 
 
 _{ Colligative Properties}  
 
_{} _{Molarity} _{} 
http://www.chalkbored.com/lessons/chemistry11/intermolecularforcescrossword.pdf
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 =  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:
1  x  (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 =  5.00 L of soln  = 0.025 mol/L 
So the molarity (M) of the solution is 0.025 mol/L.
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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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. ____________________________
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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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 semipermeable 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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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 icewater 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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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 boilingpoint elevation can be calculated by applying ClausiusClapeyron relation and Raoult's law together with the assumption of the nonvolatility of the solute.
The result is that in dilute ideal solutions, the extent of boilingpoint elevation is directly proportional to the molal concentration of the solution according to the equation:
ΔT_{b} = K_{b} · m_{B}
where
ΔT_{b}, the boiling point elevation, is defined as T_{b (solution)}  T_{b (pure solvent)}.
K_{b}, the ebullioscopic constant, which is dependent on the properties of the solvent.
m_{B} 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
m_{B} = m_{solute} · 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 CaCl_{2} into Ca^{2+} and 2Cl^{}
Values of the ebullioscopic constants K_{b} for selected solvents:
Compound  Boiling point in °C  Ebullioscopic constant K_{b} 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:
Freezing Point of solution = normal freezing point of solvent + ΔT_{f}
solvent  normal boiling point (^{o}C)  K_{b} (^{o}Cm^{1})  normal freezing point (^{o}C)  K_{f} (^{o}Cm^{1}) 

benzene  80.2  2.53  5.5  5.12 
water  100.0  0.512  0.000  1.855 
acetic acid (ethanoic acid)  118.5  3.07  16.6  3.90 
camphor  208.3  5.95  178.4  40.0 
naphthalene  218.0  5.65  80.2  6.9 
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(ii) Molality and solubity are related by the following relation.
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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.
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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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Bromides  PbBr_{2}  6.3 x 10^{6} 
AgBr  3.3 x 10^{13}  
Carbonates  BaCO_{3}  8.1 x 10^{9} 
CaCO_{3}  3.8 x 10^{9}  
CoCO_{3}  8.0 x 10^{13}  
CuCO_{3}  2.5 x 10^{10}  
FeCO_{3}  3.5 x 10^{11}  
PbCO_{3}  1.5 x 10^{13}  
MgCO_{3}  4.0 x 10^{5}  
MnCO_{3}  1.8 x 10^{11}  
NiCO_{3}  6.6 x 10^{9}  
Ag_{2}CO_{3}  8.1 x 10^{12}  
ZnCO_{3}  1.5 x 10^{11}  
Chlorides  PbCl_{2}  1.7 x 10^{5} 
AgCl  1.8 x 10^{10}  
Chromates  BaCrO_{4}  2.0 x 10^{10} 
CaCrO_{4}  7.1 x 10^{4}  
PbCrO_{4}  1.8 x 10^{14}  
Ag_{2}CrO_{4}  9.0 x 10^{12}  
Cyanides  Ni(CN)_{2}  3.0 x 10^{23} 
AgCN  1.2 x 10^{16}  
