More research

http://en.wikipedia.org/wiki/Solubility

Factors affecting solubility [edit]

Solubility is defined for specific phases. For example, the solubility of aragonite and calcite in water are expected to differ, even though they are both polymorphs of calcium carbonate and have the same chemical formula.
The solubility of one substance in another is determined by the balance of intermolecular forces between the solvent and solute, and the entropy change that accompanies the solvation. Factors such as temperature and pressure will alter this balance, thus changing the solubility.
Solubility may also strongly depend on the presence of other species dissolved in the solvent, for example, complex-forming anions (ligands) in liquids. Solubility will also depend on the excess or deficiency of a common ion in the solution, a phenomenon known as the common-ion effect. To a lesser extent, solubility will depend on the ionic strength of solutions. The last two effects can be quantified using the equation for solubility equilibrium.
For a solid that dissolves in a redox reaction, solubility is expected to depend on the potential (within the range of potentials under which the solid remains the thermodynamically stable phase). For example, solubility of gold in high-temperature water is observed to be almost an order of magnitude higher when the redox potential is controlled using a highly oxidizing Fe₃O₄-Fe₂O₃ redox buffer than with a moderately oxidizing Ni-NiO buffer.^[4]

Solubility (metastable) also depends on the physical size of the crystal or droplet of solute (or, strictly speaking, on the specific surface area or molar surface area of the solute). For quantification, see the equation in the article on solubility equilibrium. For highly defective crystals, solubility may increase with the increasing degree of disorder. Both of these effects occur because of the dependence of solubility constant on the Gibbs energy of the crystal. The last two effects, although often difficult to measure, are of practical importance.^{[citation needed]} For example, they provide the driving force for precipitate aging (the crystal size spontaneously increasing with time).

Temperature [edit]

The solubility of a given solute in a given solvent typically depends on temperature. For many solids dissolved in liquid water, the solubility increases with temperature up to 100 °C.^[5] In liquid water at high temperatures, (e.g., that approaching the critical temperature), the solubility of ionic solutes tends to decrease due to the change of properties and structure of liquid water; the lower dielectric constant results in a less polar solvent.
Gaseous solutes exhibit more complex behavior with temperature. As the temperature is raised, gases usually become less soluble in water (to minimum, which is below 120 °C for most permanent gases^[6]), but more soluble in organic solvents.^[5]
The chart shows solubility curves for some typical solid inorganic salts (temperature is in degrees Celsius).^[7] Many salts behave like barium nitrate and disodium hydrogen arsenate, and show a large increase in solubility with temperature. Some solutes (e.g., sodium chloride in water) exhibit solubility that is fairly independent of temperature. A few, such as cerium(III) sulfate, become less soluble in water as temperature increases. This temperature dependence is sometimes referred to as "retrograde" or "inverse" solubility. Occasionally, a more complex pattern is observed, as with sodium sulfate, where the less soluble decahydrate crystal loses water of crystallization at 32 °C to form a more soluble anhydrous phase.^{[citation needed]}

The solubility of organic compounds nearly always increases with temperature. The technique of recrystallization, used for purification of solids, depends on a solute's different solubilities in hot and cold solvent. A few exceptions exist, such as certain cyclodextrins.^[8]

Solubility constants are used to describe saturated solutions of ionic compounds of relatively low solubility (see solubility equilibrium). The solubility constant is a special case of an equilibrium constant. It describes the balance between dissolved ions from the salt and undissolved salt. The solubility constant is also "applicable" (i.e., useful) to precipitation, the reverse of the dissolving reaction. As with other equilibrium constants, temperature can affect the numerical value of solubility constant. The solubility constant is not as simple as solubility, however the value of this constant is generally independent of the presence of other species in the solvent.
The Flory-Huggins solution theory is a theoretical model describing the solubility of polymers. The Hansen Solubility Parameters and the Hildebrand solubility parameters are empirical methods for the prediction of solubility. It is also possible to predict solubility from other physical constants such as the enthalpy of fusion.
The partition coefficient (Log P) is a measure of differential solubility of a compound in a hydrophobic solvent (octanol) and a hydrophilic solvent (water). The logarithm of these two values enables compounds to be ranked in terms of hydrophilicity (or hydrophobicity).
The energy change associated with dissolving is usually given per mole of solute as the enthalpy of solution.

