated solution. 

If you keep adding sugar to iced tea, you eventually get to the point where the rest of the sugar just sinks to the bottom. 

When this happens, it means that the solution is saturated, because no more sugar could dissolve.

supersaturated: 

This means that MORE solute has dissolved than is possible. 

How, you might ask, does this happen?  If you have a very hot saturated solution and cool it down, the solubility of the solute decreases as the solution cools. 

(In other words, hot solutions can dissolve more solute than cold ones). 

What usually happens in this situation is that the solute starts forming crystals at the bottom of the container. 

However, under weird circumstances where there are no little grains to start crystal formation, the crystals never form - as a result, the solution is MORE concentrated than possible. 

This doesn't happen much, so you'll never run into it in real life, most likely.

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               Types of Mixtures

Homogeneous mixtures are uniform throughout.

The composition is the same in all directions in the substance.

The types of particles observed in one direction are the same that are observed in all others.

A solution of sugar in water is homogeneous.

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Heterogeneous mixtures are not uniform.

There are pockets of one substance surrounded by pockets of different substances.

A mixture of soil in water to make "mud" is heterogeneous.

Likewise a mixture of oil and water in salad dressing is heterogeneous.

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Types of Homogeneous Mixtures

Solutions

               Particle sizes distinguish one homogeneous mixture from another.

                   Solutions are mixtures with particle sizes at the molecule or ion level.

                        The particles have dimensions between 0.1 to 2 nanometers.

             Typically solutions are transparent.

            Light can usually pass through the solution.

                             If the solute is able to absorb visible light then the solution will have a color.

                       A blue liquid transmits blue light and absorbs the other colors of the spectrum.

                       A mixture of water H2O and ethanol CH3CH2OH is homogeneous.

             The particles are individual molecules of H2O and CH3CH2OH.

                            The two molecules are spread uniformly throughout the solution.

                 A mixture of water and sodium chloride is homogeneous by chemistry standards.

             The particles in the mixture are molecules of H2O and hydrated sodium cations, Na+, and chloride anions, Cl1-.

                Solutions are transparent. You can see through them.

                 The mixture remains stable and does not separate after standing for any period of time.

                     The particles are so small they cannot be separated by normal filtration.

                A solution may have a "color" but it will still be transparent.

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Colloids

Colloids are mixtures with particle sizes that consist of clumps of molecules.

The particles have dimensions between 2 to 1000 nanometers.

The colloid looks homogeneous to the naked eye.

Fog and milk are examples of colloids.

Colloids frequently appear "murky" or "opaque".

The particles are large enough to scatter light.

You have experience with the way fog interacts with the light from car headlights.

Colloids generally do not separate on standing.

They are not separated by filtration.

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Suspensions

Suspensions are homogeneous mixtures with particles that have diameters greater than 1000 nm, 0.000001 meter.

The size of the particles is great enough so they are visible to the naked eye.

Blood and aerosol sprays are examples of suspensions.

Suspensions are "murky" or "opaque".

They do not transmit light. Suspensions separate on standing.

The mixture of particles can be separated by filtration.

Examples of solutions

Solutions are mechanical combinations of materials. The physical state for the materials in the solutions is surprising varied. The combinations of states of matter that form solutions are listed below.

types of solutions:

States of matter in solution

Example

gas in gas

air ( N2, O2 , Ar, CO2 , other gases)

gas in liquid

soda pop (CO2 in water)

liquid in liquid

gasoline (a mixture of hydrocarbon compounds)

solid in liquid

sea water ( NaCl and other salts in water)

gas in solid

H2 in platinum or palladium

liquid in solid

dental amalgams ( mercury in silver)

solid in solid

alloys ( brass, (Cu/Zn), solder (Sn/Pb)

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                            Properties of Solutions

Boiling point: 

As the molarity of a solution increases, the boiling point increases. 

This is because the solute lowers the vapor pressure of the solution and solutions don't boil until the vapor pressure of the solution is equal to atmospheric pressure. 

If this doesn't make any sense, just remember that strong solutions boil at higher temperatures than weak ones.

Melting point: 

As the molarity of a solution increases, the melting point decreases. 

This is because the solute keeps the solvent molecules from forming a nice solid lattice. 

Think of this:  Ocean water doesn't freeze very easily - this is because it's got salt dissolved in it.

Osmotic pressure: 

You should have heard about this one in your biology class somewhere along the road. 

Basically, if you've got a solution that's separated from a pure solvent by a semi-permeable membrane, the pure solvent likes to move across the membrane to decrease the concentration of the solution. 

The pressure that the solvent pushes across the membrane with is called osmotic pressure. 

Not surprisingly, the more concentrated the solution, the more the pure solvent likes to push across the membrane and the higher the osmotic pressure.

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                                     HEATING AND COOLING CURVES

Phase changes are physical changes - they do not involve a change in the composition of a substance.

Instead, they involve a change in the forces of attraction among the particles of a substance.

 The role of energy during phase changes is very specific.

Consider an ice cube melting at 0.0°C.

As the ice melts, it absorbs energy from its surroundings.

All of the energy absorbed by the ice is used to overcome the intermolecular forces that hold water molecules together in a crystalline structure.

The absorbed energy is converted to potential energy.

The fact that the ice-water temperature stays at 0.0°C as the ice melts is important because it means that the kinetic energy of the water molecules is not changing during the melting process.

Temperature changes involve a change in the kinetic energy of the particles in a substance.

These changes do not involve changes in the forces of attraction between particles. It is the motion of the particles that changes when temperature changes.

As molecules absorb heat they move faster and their temperature - the indicator of their average kinetic energy - increases.

Consider this heating curve.

It consists of two kinds of lines: those indicating temperature increase (positive slope) and those indicating temperature stability (zero slope).

The positive slope lines indicate that added energy is being converted to kinetic energy.

First, solid water warms to its melting point (MP).

Then liquid water warms from the freezing point (MP) to the boiling point (BP). Finally, the water vapour (steam) is warmed.

The flat lines indicate stable, unchanging temperature - periods when phase changes occur.

The added energy is converted to potential energy as the particles in one state change to a higher potential energy state.