Zn(CN)_{2}  8.0 x 10^{12}  
Fluorides  BaF_{2}  1.7 x 10^{6} 
CaF_{2}  3.9 x 10^{11}  
PbF_{2}  3.7 x 10^{8}  
MgF_{2}  6.4 x 10^{9}  
Hydroxides  AgOH  2.0 x 10^{8} 
Al(OH)_{3}  1.9 x 10^{33}  
Ca(OH)_{2}  7.9 x 10^{6}  
Cr(OH)_{3}  6.7 x 10^{31}  
Co(OH)_{2}  2.5 x 10^{16}  
Cu(OH)_{2}  1.6 x 10^{19}  
Fe(OH)_{2}  7.9 x 10^{15}  
Fe(OH)_{3}  6.3 x 10^{38}  
Pb(OH)_{2}  2.8 x 10^{16}  
Mg(OH)_{2}  1.5 x 10^{11}  
Mn(OH)_{2}  4.6 x 10^{14}  
Ni(OH)_{2}  2.8 x 10^{16}  
Zn(OH)_{2}  4.5 x 10^{17}  
Iodides  PbI_{2}  8.7 x 10^{9} 
AgI  1.5 x 10^{16}  
Oxalates  BaC_{2}O_{4}  1.1 x 10^{7} 
CaC_{2}O_{4}  2.3 x 10^{9}  
MgC_{2}O_{4}  8.6 x 10^{5}  
Phosphates  AlP0_{4}  1.3 x 10^{20} 
Ba_{3}(P0_{4})_{2}  1.3 x 10^{29}  
Ca_{3}(P0_{4})_{2}  1.0 x 10^{25}  
CrP0_{4}  2.4 x 10^{23}  
Pb_{3}(P0_{4})_{2}  3.0 x 10^{44}  
Ag_{3}P0_{4}  1.3 x 10^{20}  
Zn_{3}(P0_{4})_{2}  9.1 x 10^{33}  
Sulfates  BaS0_{4}  1.1 x 10^{10} 
CaS0_{4}  2.4 x 10^{5}  
PbS0_{4}  1.8 x 10^{8}  
Ag_{2}S0_{4}  1.7 x 10^{5}  
Sulfides  CaS  8 x 10^{6} 
CoS  5.9 x 10^{21}  
CuS  7.9 x 10^{37}  
FeS  4.9 x 10^{18}  
Fe_{2}S_{3}  1.4 x 10^{88}  
PbS  3.2 x 10^{28}  
MnS  5.1 x 10^{15}  
NiS  3.0 x 10^{21}  
Ag_{2}S  1.0 x 10^{49}  
ZnS  2.0 x 10^{25}  
Sulfites  BaS0_{3}  8.0 x 10^{7} 
CaS0_{3}  1.3 x 10^{8}  
Ag_{2}S0_{3}  1.5 x 10^{14} 
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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 X_{i} is the fraction of moles of component i of the total number of moles of all components in solution.
moles of component i  
X_{i}  =   
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 H_{2}O  
25.0 g H_{2}O  x    =  1.39 mole H_{2}O 
18.02 g H_{2}O 
Substitute these two quantities into the defining equation for mole fraction.
0.0611 mol NaCl  
X_{NaCl}  =    =  0.0421 
(0.0611 + 1.39) 
1.39 mole H_{2}O  
X_{water}  =    =  0.958 
(0.0611 + 1.39) 
Note that within the limits of the significant figures that X_{NaCl} + X_{water} = 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
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 (18301901).
His results are described by the equation known as Raoult's law:
P_{solution} = X_{solvent}P^{o}_{solvent}
where P_{solution} is the observed vapor pressure of the solution, X_{solvent} is the mole fraction of solvent, and P^{o}_{solvent }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 solutesolvent 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 liquidliquid solutions with both components being volatile, a modified form of Raoult's law applies:
P_{total} = P_{a} + P_{b} = X_{a}P^{o}_{a} + X_{b}P^{o}_{b}
where P_{total} represents the total vapor pressure of a solution containing A and B. X_{a} and X_{b} are the mole fractions of A and B. P^{o}_{a} and P^{o}_{b} are the vapor
pressures of pure A and B, and P_{a} and P_{b} are the partial pressures resulting from molecules of A and B in the vapor above the solution.
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Solubility Product Constants, K_{sp}
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.
The general equilibrium constant for such processes can be written as:
K_{c} = [M^{y+}]^{x}[A^{x}]^{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 K_{sp}. Solubility product constants can be calculated, and used in a variety of applications.