Research for Hypothesis/Introduction finally added

http://www.chem.tamu.edu/class/majors/tutorialnotefiles/ksp.htm

Solubility equilibrium is base on the assumption that solids dissolve in water to give the basic particles from which they are formed.

·  Molecular solids dissolve to give individual aqueous molecules.

·  Ionic solids dissociate to give their respective positive and negative ions:

The ions in formed from the dissociation of ionic solids can carry an electrical current. Salt solutions, therefore, are good conductors of electricity. Molecular solids, however, do not dissociate in water to give ions, so no electrical current can be carried.

Solubility

1.      The ratio of the maximum amount of solute to the volume of solvent in which this solute can dissolve.

1.      Generally expressed in two ways:

1.      grams of solute per 100 g of water

2.      moles of solute per Liter of solution

Soluble: Dissolve - Do NOT form a solid precipitate. **alkali metal ions and ammonium ion: Li+, Na+, K+, NH4+ acetate ion: C2H3O21- nitrate ion: NO31- halide ions (X): Cl-, Br-, I- (Exceptions: AgX, HgX, and PbX2 are insoluble) sulfate ion: SO42- (Exceptions: SrSO4, BaSO4, and PbSO4 are insoluble; AgSO4, CaSO4, and Hg2SO4 are slightly soluble)

 

Insoluble: Do NOT Dissolve - Do form a solid precipitate. carbonate ion: CO32- chromate ion: CrO42- phosphate ion: PO43- sulfide ion: S2- (Exceptions: CaS, SrS, and BaS are soluble) hydroxide ion: OH- (Exceptions: Sr(OH)2 and Ba(OH)2 are soluble; Ca(OH)2 is slightly soluble)

• A salt is considered soluble if it dissolves in water to give a solution with a concentration of at least 0.1 M at room temperature.
• A salt is considered insoluble if the concentration of an aqueous solution is less than 0.0001 M at room temperature.
• Salts with solubilities between 0.0001 M and 0.1 M are considered to be slightly soluble.

Salts that have extremely low solubilities dissociate in water according to the principles of equilibrium. For example, the reaction for the dissociation of the salt AgCl is:

The reverse reaction for the dissolving of the salt would be the precipitation of the ions to form a solid:

The system has reached equilibrium when the rate at which AgCl dissolves is equal to the rate at which AgCl precipitates.

• Saturated solution - Contains the maximum concentration of ions that can exist in equilibrium with the solid salt at a given temperature.

The equilibrium reaction for the dissociation of AgCl is:

Solubility product equilibrium constant (K_sp) - The product of the equilibrium concentrations of the ions in a saturated solution of a salt. Each concentration is raised to the power of the respective coefficient of ion in the balanced equation.

• NOTE: There is no denominator in the solubility product equilibrium constant. The key word to remember is PRODUCT which can remind you that you should have a multiplication (or product) of the concentrations of the ions. The reason that the solid reactant is not written is because its concentration effectively remains constant.

For example, the solubility product equilibrium constant for the dissociation of AgCl is:

Let's try another example of a solubility product equilibrium constant. Consider the reaction for the dissociation of CaF₂ in water:

The solubility product equilibrium constant for this reaction would be the product of the concentration of Ca²⁺ ion and the concentration of the F^- ion raised to the second power (squared):

NOTE: Unlike K_a and K_b for acids and bases, the relative values of K_sp cannot be used to predict the relative solubilities of salts if the salts being compared produce a different number of ions.

The solubility product is literally the product of the solubilities of the ions in units of molarity (mol/L)

Sample Calculations

• K_sp can be calculated from the solubility of a salt. Conversely, the solubility of a salt can be calculated from K_sp.

Let's try an example calculation problem to demonstrate the relationship between the solubility and the solubility product of a salt.

1) Calculate the solubility of CaF₂ in g/L (K_sp = 4.0 x 10^-8)

First, write the BALANCED REACTION:

Next, set up the SOLUBILITY PRODUCT EQUILIBRIUM EXPRESSION:

In the above equation, however, we have two unknowns, [Ca²⁺] and [F^-]². So, we have to write one in terms of the other using mole ratios. According to the balanced equation, for every one mole of Ca²⁺ formed, 2 moles of F^- are formed. To simplify things a little, let's assign the the variable X for the solubility of the Ca²⁺:

If we SUBSTITUTE these values into the equilibrium expression, we now only have one variable to worry about, X:

We can now SOLVE for X:

We assigned X as the solubility of the Ca²⁺ which is equal to the solubility of the salt, CaF₂. However, our units right now are in molarity (mol/L), so we have to convert to grams:

Now, let's try to do the opposite, i.e., calculate the K_sp from the solubility of a salt.