The absorbed energy weakens or breaks the forces of attraction that hold the particles in a certain state.

The flat line at the melting point indicates that all of the added energy is used to convert solid ice to liquid water.

The flat line at the boiling point indicates that all of the added energy is used to convert liquid water to water vapour.

At any point on a heating curve, just one of the two types of energy change is occurring.

Either kinetic energy is increasing or potential energy is increasing, but never both at the same time. (Roll your mouse over the heating curve above.)

Now consider this cooling curve.

The regions of a cooling curve can be interpreted in the same way as those of a heating curve.

The essential difference is that energy is constantly being lost to the surroundings by the system.

The result is either a decrease in kinetic energy or a decrease in potential energy.

The first temperature decrease occurs as steam cools (loses kinetic energy).

At the condensation point (BP), potential energy is lost as the as the gas system condenses to the liquid state.

The next temperature decrease represents the cooling of the liquid phase.

At the freezing point (MP) liquid water loses potential energy as bonds form to hold molecules in fixed positions in the solid state.

Finally, kinetic energy decreases once more as the solid cools. (Roll your mouse over the cooling curve.)

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                                                               Colligative Properties

video:  colligative properties

Not surprisingly, when you dissolve something in a liquid, it has different properties than the pure liquid. 

Any properties that change when the concentration of a solution changes are called colligative properties.

People always list boiling point elevation, melting point depression, and osmotic pressure as the main colligative properties of liquids. 

Think about Kool Aid.  Let's make a weak Kool Aid solution by dissolving one grain of Kool Aid in a glass of water. 

Let's also make a strong Kool Aid solution by dissolving a cup of Kool Aid powder in a glass of water. 

The properties that are different between the two glasses of Kool Aid are "colligative properties".

In our example, we would find that:

Strong Kool Aid is darker than weak Kool Aid.  As a result, we would say that "color" is a colligative property of liquids.

This is the basis for a field called spectophotometry, where you can figure out how concentrated a solution is by looking at the color.

Strong Kool Aid is denser than weak Kool Aid.  Density is a colligative property.

Strong Kool Aid boils at a higher temperature than weak Kool Aid.  Boiling point is a colligative property.

Strong Kool Aid freezes at a lower temperature than weak Kool Aid (if you've ever tried to make homemade popsicles, you know this to be true). 

Melting / freezing point is a colligative property.

 The extent of boiling-point elevation can be calculated by applying Clausius-Clapeyron relation and Raoult's law together with the assumption of the non-volatility of the solute.

The result is that in dilute ideal solutions, the extent of boiling-point elevation is directly proportional to the molal concentration of the solution according to the equation:

ΔTb = Kb · mB

 where

ΔTb, the boiling point elevation, is defined as Tb (solution) - Tb (pure solvent).

Kb, the ebullioscopic constant, which is dependent on the properties of the solvent.

mB is the molality of the solution, calculated by taking dissociation into account since the boiling point elevation is a colligative property, dependent on the number of particles in solution.

This is most easily done by using the van 't Hoff factor i as

mB = msolute · i. The factor i accounts for the number of individual particles (typically ions) formed by a compound in solution. Examples:

i = 1 for sugar in water

i = 2 for sodium chloride in water, due to the full dissociation of NaCl into Na+ and Cl-

i = 3 for calcium chloride in water, due to dissociation of CaCl2 into Ca2+ and 2Cl-

Values of the ebullioscopic constants Kb for selected solvents:

CompoundBoiling point in °CEbullioscopic constant Kb in units of [(°C·kg)/mol] or [°C/molal]
Acetic acid         118.1                          3.07
Benzene         80.1                          2.53
Carbon disulfide         46.2                          2.37
Carbon tetrachloride         76.8                          4.95
Naphthalene       217.9                          5.8
Phenol       181.75                          3.04
Water       100

                          0.512

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Freezing Point Depression:

 A solute lowers the freezing point of a solvent.

In dilute solutions, the freezing point depression is proportional to the molality of the solute particles:

ΔTf = -Kf x m x i

ΔTf = the amount by which the freezing point is lowered
 
m = molality (moles solute particles per kg of solution)
 
Kf = molal freezing-point depression constant (solvent dependent)

Freezing Point of solution = normal freezing point of solvent + ΔTf

Some Boiling-Point Elevation and Freezing-Point Depression Constants

solventnormal boiling
point (oC)
Kb (oCm-1)normal freezing
point (oC)
Kf (oCm-1)
benzene80.22.535.55.12

water100.00.5120.0001.855

acetic acid
(ethanoic acid)
118.53.0716.63.90

camphor208.35.95178.440.0

naphthalene218.05.6580.26.9

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                                     Molality

It is defined as the number of the moles of the solute present in 1 kg of the solvent, It is denoted by m.
 
Molality (m) = Number of moles of solute/Number of kilo/grams of the solvent
 
Let wA grams of the solute of molecular mass mA be present in wB grams of the solvent, then
 
                                   Molality (m) = Moles of solute/Kg or solvent
 
 Note :
(i)   Molality is the most convenient method to express the concentration because it involves the mass of liquids rather than their volumes.
 
It is also independent of the variation in temperature.

(ii) Molality and solubity are related by the following relation.

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                                       Parts Per Million (ppm)

 
Milligrams Per Liter
(Parts Per Million)
 

Another popular way of recording the concentration of chemicals in the water is called parts per million (ppm).

milligrams per liter (mg/L) is the same amount as ppm.

Using a sodium nitrate example, (0.01 g/1000 + 0.01 gram) x 1000000 is equivalent to 10 ppm, meaning that there are 10 units of sodium nitrate in every million units tested.

This is the same as 10 particles nitrate within one million particles.  

A part per million is equal to a milligram per liter. 
 
Parts per million (abbreviated ppm) is one part in a million parts. 
  
One part per million is one ounce in a million ounces, one gram in a million grams, etc. 
                                        http://www.sciencegeek.net/Chemistry/Video/Unit6/SolnConc9.shtml

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        General Solubility (Regents Level)

The New York State Board of Regents require very little knowledge about solubiliy. 

However, students should know how to use table G and figure out the solubility of certain substances with changing temperature. 