for an excellent explaination of solubility try the link below:
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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: SrSO_{4}
b. write the ionization: SrSO_{4(s) }→ Sr^{+2}_{(aq)} + SO_{4}^{2}_{(aq)}
c. write the expression: K_{sp} = [Sr^{+2}][ SO_{4}^{2}]
2. Write the expression for the solubility product constant for barium nitrate.
a. write the formula: Ba(NO_{3})_{2}
b. write the ionization: Ba(NO_{3})_{2(s) }→ Ba^{+2}_{(aq)} + 2NO_{3}^{1}_{(aq)}
c. write the expression: K_{sp} = [Ba^{+2}][NO_{3}^{}]^{2}
3. Write the expression for the solubility product constant for aluminum sulfate.
a. write the formula: Al_{2}(SO_{4})_{3}
b. write the ionization: Al_{2}(SO_{4})_{3(s) }→ 2Al^{+3}_{(aq)} + 3SO_{4}^{2}_{(aq)}
c. write the expression: K_{sp} = [Al^{+3}]^{2}[SO_{4}^{2}]^{3}
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In order to calculate the K_{sp} 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 25^{o}C, the concentration of Ag^{+} ions in a saturated solution of AgBr is 7.07 x 10^{7} M. What is the value of K_{sp} 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: K_{sp} = [Ag^{+}][ Br^{}]
Initial Concentration  
Change in Concentration  +0.000000707 M  
Equilibrium Concentration 
Calculate K_{sp} = [0.000000707M][0.000000707M]
K_{sp}=[0.000000707M]^{2}
K_{sp}= 4.9985 x 10^{ 13}
Easy calculations:
At 25^{o}C, 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 K_{sp} 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: K_{sp} = [Sr^{+2}][OH^{}]^{2}
Initial Concentration  
Change in Concentration  +0.0431M  
Equilibrium Concentration 
Calculate K_{sp} = [0.0431][0.08621]^{2}
K_{sp}=[0.0431][[0.00743]
K_{sp}= 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.
K_{sp} = [Pb^{2+}][Cl^{}]^{2}
Initial Concentration  
Change in Concentration  
Equilibrium Concentration 
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Example: Estimate the solubility of Ag_{2}CrO_{4} in pure water if the solubility product constant for silver chromate is 1.1 x 10^{12}.
K_{sp} = [Ag^{+}]^{2}[CrO_{4}^{2}]
Initial Concentration  
Change in Concentration  
Equilibrium Concentraion 
K_{sp} = [Ca^{+2}]^{3}[PO_{4}^{3}]^{2}
Initial Concentration  
Change in Concentration  
Equilibrium Concentraion 
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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}.
Ksp = [Ba^{2+}][SO_{4}^{2}]
Initial Concentration  
Change in Concentration  
Equilibrium Concentration 
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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 (K_{sp} tables will also do).
Second, determine if the concentrations of the ions are great enough so that the reaction quotient Q exceeds the K_{sp} 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, K_{sp} = 2.1 x 10^{7} (The ionization of the compound is written as follows:
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^{4}M lead (II) nitrate with 20 mls of 2.0 x 10^{5}M of iron III iodide.
1. write the balanced equation:
3Pb(NO_{3})_{2} + 2FeI_{3} → 3PbI_{2} + 2Fe(NO_{3})_{3}
_{}
2. write the ionic equation:
3Pb^{+2}_{(aq)} + 6NO_{3}^{}_{(aq)} + 2Fe^{+3}_{(aq)} + 6I_{(aq)}^{} →
3PbI_{2(s)} + 2Fe^{+3}_{(aq)} + 6NO_{3}^{}_{(aq)}
_{}
_{}
3. write the net ionic equation: This will be a precipitate and has a Ksp value.
3Pb^{+2}_{(aq)} + 6I^{}_{(aq)}^{ } → 3PbI_{2(s)}
_{}
_{}
4. show the ionizstion of silver iodide which is the reverse of the net ionic
equation.
3PbI_{2(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 PbI_{2(s)}:
PbI_{2(s)} → Pb^{+2}_{(aq)} + 2I^{}_{(aq)}
_{}
_{}
6. determine the K_{sp} value from the Table of K_{sp }values:
K_{sp} = 8.7 x 10^{9}
^{}
7. find moles of each species from the initial concentrations:
50.0 mls of 1 x 10^{4}M lead (II) nitrate = 0.05 liter x 0.0001 mol/liter =
5 x 10^{6} mol
20 mls of 2 x 10^{5}M 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 105M
[I^{}] = 4 x 10^{7} mol/0.07 = 5.7 x 10^{6}M
9. write the K_{sp} expression: K_{sp} = [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^{}]^{2 }= [7.1 x 10^{5}] [5.7 x 10^{6}]^{2 }
=^{ } [7.1 x 10^{5}] [3.25 x 10^{11}] = 2.3 x 10^{15}
^{}
10. compare Q to Ksp
K_{sp} = 8.7 x 10^{9} and Q = 2.3 x 10^{15 }
Q is much small than the K_{sp}, therefore, no precipitate will form.