2) The solubility of AgCl in pure water is 1.3 x 10^-5 M. Calculate the value of K_sp.

First, write the BALANCED REACTION:

Next, set up the SOLUBILITY PRODUCT EQUILIBRIUM EXPRESSION:

It is given in the problem that the solubility of AgCl is 1.3 x 10^-5. Since the mole ratio of AgCl to both Ag⁺ and Cl^- is 1:1, the solubility of each of the ions is equal to the solubility of AgCl. SUBSTITUTE the solubility given into the equilibrium expression to get K_sp:

http://pharmlabs.unc.edu/labs/solubility/structure.htm

Solute and Solvent Structure/Polarity

Solute molecules are held together by certain intermolecular forces (dipole-dipole, induced dipole-induced dipole, ion-ion, etc.), as are molecules of solvent. In order for dissolution to occur, these cohesive forces of like molecules must be broken and adhesive forces between solute and solvent must be formed.

The solubility of a drug in a given solvent is largely a function of the polarity of the solvent. Solvents may be considered polar, semi-polar or non-polar. Polar solvents will dissolve ionic and other polar solutes (i.e. those with an asymmetric charge distribution [like dissolves like]), whereas, non-polar solvents will dissolve non-polar molecules. Semi-polar solvents (eg. alcohols and ketones) may induce a certain degree of polarity in non-polar molecules and may thus act to improve the miscibility of polar and non-polar liquids. The relationship between polarity and solubility may be used in practice to alter the solubility of a drug in a pharmaceutical solution.

One approach is to alter the polarity of the solute by shifting it between its molecular (undissociated) and ionic (dissociated) states. A shift toward the ionic form improves solubility of the solute in water and other polar solvents. A shift toward the molecular species improves solute solubility in non-polar solvents. Such shifts may be produced by altering the pH of the solution (or using the salt form of the compound).

Another approach is to mix solvents of different polarities to form a solvent system of optimum polarity to dissolve the solute. Such solvents must, obviously, be miscible. This method is referred to as solvent blending or cosolvency and uses the dielectric constant as a guide to developing the cosolvent system. Since many solvents may be toxic when ingested, most solvent blends are limited to mixtures containing water, ethanol, glycerin, propylene glycol, polyethylene glycol 400 or sorbitol solution. The list is somewhat expanded for solutions for external application.

The dielectric constant (δ) of a compound is an index of its polarity. A series of solvents of increasing polarity will show a similar increase in dielectric constant.

http://aceyourchemistry.blogspot.com.au/2013/01/part-1-like-dissolves-like-to-dissolve.html

Part 1 : "Like dissolves Like" - To dissolve or not to dissolve

To dissolve or not to dissolve, that is the question. ''Like dissolves like'' is an expression used by chemists to remember how solvents work. It refers to ''polar'' and ''non-polar'' solvents and solutes. For instance, water is polar while oil is non-polar. Therefore, water will not dissolve oil. In other words, they are immiscible. Meanwhile, ionic salt such as sodium chloride is ionic (which is considered to be extremely polar). Since like dissolves like, that means polar dissolves polar, thus the ionic salt is soluble in water.

 

However, how do I know whether a molecule is polar? There are two conditions that you need to consider before you determine if the molecule is polar.

 Condition 1: Is the bond polar? As long as the bonding atoms are different, they will have different electronegativity, which leads to a dipole moment in the bond i.e. the bond is polar. For example, H-Br is a polar bond whereas Br-Br is not.

 Condition 2: What is the molecular geometry?

 Consider the molecule CO2, which is carbon dioxide. The displayed structure looks like this i.e. O=C=O.  The C=O bond is polar because the bonding atoms are different and they have different electronegativity.  However, the molecule is linear in shape. As a result, the dipoles (treated as vectors) will cancel out each other. In the linear molecules, these dipole moments have the same magnitude, but since they are pointing in opposite directions, there is cancellation of dipoles that renders the molecule non-polar.