At the honor's level, the story changes dramatically!

Consider the partial solubility graph below:

All of these substances represent salts that are dissolving in water at different temperatures. 

All of the lines are rising as the temperature increases.

Be aware, the the curves for gases are DECENDING.

In other words, as the temperature of water increases, the solubility of gases decreases.

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     Table of Solubility Product Constants (Ksp at 25o C)

Bromides PbBr26.3 x 10-6

 AgBr 3.3 x 10-13
Carbonates BaCO38.1 x 10-9

CaCO33.8 x 10-9

CoCO38.0 x 10-13

CuCO32.5 x 10-10

FeCO33.5 x 10-11

PbCO31.5 x 10-13

MgCO34.0 x 10-5

MnCO31.8 x 10-11

NiCO36.6 x 10-9

Ag2CO38.1 x 10-12

ZnCO31.5 x 10-11
ChloridesPbCl21.7 x 10-5

AgCl 1.8 x 10-10
Chromates  BaCrO42.0 x 10-10

CaCrO47.1 x 10-4

PbCrO41.8 x 10-14

Ag2CrO49.0 x 10-12
CyanidesNi(CN)23.0 x 10-23

AgCN1.2 x 10-16

Zn(CN)28.0 x 10-12
Fluorides BaF21.7 x 10-6

CaF23.9 x 10-11

PbF23.7 x 10-8

MgF26.4 x 10-9
Hydroxides AgOH2.0 x 10-8

Al(OH)31.9 x 10-33

Ca(OH)27.9 x 10-6

Cr(OH)36.7 x 10-31

Co(OH)22.5 x 10-16

Cu(OH)21.6 x 10-19

Fe(OH)27.9 x 10-15

Fe(OH)36.3 x 10-38

Pb(OH)2 2.8 x 10-16

Mg(OH)2 1.5 x 10-11

Mn(OH)24.6 x 10-14

Ni(OH)22.8 x 10-16

Zn(OH)24.5 x 10-17
Iodides PbI28.7 x 10-9

AgI1.5 x 10-16
Oxalates BaC2O41.1 x 10-7

CaC2O42.3 x 10-9

MgC2O48.6 x 10-5
PhosphatesAlP041.3 x 10-20

Ba3(P04)21.3 x 10-29

Ca3(P04)21.0 x 10-25

CrP042.4 x 10-23

Pb3(P04)23.0 x 10-44

Ag3P041.3 x 10-20

Zn3(P04)29.1 x 10-33
Sulfates BaS041.1 x 10-10

CaS042.4 x 10-5

PbS041.8 x 10-8

Ag2S041.7 x 10-5
Sulfides CaS 8 x 10-6

CoS 5.9 x 10-21

CuS 7.9 x 10-37

FeS4.9 x 10-18

Fe2S31.4 x 10-88

PbS 3.2 x 10-28

MnS5.1 x 10-15

NiS3.0 x 10-21

Ag21.0 x 10-49

ZnS 2.0 x 10-25
SulfitesBaS038.0 x 10-7

CaS031.3 x 10-8

Ag2S031.5 x 10-14

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                                Mole Fraction

The concentration of a solution refers to the amount of solute in a given amount of solvent. 

There are many ways to express, quantatively, the concentration of a solution.  

Some solution properties depend on the relative amounts of all the solution components in terms of moles.  

The mole fraction of a solution component Xi  is the fraction of moles of component i of the total number of moles of all components in solution.  

moles of component i

Xi

=

------------------------

total moles of solution

Lets work through an example problem that will illustrate this concept.

Example problem

What is the mole fraction of each component in a solution in which 3.57 g of sodium chloride, NaCl, is dissolved in 25.0 g of water?

Solution

First, convert from mass of NaCl to moles of NaCl.

1 mole NaCl

3.57 g NaCl

x

----------------------

=

0.0611 mole NaCl

58.44 g NaCl

 

Next, convert from mass of water to moles of water.

1 mole H2O

25.0 g H2O

x

-----------------------

=

1.39 mole H2O

18.02 g H2O

Substitute these two quantities into the defining equation for mole fraction.

 

0.0611 mol NaCl

XNaCl

=

--------------------------------

=

0.0421

(0.0611 + 1.39)


1.39 mole H2O

Xwater

=

--------------------

=

0.958

(0.0611 + 1.39)

 Note that within the limits of the significant figures that XNaCl + Xwater = 1.  The sum of the mole fractions for a solution will equal 1.  In our example above, after we have calculated one mole fraction we could have subtracted it from 1 to obtain the other.

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               Vapor Pressure Lowering of Solutions

Vapor Pressure Lowering of Solutions

In a solution, a fraction of the molecules with energy in excess of the intermolecular force are nonvolatile solute molecules, or another way of saying this is that the number of solvent molecules with energies above the intermolecular force has been reduced. 

Because of this the solution has a lower vapor pressure than the pure solvent. 

Experiments on the vapor pressures of solutions containing nonvolatile solutes were carried out by Francois Marie Raoult (1830-1901).

His results are described by the equation known as Raoult's law:

Psolution  =  XsolventPosolvent

where Psolution is the observed vapor pressure of the solution, Xsolvent is the mole fraction of solvent, and Posolvent is the vapor pressure of the pure solvent.

Any solution that obeys Raoult's law is called an ideal solution.

Raoult's law is to solutions what the ideal gas law is to gases.

Yet, some strong solute-solvent interactions do not behave ideally, and gives a vapor pressure lower than that predicted by Raoult's law. 

When this happens, a negative deviation from Raoult's law results.

When liquid-liquid solutions with both components being volatile, a modified form of Raoult's law applies:

Ptotal = Pa + Pb = XaPoa + XbPob

where Ptotal represents the total vapor pressure of a solution containing A and B. Xa and Xb are the mole fractions of A and B. Poa and Pob are the vapor

pressures of pure A and B, and Pa and Pb are the partial pressures resulting from molecules of A and B in the vapor above the solution.

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           Solubility Product Constants

Ksp video 

Solubility Product Constants, Ksp

Solubility product constants are used to describe saturated solutions of ionic compounds of relatively low solubility. 