^{ }ok, hopefully this will help you to understand better.
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K_{sp} = [Pb^{2+}][Cl^{}]^{2}
Initial Concentration  
Change in Concentration  
Equilibrium Concentration 
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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. K_{sp} of lead(II) chromate is 1.8 x 10^{14}.
Q is greater than K_{sp} so a precipitate of lead(II) chromate will form.
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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?
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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?
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7. At what temperature will the solubility of KNO3 be 25 g per 100 g of water?
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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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Here are some practice problems for writing K_{sp} expressions. Write the chemical equation showing how the substance dissociates and write the K_{sp} expression:
1) AlPO_{4}
_{}
2) BaSO_{4}3) CdS
_{}
4) Cu_{3}(PO_{4})_{2}
5) CuSCN_{}
6) Hg_{2}Br_{2}
7) AgCN
8) Zn_{3}(AsO_{4})_{2}9) Mn(IO_{3})_{2}
_{}_{}
10) PbBr_{2}
11) SrCO_{3}
12) Bi_{2}S_{3}
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Dissociation Equation  K_{sp} expression 
AlPO_{4} <===> Al^{3+} (aq) + PO_{4}^{3}¯ (aq)  K_{sp} = [Al^{3+}] [PO_{4}^{3}¯] 
BaSO_{4} <===> Ba^{2+} (aq) + SO_{4}^{2}¯ (aq)  K_{sp} = [Ba^{2+}] [SO_{4}^{2}¯] 
CdS <===> Cd^{2+} (aq) + S^{2}¯ (aq)  K_{sp} = [Cd^{2+}] [S^{2}¯] 
Cu_{3}(PO_{4})_{2} <===> 3 Cu^{2+} (aq) + 2 PO_{4}^{3}¯ (aq)  K_{sp} = [Cu^{2+}]^{3} [PO_{4}^{3}¯]^{2}

CuSCN <===> Cu^{+} (aq) + SCN¯ (aq)  K_{sp} = [Cu^{+}] [SCN¯] 
Hg_{2}Br_{2} <===> Hg_{2}^{2+} (aq) + 2 Br¯ (aq)  K_{sp} = [Hg_{2}^{2+}] [Br¯]^{2} 
AgCN <===> Ag^{+} (aq) + CN¯ (aq)  K_{sp} = [Ag^{+}] [CN¯] 
Zn_{3}(AsO_{4})_{2} <===> 3 Zn^{2+} (aq) + 2 AsO_{4}^{3}¯ (aq)  K_{sp} = [Zn^{2+}]^{3} [AsO_{4}^{3}¯]^{2}

Mn(IO_{3})_{2} <===> Mn^{2+} (aq) + 2 IO_{3}¯ (aq)  K_{sp} = [Mn^{2+}] [IO_{3}¯]^{2} 
PbBr_{2} <===> Pb^{2+} (aq) + 2 Br¯ (aq)  K_{sp} = [Pb^{2+}] [Br¯]^{2} 
SrCO_{3} <===> Sr^{2+} (aq) + CO_{3}^{2}¯ (aq)  K_{sp} = [Sr^{2+}] [CO_{3}^{2}¯]

Bi_{2}S_{3} <===> 2 Bi^{3+} (aq) + 3 S^{2}¯ (aq)  K_{sp} = [Bi^{3+}]^{2} [S^{2}¯]^{3} 
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PbI_{2 (s)} Pb^{+2}_{ (aq)} + 2I ^{1}_{ (aq)}
Cu_{3}(PO_{4})_{2 (s)} 3Cu^{+2}_{ (aq)} + 2PO_{4} ^{3}_{ (aq)}
2. A saturated solution of PbI_{2 }has a lead ion concentration of 1.21X10 ^{3}M. What is the K_{sp} for PbI_{2 }?
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Solubility Product Worksheet
What is the concentration of a saturated silver (I) acetate solution?