 Now, what about water? Isn't water a non-polar molecule? The answer is NO! If the water molecule is linear too, then it will not be polar. However if you were to draw the dot and cross diagram for water, you will find that the central oxygen atom has 2 lone pairs of electrons.

 As a result, the molecular geometry is actually bent. As a result, there is a collective effect of the individual polar O-H bond that give rise to a net dipole moment, and hence water is a polar molecule.

http://www.elmhurst.edu/~chm/vchembook/174temppres.html

Temperature and Pressure Effects on Solubility

Effect of Temperature on Solubility:

The solubility of solutes is dependent on temperature. When a solid dissolves in a liquid, a change in the physical state of the solid analogous to melting takes place. Heat is required to break the bonds holding the molecules in the solid together. At the same time, heat is given off during the formation of new solute -- solvent bonds.

CASE I: Decrease in solubility with temperature:

If the heat given off in the dissolving process is greater than the heat required to break apart the solid, the net dissolving reaction is exothermic (energy given off). The addition of more heat (increases temperature) inhibits the dissolving reaction since excess heat is already being produced by the reaction. This situation is not very common where an increase in temperature produces a decrease in solubility.

CASE II: Increase in solubility with temperature:

If the heat given off in the dissolving reaction is less than the heat required to break apart the solid, the net dissolving reaction is endothermic (energy required). The addition of more heat facilitates the dissolving reaction by providing energy to break bonds in the solid. This is the most common situation where an increase in temperature produces an increase in solubility for solids.

The use of first-aid instant cold packs is an application of this solubility principle. A salt such as ammonium nitrate is dissolved in water after a sharp blow breaks the containers for each. The dissolving reaction is endothermic - requires heat. Therefore the heat is drawn from the surroundings, the pack feels cold.

Further re-evaluated method

Procedure-

NaCl in water:

1) 135mL of water was poured into a 250mL beaker. 35.7g of sodium chloride was weighed in a beaker upon the electronic scales and added to the water. The solution was stirred until dissolved.

2) After protective bench paper had been laid across the workspace and safety equipment collected the titration apparatus was set up. The burette clamp was fastened to the retort stand, along with the burette after it had been rinsed through with both distilled water and silver nitrate.  

3) The burette was filled with silver nitrate, and 10mL of the saturated NaCl solution with three drops of the potassium dichromate indicator was placed in a conical flask beneath the apparatus.

4) The silver nitrate was slowly titrated into the salt solution until a brick red precipitate was formed. The number of moles of silver nitrate used was then calculated to determine the chloride ion concentration and consequently the solubility product of the salt.

5) The above procedure was then repeated; however, the sodium chloride solution was heated to 50 degrees Celsius, before 10mL was measured out for the titration.  

NaCl in ethanol:

6) A known amount of salt, in this instance 35.7g, was placed into 135mL of ethanol.  The solution was stirred for approximately three minutes. In order to determine how much of the sodium chloride was dissolved, the solution was filtered through filter paper.

7) When the filter paper was dry, the amount of salt present on the filter paper was weighed and comparisons made to the original amount of salt placed in the ethanol. Both weights were recorded.

8) The ethanol and salt solution was titrated with silver nitrate to determine the amount of chloride ions according to the same procedure as the NaCl and water solution. The solubility product was then calculated of salt in ethanol.

9) The same procedure for ethanol was repeated, though over a heat of 50 degrees celcius before the addition of NaCl.

NaCl in Oil:

10) A known amount of salt (in this case, 35.7g) was added to weighed flask containing 135mL of olive oil and mixed well. 100mL of distilled water was added to the solution and stirred for approximately three minutes.

11) Using a separating funnel, the water was filtered off, leaving only the olive oil. This liquid was weighed in order to determine whether any of the NaCl had been dissolved into the oil. If the weight was changed, then a titration was used to determine the chloride concentration. If not, no titration was used due to the absence of any chloride ions.

12) Regardless of whether or not any sodium chloride dissolved, the test was repeated over a heat of 50 degrees Celsius before the water was added to the oil and salt.

Hypothesis and adjusted method

Hypothesis: The more polar the substance, the higher the solubility product of salt. If temperature is increased, then the solubility will increase, more effectively for the polar substances.