A saturated solution is in a state of dynamic equilibrium between the dissolved, dissociated, ionic compound and the undissolved solid.

 

MxAy(s) --> x My+ (aq) + y Ax- (aq)
 

The general equilibrium constant for such processes can be written as:

Kc = [My+]x[Ax-]y

Since the equilibrium constant refers to the product of the concentration of the ions that are present in a saturated solution of an ionic compound, it is given the name solubility product constant, and given the symbol Ksp.  Solubility product constants can be calculated, and used in a variety of applications.  

for an excellent explaination of solubility try the link below:

http://rds.yahoo.com/_ylt=A0geurVkm7hLO8IANoVXNyoA;_ylu=X3oDMTE1dnYycDlyBHNlYwNzcgRwb3MDOARjb2xvA2FjMgR2dGlkA01BUDAxNF8xMTk/SIG=12vt5oudm/EXP=1270476004/**http3a//www.brynmawr.edu/Acads/Chem/sburgmay/chem104/solubilityAnimation.jar

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              Writing Solubility Product Expressions

The following are examples of how to write solublity product expressions. 

1.  Write the expression for the solubility product constant for strontium sulfate.

a.  write the formula:  SrSO4

 b.  write the ionization:  SrSO4(s)     →  Sr+2(aq)    +   SO4-2(aq)

 c.  write the expression:  Ksp = [Sr+2][ SO4-2]

2.  Write the expression for the solubility product constant for barium nitrate.

a.  write the formula:  Ba(NO3)2

b.  write the ionization:  Ba(NO3)2(s)     →  Ba+2(aq)    +   2NO3-1(aq)

c.  write the expression:  Ksp = [Ba+2][NO3-]2

3.  Write the expression for the solubility product constant for aluminum sulfate.

a.  write the formula:  Al2(SO4)3

b.  write the ionization:  Al2(SO4)3(s)     →  2Al+3(aq)    +   3SO4-2(aq)

 c.  write the expression:  Ksp = [Al+3]2[SO4-2]3

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                                            Calculating Ksp's

In order to calculate the Ksp for an ionic compound you need the equation for the dissolving process so the equilibrium expression can be written. 

You also need the concentrations of each ion expressed in terms of molarity, or moles per liter, or the means to obtain these values.

Easy calculations:

At 25oC, the concentration of Ag+ ions in a saturated solution of AgBr is 7.07 x 10-7 M.  What is the value of Ksp for AgBr?

Write the expression for the solubility product constant for silver bromide.

a.  write the formula:  AgBr

 b.  write the ionization:  AgBr(s)     →  Ag+(aq)    +   Br-(aq)

c.  write the expression:  Ksp = [Ag+][ Br-]

AgBr (s) 
Ag+(aq)
Br-(aq)
Initial Concentration
All solid
0
0
Change in Concentration
- 0.000000707 M (dissolves)
+0.000000707 M
+0.000000707  M
Equilibrium Concentration
Less solid
0.000000707M
0.000000707M

Calculate Ksp = [0.000000707M][0.000000707M]

                Ksp=[0.000000707M]2

                Ksp= 4.9985 x 10 -13

 Easy calculations:

At 25oC, the concentration of Sr+2 ions in a saturated solution of Sr(OH)2 is 4.31 x 10-2 M.  What is the value of Ksp for Sr(OH)2?

Write the expression for the solubility product constant for strontium hydroxide.

a.  write the formula:  Sr(OH)2

b.  write the ionization:  Sr(OH)2(s)     →  Sr+2(aq)    +   2OH-(aq)

c.  write the expression:  Ksp = [Sr+2][OH-]2

 

Sr(OH)2 (s) 
Sr+2(aq)
2OH-(aq)
Initial Concentration
All solid
0
0
Change in Concentration
- 0.0431M (dissolves)
   +0.0431M
+0.0862M
Equilibrium Concentration
Less solid
 0.0431M
0.08621M
see the molar ratio above

Calculate Ksp = [0.0431][0.08621]2

                Ksp=[0.0431][[0.00743]

                Ksp= 3.20 x 10 -4

Easy calculations starting with mass:

 

Example:  Calculate the solubility product constant for lead(II) chloride,

if 50.0 mL of a saturated solution of lead(II) chloride was found to contain

0.2207 g of lead(II) chloride dissolved in it.

  • First, write the equation for the dissolving of lead(II) chloride and the equilibrium expression for the dissolving process.
PbCl2(s) --> Pb2+(aq) + 2 Cl-(aq)

Ksp = [Pb2+][Cl-]2

  • Second, convert the amount of dissolved lead(II) chloride into moles per liter.
(0.2207 g PbCl2)(1/50.0 mL solution)(1000 mL/1 L)(1 mol PbCl2/278.1 g PbCl2) = 0.0159 M PbCl2
  • Third, create an "ICE" table.

PbCl2 (s) 
Pb2+(aq)
2Cl-(aq)
Initial Concentration
All solid
0
0
Change in Concentration
- 0.0159 M (dissolves)
+ 0.0159 M
+ 0.0318 M
Equilibrium Concentration
Less solid
0.0159 M
0.0318 M

  • Fourth, substitute the equilibrium concentrations into the equilibrium expression and solve for Ksp.
Ksp = [0.0159][0.0318]2 = 1.61 x 10-5

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                     Ionic Compound in Pure Water

 Video: Calculating Ksp

Example: Estimate the solubility of Ag2CrO4 in pure water if the solubility product constant for silver chromate is 1.1 x 10-12.

  • Write the equation and the equilibrium expression.
Ag2CrO4(s) --> 2 Ag+(aq) + CrO42-(aq)

Ksp = [Ag+]2[CrO42-]

  • Make an "ICE" chart.
Let "x" be the number of moles of silver chromate that dissolves in every liter of solution (its solubility).
Ag2CrO4(s)
Ag+(aq)
CrO42-(aq)
Initial Concentration
All solid
0
0
Change in Concentration
- x dissolves
+ 2 x
+ x
Equilibrium Concentraion
Less solid
2 x
x
  • Substitute the equilibrium amounts and the Ksp into the equilibrium expression and solve for x.
1.1 x 10-12 = [2x]2[x]
x = 6.50 x 10-5 M
 
[Ag+] = 2 x 6.50 x 10-5 M = 1.30 x 10-4 M
 
[CrO42-] = 6.50 x 10-5 M
 
Little more difficult problem:
 
What are the equilibrium concentrations of the dissolved ions in a saturated solution of Ca3(PO4)2, Ksp = 2.0 x 10-29?
 