K_{sp}(AgC_{2}H_{3}O_{2}) = 1.94 x 10^{3}.
What is the concentration of a saturated lead chloride solution?
K_{sp}(PbCl_{2}) = 1.17 x 10^{5}.
I have discovered a new chemical compound with the formula A_{2}B. If a saturated solution of A_{2}B has a concentration of 4.35 x 10^{4} M, what is the solubility product constant for A_{2}B?
Solubility product constants are usually specified for 25^{0} C. Why does the K_{sp} value for a chemical compound depend on the temperature?
The K_{sp} 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? K_{sp}(AgC_{2}H_{3}O_{2}) = 1.94 x 10^{3}.
Since K_{sp} = [Ag^{+}][C_{2}H_{3}O_{2}^{}], 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} = x^{2}
x = 0.0440 M
What is the concentration of a saturated lead chloride solution? K_{sp}(PbCl_{2}) = 1.17 x 10^{5}.
K_{sp} = [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} = 4x^{3}
x = 0.0143 M
I have discovered a new chemical compound with the formula A_{2}B. If a saturated solution of A_{2}B has a concentration of 4.35 x 10^{4} M,
what is the solubility product constant for A_{2}B?
K_{sp} = [A^{+}]^{2}[B^{2}]. 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:
K_{sp} = [2(4.35 x 10^{4} M)]^{2 }[4.35 x 10^{4} M]
K_{sp} = 3.29 x 10^{10}
Solubility product constants are usually specified for 25^{0} C.
Why does the K_{sp} value for a chemical compound depend on the temperature?
K_{sp} depends on temperature because solubility depends on temperature.
Generally, solids become more soluble as the temperature of the solution increases.
As a result, K_{sp} values of solids tend to increase as the temperature increases.
The K_{sp} 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 K_{sp} and K_{b} values are identical.
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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:
1. How many grams of CaCl_{2} 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 10^{2} 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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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 NaHCO_{3} 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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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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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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When a solution is diluted: M_{1} V_{1} = M_{2} V_{2} _{} 
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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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 (C_{6}H_{5}N_{2}) is dissolved in 25 grams of acetic acid. K_{b} of acetic acid is 2.93 ˚C/m.
5. How many grams of sucrose (C_{12}H_{22}O_{11}) need to be dissolved in 600 grams of water to increase the boiling point by 3.5˚C. K_{b }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. K_{f} of water is 1.86˚C/m. What is the molar mass of the unknown solute?
back to colligative properties
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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.
Plan a Strategy.
Steps 14: 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, dipoledipole 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:
Given molar heats of phase change and specific heat capacity data, you can calculate the total heat change for a system using and .
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.
Plan a strategy.
Step 1: Sketch a curve
Step 2: Locate and list the reference data.
 c_{ice} = 2.01 J/g°C (honors only)
 c_{water} = 4.184 J/g°C
 ΔH_{fus} = +6.03 kJ/mol or 334 J/g
 ΔH_{vap} = +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: Roundoff 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.
MgF_{2}(s) => Mg^{2+}(aq) + 2 F^{}(aq)
In a saturated solution of MgF_{2} at 18ºC, the concentration of Mg^{2+} is 1.21 x 10^{3} molar. The equilibrium is represented by the equation above.
(a) Write the expression for the solubilityproduct constant, K_{sp}, and calculate its value at 18ºC.
(b) Calculate the equilibrium concentration of Mg^{2+} in 1.000 liter of saturated MgF_{2} 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 K_{sp} expression and solve for [Mg^{2+}].
(c) Predict whether a precipitate of MgF_{2 }will form when 100.0 milliliters of a 3.00 x 10^{3 }molar Mg(NO_{3})_{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.