 

Method:

1) Pour 100mL of water into a 250mL conical flask. Measure out 38g of salt and add to the solution. Stir until dissolved.

2) Fill a burette with silver nitrate and add a potassium permanganate indicator to the NaCl solution. Slowly titrate the silver nitrate into the solution, stopping when the precipitate turns brick red. Record the amount of silver nitrate used, and consequently discover the amount of chlorine ions.

NOTE: The silver nitrate should react with the chloride ions, which will then react with the potassium permanganate indicator to create a brick red color. The titration will then be complete, because the chlorine will only react with the indicator when all the ions in the solution have been used up

3) Calculate the Ksp, knowing the chlorine concentration.

4) Repeat procedure for ethanol and oil.

5) Repeat the procedure for same substances, though heat over a temperature of 80 degrees before titration.

New Investigation-Risk Assesment Complete

Alteration of previous Idea:

Chemistry EEI Take 2

Theory: Investigating the extent to which like dissolves like.

Investigate: How does the Ksp of salt in water compare to the Ksp of salt in slightly polar and non-polar solutions.

1) Ksp of salt in water

2) Ksp of salt in ethanol

3) Ksp of salt in oil

Then investigate temperature dependence of the Ksp of salt in the same solutions, at a constant temperature.

How do the results of this experiment contribute to the theory ‘like dissolves like’?

General Equipment list/method:

bubette

retort stand

clamp

ethanol

oil

distilled water

Therefore, I need to investigate the cause of separation of the Na ions from the Cl ions

Procedure:

1.         Select “Hot Plate & Mag Stir” from the Equipment menu.  Place 25.00 mL of an unknown brine solution into a 100 mL beaker.  Place the beaker on the magnetic stirrer.  Add 0.100 mL of potassium chromate. Select the magnetic stirrer and use the context menu to set the stirring rate.

 

2.         Carefully fill a 50.00 mL burette with standardized silver nitrate solution. Zero the burette to 0.00 mL and position  it over the 100 mL beaker.

 

3.         Titrate the contents of the beaker with silver nitrate from the burette.  Initially the silver nitrate may be added rapidly with continuous stirring.  At this point, large amounts of white solid should be forming.  Slow down subsequent additions of silver nitrate until a faint brick red precipitate forms indicating the end of the titration.

 

4.                  Record all pertinent data on the data sheet.  Read the burette volume to +/- 0.1 mL .

 

5.                  Repeat the titration until the final volumes agree to +/- 0.2 mL.

 

6.                  Fill in the remainder of the data sheet and perform all calculations.

Overall Theory + Hypotheses

Overall Theory/Research Question investigating: To what extent is the statement ‘like dissolves like’ true?

Part 1 Working Hypothesis:  If the temperature of the water and either one of the four other substances is increased, then the solubility and/or the rate of solubility of that substance in water will increase.

Part 2 Working Hypothesis:  If substances are mixed together over heat before being added to water, the hydrocarbons and alcohols will mix along with the alcohol and salt, whereas the salt and oil will not mix, though the salt will still dissolve in water.

Part 3 Working Hypothesis:  The addition of both the liquid soap and dishwashing fluid will cause all substances to dissolve in water, however, the dishwashing liquid will be more effective in increasing solubility.

Risk Assessment

Risk Assessment complete- will be attached at end of investigation

Materials Request Form

ST STEPHEN’S CATHOLIC COLLEGE

                              YEAR 12 CHEMISTRY

                  Materials Request Form

Student Name: ……Kate Donovan………..

Item	Quantity	Conditions	Date required
Olive Oil	90mL		Monday May 6th
Ethanol	90mL		Monday May 6th
Butanol	90mL		Monday May 6th
Table Salt	90g		Monday May 6th
Liquid Soap	100mL		Monday May 6th
Dishwashing Liquid	100mL		Monday May 6th
Electronic Temp. Apparatus (hotplate)		To heat beakers with liquid in them.	Monday May 6th
100mL Beakers	8		Monday May 6th
100mL Measuring Cylinder	1		Monday May 6th
Stopwatch	1		Monday May 6th
Stirrer	4		Monday May 6th
Spoon	1		Monday May 6th
Thermometer	1		Monday May 6th
			Monday May 6th