  • Write the equation and the equilibrium expression.
Ca3(PO4)2(s) --> 3Ca+2(aq) + 2PO4-3(aq)

Ksp = [Ca+2]3[PO4-3]2

  • Make an "ICE" chart.
Let "x" be the number of moles of silver chromate that dissolves in every liter of solution (its solubility).
Ca3(PO4)2(s)
3Ca+2(aq)
2PO4-3(aq)
Initial Concentration
All solid
0
0
Change in Concentration
- x dissolves
+ 3x
+ 2x
Equilibrium Concentraion
Less solid
3x
2x
  • Substitute the equilibrium amounts and the Ksp into the equilibrium expression and solve for x.
2.0 x 10-29 = [3x]3[2x]2
 
2.0 x 10-29 = [27x3][4x2]
2.0 x 10-29 = 108x5

x = 7.13x 10-7M
 
[Ca+2-] = 3 x 7.13x 10-7M = 2.14x 10-6M
 
[PO4-3] = 2 x 7.13x 10-7M = 1.42 x 10-6 M
  

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Calculating the Solubility of an Ionic Compound in a Solution that Contains a Common Ion

The solubility of an ionic compound decreases in the presence of a common ion.

A common ion is any ion in the solution that is common to the ionic compound being dissolved.

For example, the chloride ion in a sodium chloride solution is common to the chloride in lead(II) chloride.

The presence of a common ion must be taken into account when determining the solubility of an ionic compound.

To do this, simply use the concentration of the common ion as the initial concentration.

Example: Estimate the solubility of barium sulfate in a 0.020 M sodium sulfate solution. The solubility product constant for barium sulfate is 1.1 x 10-10.

  • Write the equation and the equilibrium expression for the dissolving of barium sulfate.
BaSO4(s) --> Ba2+(aq) + SO42-(aq)

Ksp = [Ba2+][SO42-]

  • Make an "ICE" chart.
Let "x" represent the barium sulfate that dissolves in the sodium sulfate solution expressed in moles per liter.


 
BaSO4(s)
Ba2+(aq)
SO42-(aq)
Initial Concentration
All solid
0
0.020 M (from Na2SO4)
Change in Concentration
 - x dissolves
+ x
+ x
Equilibrium Concentration
Less solid
x
0.020 M + x

  • Substitute into the equilibrium expression and solve for x. 

  •  We will make the assumption that since x is going to be very small (the solubility is reduced in the presence of a common ion), the term "0.020 + x" is the same as "0.020." 

  • (You can leave x in the term and use the quadratic equation or the method of successive approximations to solve for x, but it will not improve the significance of your answer.)
 1.1 x 10-10 = [x][0.020 + x] = [x][0.020]
 
x = 5.5 x 10-9
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Determining Whether a Precipitate will, or will not Form When Two Solutions are Combined

 Ksp video

When two electrolytic solutions are combined, a precipitate may, or may not form.  In order to determine whether or not a precipitate will form or not, one must examine two factors. 

First, determine the possible combinations of ions that could result when the two solutions are combined to see if any of them are deemed "insoluble" base on solubility tables (Ksp tables will also do).  

Second, determine if the concentrations of the ions are great enough so that the reaction quotient Q exceeds the Ksp value. 

One important factor to remember is there is a dilution of all species present and must be taken into account.

Easy calculations starting with mass:

Example: A solution is prepared by dissolving 0.003 mol of ferrous oxalate in 5 liters of hot water, Ksp = 2.1 x 10-7 (The ionization of the compound is written as follows: 

FeC2O4(s) -->Fe2+(aq) + C2O42-(aq)
 
Ksp = [Fe2+][C2O42-]
 
find molarity: 0.003mol/5 liters =  6.4 x 10-4
 
this represents both [Fe2+] and [C2O42-]
 
Ksp = [6.4 x 10-4][6.4 x 10-4]
or
Ksp = [6.4 x 10-4]2
 
Ksp = 4.0 x 10-7

 

This value is greater than the original Ksp,  therefore, the substance will precipitate.

Example:

Consider the following reaction:

Determine if a precipitate will form if you mix 50.0 mls of 1.0 x 10-4M  lead (II) nitrate with 20 mls of 2.0 x 10-5M of iron III iodide.

1.  write the balanced equation: 

 3Pb(NO3)2 + 2FeI3 → 3PbI2  +  2Fe(NO3)3

2.  write the ionic equation:

3Pb+2(aq)  + 6NO3-(aq) +   2Fe+3(aq) +  6I(aq)-

3PbI2(s)  +  2Fe+3(aq) +  6NO3-(aq)

3.  write the net ionic equation: This will be a precipitate and has a Ksp value.

3Pb+2(aq)  + 6I-(aq)  → 3PbI2(s)

4.  show the ionizstion of silver iodide which is the reverse of the net  ionic

 equation.

3PbI2(s)    → 3Pb+2(aq)  + 6I-(aq)

5.  Ksp is based on 1 mole of the insoluble compound so rewrite the net ionic equation in terms of 1 mole of PbI2(s):

PbI2(s) → Pb+2(aq)  + 2I-(aq)

6.  determine the Ksp value from the Table of Ksp values:   

Ksp = 8.7 x 10-9

7.  find moles of each species from the initial concentrations:

50.0 mls of 1 x 10-4M  lead (II) nitrate = 0.05 liter x 0.0001 mol/liter =

5 x 10-6 mol

20 mls of 2 x 10-5M of iron III iodide  = 0.02 liter x 0.00002 mol/liter =

4 x 10-7 mol

8.  adjust for new total volume which is 70 total mls.

[Pb+2]  = 5 x 10-6 mol/0.070 L = 7.1 x 10-5M

[I-] = 4 x 10-7 mol/0.07 =  5.7  x 10-6M

9.  write the Ksp expression:   Ksp = [Pb+2] [I-]2

10. insert new molarities and solve for Ksp, do not use the molar ration for

 adjust the amount of iodide ion.

Q = [Pb+2] [I-]=  [7.1 x 10-5] [5.7  x 10-6]

=  [7.1 x 10-5] [3.25 x 10-11] = 2.3 x 10-15

10.  compare Q to Ksp

Ksp = 8.7 x 10-9 and Q = 2.3 x 10-15 

Q is much small than the Ksp, therefore, no precipitate will form.

 ok, hopefully this will help you to understand better.

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PRACTICE PROBLEMS

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A little more difficult problem
 
if 50.0 mL of a saturated solution of lead(II) chloride was found to contain 0.2207 g of lead(II) chloride dissolved in it.
  • First, write the equation for the dissolving of lead(II) chloride and the equilibrium expression for the dissolving process.
PbCl2(s) --> Pb2+(aq) + 2 Cl-(aq)

Ksp = [Pb2+][Cl-]2

  • Second, convert the amount of dissolved lead(II) chloride into moles per liter.
(0.2207 g PbCl2)(1/50.0 mL solution)(1000 mL/1 L)(1 mol PbCl2/278.1 g PbCl2) = 0.0159 M PbCl2
  • Third, create an "ICE" table.

PbCl2 (s) 
Pb2+(aq)
2Cl-(aq)
Initial Concentration
All solid
0
0
Change in Concentration
- 0.0159 M (dissolves)
+ 0.0159 M
+ 0.0318 M
Equilibrium Concentration
Less solid
0.0159 M
0.0318 M

  • Fourth, substitute the equilibrium concentrations into the equilibrium expression and solve for Ksp.
Ksp = [0.0159][0.0318]2 = 1.61 x 10-5

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Example:  25.0 mL of 0.0020 M potassium chromate are mixed with 75.0 mL of 0.000125 M lead(II) nitrate.  Will a precipitate of lead(II) chromate form.  Ksp of lead(II) chromate is 1.8 x 10-14.

  • First, determine the overall and the net-ionic equations for the reaction that occurs when the two soltutions are mixed.
K2CrO4(aq) + Pb(NO3)2(aq) --> 2 KNO3(aq) + PbCrO4(s)

Pb2+(aq) + CrO42-(aq) --> PbCrO4(s)
The latter reaction can be written in terms of Ksp as:
 
PbCrO4(s) --> Pb2+(aq) + CrO42-(aq)
 
Ksp = [Pb2+][CrO42-]
  • Using the dilution equation, C1V1 = C2V2, determine the initial concentration of each species once mixed (before any reaction takes place).
(0.0020 M K2CrO4)(25.0 mL) = (C2)(100.0 mL)
 
C2 for K2CrO4 = 0.00050 M

Similar calculation for the lead(II) nitrate yields:
C2 for Pb(NO3)2 = 0.0000938 M
  • Using the initial concentrations, calculate the reaction quotient Q, and compare to the value of the equilibrium constant, Ksp.
Q = (0.0000938 M Pb2+)(0.00050 M CrO42-) = 4.69 x 10-8

Q is greater than Ksp so a precipitate of lead(II) chromate will form.

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                                General Solubility

print me.

1. In your own words, what is the relationship shown on this solubility curve for each of the three substances?

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2. Notice the solubility graph for sodium chloride (NaCl) is almost horizontal. What does this tell us about the solubility of the substance as temperature increases?

___________________________________

3. At 50° C, how much potassium nitrate (KNO3) will dissolve in 100 grams of water to form a saturated solution?

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4. Consider a solution containing 80 grams of KNO3 in 100 grams of water at 60° C. Locate this point on the graph. What does this tell you about the level of saturation of this solution? 

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5. Considering the information from Question 4, what would happen if you cooled this solution to 40° C?

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6. What is the solubility of KNO3 at 70° C?

___________________________________

7. At what temperature will the solubility of KNO3 be 25 g per 100 g of water?

 ___________________________________

8. How much KNO3 will dissolve in 150 g of water at 40° C?

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9. What mass (in grams) of KNO3 will dissolve in 25 g of water at 60° C?

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 10. What mass of KCl will dissolve in 200g of water at 60° C?

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               Writing Solubility Product Expressions

Here are some practice problems for writing Ksp expressions. Write the chemical equation showing how the substance dissociates and write the Ksp expression:

1) AlPO4


2) BaSO4

3) CdS


4) Cu3(PO4)2


5) CuSCN


6) Hg2Br2


7) AgCN


8) Zn3(AsO4)2

9) Mn(IO3)2


10) PbBr2


11) SrCO3


12) Bi2S3

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                  Writing Solubility Expressions Answers

Dissociation Equation Ksp expression
AlPO4 <===> Al3+ (aq) + PO43¯ (aq)

Ksp = [Al3+] [PO43¯]

BaSO4 <===> Ba2+ (aq) + SO42¯ (aq)

Ksp = [Ba2+] [SO42¯]

CdS <===> Cd2+ (aq) + S2¯ (aq)

Ksp = [Cd2+] [S2¯]

Cu3(PO4)2 <===> 3 Cu2+ (aq) + 2 PO43¯ (aq)

Ksp = [Cu2+]3 [PO43¯]2

 

CuSCN <===> Cu+ (aq) + SCN¯ (aq)

Ksp = [Cu+] [SCN¯]

Hg2Br2 <===> Hg22+ (aq) + 2 Br¯ (aq)

Ksp = [Hg22+] [Br¯]2

AgCN <===> Ag+ (aq) + CN¯ (aq)

Ksp = [Ag+] [CN¯]

Zn3(AsO4)2 <===> 3 Zn2+ (aq) + 2 AsO43¯ (aq)

Ksp = [Zn2+]3 [AsO43¯]2

 

Mn(IO3)2 <===> Mn2+ (aq) + 2 IO3¯ (aq)

Ksp = [Mn2+] [IO3¯]2

PbBr2 <===> Pb2+ (aq) + 2 Br¯ (aq)

Ksp = [Pb2+] [Br¯]2

SrCO3 <===> Sr2+ (aq) + CO32¯ (aq)

Ksp = [Sr2+] [CO32¯]

 

Bi2S3 <===> 2 Bi3+ (aq) + 3 S2¯ (aq) Ksp = [Bi3+]2 [S2¯]3

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                                          Ksp

1.  Write the solubility product constant expression for these equations:

PbI2 (s) Pb+2 (aq) + 2I -1 (aq)

Cu3(PO4)2 (s) 3Cu+2 (aq) + 2PO4 -3 (aq)

 

2.  A saturated solution of PbI2 has a lead ion concentration of 1.21X10 -3M. What is the Ksp for PbI2 ?

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Solubility Product Worksheet

What is the concentration of a saturated silver (I) acetate solution? 

Ksp(AgC2H3O2) = 1.94 x 10-3.

 What is the concentration of a saturated lead chloride solution? 

Ksp(PbCl2) = 1.17 x 10-5.

I have discovered a new chemical compound with the formula A2B.  If a saturated solution of A2B has a concentration of 4.35 x 10-4 M, what is the solubility product constant for A2B?

Solubility product constants are usually specified for 250 C.  Why does the Ksp value for a chemical compound depend on the temperature?

The Ksp for nickel (II) hydroxide is 5.47 x 10-16.  What is the base dissociation constant for nickel (II) hydroxide?

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           Solubility Product Worksheet - Answers

 

 What is the concentration of a saturated silver (I) acetate solution?  Ksp(AgC2H3O2) = 1.94 x 10-3.

Since Ksp = [Ag+][C2H3O2-], and the concentration of silver ions is the same as the concentration of acetate ions, we can set up the following equation:

1.94 x 10-3 = x2

x = 0.0440 M

What is the concentration of a saturated lead chloride solution?   Ksp(PbCl2) = 1.17 x 10-5.

Ksp = [Pb+2][Cl-]2

Since the concentration of chloride ions is twice that of lead (II) ions, this boils down to the following equation:

1.17 x 10-5 = (x)(2x)2

1.17 x 10-5 = 4x3

x = 0.0143 M

I have discovered a new chemical compound with the formula A2B.   If a saturated solution of A2B has a concentration of 4.35 x 10-4 M,

what is the solubility product constant for A2B?

Ksp = [A+]2[B2-].  Since the concentration of A is twice that of B, and the concentration of B is 4.35 x 10-4 M, we can set up the following equation:

Ksp = [2(4.35 x 10-4 M)]2 [4.35 x 10-4 M]

Ksp­ = 3.29 x 10-10

Solubility product constants are usually specified for 250 C. 

Why does the Ksp value for a chemical compound depend on the temperature?

Ksp depends on temperature because solubility depends on temperature.

Generally, solids become more soluble as the temperature of the solution increases. 

As a result, Ksp values of solids tend to increase as the temperature increases.

The Ksp for nickel (II) hydroxide is 5.47 x 10-16

What is the base dissociation constant for nickel (II) hydroxide? 

5.47 x 10-16

Because nickel (II) hydroxide dissociates to become a base, the Ksp and Kb values are identical.

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                          Solubility Curves Practice

PRACTICE:

1. You dissolve 25 g of KNO3 in 100 g of water at 30° C producing an     unsaturated solution.

a. How much more KNO3 do you need to add in order for the solution to be saturated?

b. What is the minimum mass of 30° C water to dissolve 25 g of KNO3?

2. A supersaturated solution of KNO3 is formed by adding 150 g of KNO3 to 100 g of water, heating until the solute completely dissolves, and then cooling the solution to 55° C.

a. If the solution is agitated, how much KNO3 will precipitate out?

b. How much 55° C water would have to be added (to the original 100 g) to dissolve all of the KNO3?

3.  Use the solubility curve graph to determine how many grams of sodium nitrate can be dissolved in 100 grams of water at 80°C?

 

4.  Use the solubility curve graph to determine how many grams of sodium nitrate can be dissolved in 235 grams of water at 30˚C

 

5.  Use the solubility curve graph to determine if a solution is unsaturated, saturated, or supersaturated, when 80 grams of KI are dissolved in 100 grams of water at 20°C.

 

 

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                                Molarity Practice

Molarity Practice:

1.  How many grams of CaCl2 would be required to produce a 3.5 M (molar) solution with a volume of 2.0 L?

 

2.  What is the molarity of a 5.00 x 102 ml solution containing 249 g of KI?

 

3.  How many moles of LiF would be required to produce a 2.5 M solution with a volume of 1.5 L?

 Answers     1)  780 g                                           2) 3.00 M                              3)   3.75 moles

 

 

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                            Mole Fraction Practice

1.  What is the mole fraction of xenon in a mixture of 0.584 grams xenon, 8.6 grams of argon, and 3.2 grams of neon?

 2.  How many grams of NaHCO3 are there in 150 ml of a 0.30 M solution?

3.  What is the mass percentage of a solution that is 0.3 moles of KBr dissolved in 750 mL of aqueous solution?  (Assume the aqueous solution has the same density as water)

 4.  What is the mole fraction of oxygen in a mixture that contains 66.8 grams of oxygen, 41 grams of nitrogen, and 21.5 grams of hydrogen?

 5.  What is the mole fraction of carbon dioxide in a solution that contains 15 grams of carbon dioxide dissolved in 54 grams of water?

 6.  Calculate the mole fraction of silver in an alloy that is 5.6 grams of silver and 10.2 grams of copper.

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                              Molality Practice

1.  In order to prepare a 0.523 m aqueous solution of potassium iodide, how many grams of potassium iodide must be added to 2.00 kg of water?

 

2.   What is the molality of a solution made by dissolving 8.11 grams of potassium sulfide in 47.6 grams of ethanol?

3.  What is mass percent of a solution made by dissolving 4.5 grams of calcium chloride in 27 grams of water?

 

4.  What is the molality of a solution that contains 15.2 grams of calcium chloride dissolved in 345 grams of methanol?

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                         Part per Million practice

1.   What is the concentration in parts per million of mercury in drinking water if there are 0.002 grams of mercury dissolved in 300 kg of water?

2.   A 20 L sample of drinking water contained 0.03 grams of lead.  Calculate the concentration of lead in parts per million. 

 

3.   What is the concentration in parts per million of a solution that is 0.03 moles of magnesium bromide dissolved in 700 grams of solution?

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                              Dilution Practice

 

When a solution is diluted:   M1 V1 = M2 V2

1.  If you have 250 ml of 0.57 M NaCl and add 430 ml of water what is the molarity of the new solution?

 

2.  How much 1.5 M solution is needed to make 500 ml of 0.35 M solution?

 

3.  If you have 360 ml of 0.87 M NaOH solution and want to dilute it to a concentration of 0.35 M, how much water do you need to add?

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                        Colligative Properties Practice

1.  What happens to the solubility of an ionic solid in water as temperature is increased?  What happens to the solubility of a gas as temperature is increased?

 2.  What is an alloy?  Give two examples of alloys.  Why are alloys used?

3.  Calculate the molality (m) of a solution made by dissolving 25 grams of magnesium chloride in 850 grams of water.

4.  How much will the boiling point of acetic acid increase when 2.69 grams of picolinic acid (C6H5N2) is dissolved in 25 grams of acetic acid.  Kb of acetic acid is 2.93 ˚C/m.

5.  How many grams of sucrose (C12H22O11) need to be dissolved in 600 grams of water to increase the boiling point by 3.5˚C.  Kb of water is 0.52˚C/m.

6.  4.5 grams of an unknown solute are dissolved in 750 grams of water and the freezing point of the water is decreased by 0.88˚C.  Kf of water is 1.86˚C/m.  What is the molar mass of the unknown solute?

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                                       HEATING CURVES PRACTICE

Sample Exercise 1

Draw and interpret a graph for the conversion of methanol at –120°C to vapour at 120°C. The melting and boiling points of methanol are -97.7°C and 64.0°C.

Answer

Plan a Strategy.

  1. The initial temperature is low and the final temperature is high, so a heating curve should be drawn.
  2. The increments on the y-axis should begin at -60 and end at 120 with equal divisions in between.
  3. Draw horizontal lines to indicate the melting and boiling points.
  4. Beginning at -120°C, draw lines to show the warming, melting, warming evaporation and warming of methanol.
  5. Interpret each of the five regions of the curve.

Steps 1-4: Construct the heating curve.

Step 5: Interpret the curve.

The kinetic energy of the solid methanol increases as energy is added. At -97.7°C, the intermolecular forces (London forces, dipole-dipole forces, and hydrogen bonds) weaken and solid methanol becomes liquid methanol. Liquid methanol gains kinetic energy and warms to the boiling point. At 64°C, the intermolecular forces are completely overcome as molecules pass from the liquid state to the vapour state. The vapour then gains kinetic energy as it warms to 120°C.

Important Notes:

  1. The manipulated variable (energy added) is on the x-axis and temperature, which is the responding variable, is located on the y-axis.
  2. Pressure is assumed to be constant at all times.
  3. Total energy increases as energy enters the system from the surroundings. Either kinetic energy of potential energy increases, but not at the expense of the other.
  4. The type of attractive force weakened or broken during the phase changes is a function of the nature of the system.
    • Molecular substances in the solid and liquid states possess intermolecular forces like van der Waals forces and hydrogen bonding.
    • Metal atoms are held together as a solid or a liquid by metallic bonds.
    • Ionic solids and liquids are held together by ionic bonds.
  5. Some substances sublimate - they pass from the solid phase to the gas phase without passing through the liquid phase. There is only one phase change in the heating curve for these systems. For example, a heating curve for iodine sublimating at 70°C under normal pressure conditions would show one flat region.

Calculating the Total Heat Change for a System

Given molar heats of phase change and specific heat capacity data, you can calculate the total heat change for a system using  and .

Sample Exercise 2

Calculate the total heat change for a 35.7 g ice cube at -25°C that is heated to become water vapour at 125°C.

Answer

Plan a strategy.

  1. Sketch a heating curve.
  2. Look-up the reference data.
  3. Use to calculate heating changes associated with temperature changes and to calculate the heat changes associated with phase changes.
  4. Add the heat values for each region of the curve.
  5. Round-off and communicate the final answer.

Step 1: Sketch a curve

Step 2: Locate and list the reference data.

  • cice = 2.01 J/g°C (honors only)
  • cwater = 4.184 J/g°C
  • ΔHfus = +6.03 kJ/mol or 334 J/g
  • ΔHvap = +40.8 kJ/mol  or 2260 J/g
  • REMEMBER: THE NEW YORK STATE REGENTS EXAM WILL NOT REQUIRE CALCULATIONS IN kJ/mol.

Step 3: Calculate the individual heat changes.

The warming of ice:

The melting of ice:

The warming of water:

The evaporation of water:

The warming of steam:

Step 3: Add the q values.

Step 4: Round-off and communicate the answer.

The total heat change for the ice cube is 111 kJ.

Total heat change calculations may involve as many has five steps; however some may involve fewer steps. You need to carefully consider the given information and the melting/boiling point data. Sketching the heating (or cooling) curve will help you determine the number of steps.

 

                            FORCED PRECIPITATION PRACTICE PROBLEM

MgF2(s)    => Mg2+(aq) + 2 F-(aq)

In a saturated solution of MgF2 at 18ºC, the concentration of Mg2+ is 1.21 x 10-3 molar. The equilibrium is represented by the equation above.

(a)     Write the expression for the solubility-product constant, Ksp, and calculate its value at 18ºC.

(b)     Calculate the equilibrium concentration of Mg2+ in 1.000 liter of saturated MgF2 solution at 18ºC to which 0.100 mole of solid KF has been added. The KF dissolves completely. Assume the volume change is negligible.  Hint:  you do this problem as if the KF were there from the beginning.  Therefore, you simply plug in a known value for [F-] into your Ksp expression and solve for [Mg2+].

(c)     Predict whether a precipitate of MgF2 will form when 100.0 milliliters of a 3.00 x 10-3 molar Mg(NO3)2 solution is mixed with 200.0 milliliters of a 2.00 x l0-3 molar NaF solution at 18ºC. Calculations to support your prediction must be shown. 

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