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Discover the fundamentals of solutions, including various types, solubility, and factors affecting the dissolving process. Explore how polarity influences molecular interactions to understand "like dissolves like" concept.
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Solutions • A solution is a homogeneous (uniform) mixture • CONSISTING of a solvent and solute. • Solvent-the dissolving medium • Solute-the substance that is dissolved • Aqueous Solution-solution with H2O as the solvent
Solubility Salt-solute Water -solvent
Solubility of Ionic Substances • Example: • NaCl separates into cations and anions when placed in water • We know this because it conducts electricity • Because water has a strong dipole-dipole attraction, it interacts with the positive and negative ions of NaCl • Na+ attracts to water’s negative end • Cl- attracts to water’s positive end • This is possible because the attractive forces of the water overcome the ionic forces of the salt crystal
SolubilityPolar Vs. Nonpolar Hexane molecule Polar molecule-dissolves in H2O Nonpolar molecule-does Not dissolve in H2O
Polar Molecules • Water dissolves non-ionic substances like sucrose because it contains O-H molecules like water. • Each O-H group can Hydrogen bond to a water molecule. • These attractions are very similar in both molecules
Nonpolar Molecules • Hexane does not uniformly disperse in water. • Since Carbon and Hydrogen have similar electronegativites, the bonding electrons are shared almost equally and there is essentially little or no temporary charge created. • Since the polarities of the two molecules are not similar, the hexane is not able to attract the hydrogen bonds in water to break them and make a “hole” for the solute particles. • Molecules with similar polarities dissolve, “like dissolves like”
“Like-Dissolves-Like” • Polar (water) Dissolves in Polar • Nonpolar (oil) Dissolves in Nonpolar (oil Oil Nonpolar Water Polar
Nonpolar Molecules • Hexane does not uniformly disperse in water. • Since Carbon and Hydrogen have similar electronegativites, the bonding electrons are shared almost equally and there is essentially little or no temporary charge created. • Since the polarities of the two molecules are not similar, the hexane is not able to attract the hydrogen bonds in water to break them and make a “hole” for the solute particles. • Molecules with similar polarities dissolve, “like dissolves like”
Solutions of Solids “Alloys” • ALLOYS Mixture of metals BRONZE BRASS Magnesium + Aluminum Copper + Zinc Copper + Tin
Factors Affecting the Rate of Dissolving • SURFACE AREA • STIRRING • TEMPERATURE
Surface Area • The greater the surface area exposed to the solvent, the faster the dissolving process occurs. • If we wish to speed up the process of dissolving a sugar cube, we would grind up the crystal into powder form to expose more of the surface of the sugar at any one time.
Stirring • Stirring removes newly dissolved particles from the surface of the solute and increases the amount of interaction between the solute and solvent by continually exposing the solute to fresh solvent. • Therefore, more molecular interactions are happening at any one time.
Temperature • SOLIDS VS. GASES • Dissolving in solids occurs faster at higher temperatures. • Since temperature is a measure of how quickly molecules move, the faster the solid solute moves in a liquid solvent, the more surface area of the solute is exposed to the solvent at any one time, thus increasing the rate of dissolution.
Sugar in ice Sugar exposed to heat
Solubility • How much can dissolve • Depends on TEMPERATURE!!! • For crystals : T Solubility
Temperature • GASES • For gases, as temperature increases, solubility decreases. • As temperature increases, molecules move faster. As gas solutes move faster within a liquid solvent, they escape the liquid into the surroundings.
SOLUBILITY CURVES - Gases For Gases T Solubility T Solubility
Pressure on Solubility • SOLIDS • If pressure is increased above standard (1 atmosphere), molecules become more compact, (closer together), and dissolve at a faster rate in liquids. More solid solute can come in contact with the solvent at any one time in a smaller amount of space. • GASES • Likewise, if pressure is increased above standard (1 atmosphere), molecules become more compact, (closer together), and dissolve at a faster rate in liquids. More solid solute can come in contact with the solvent at any one time in a smaller amount of space.
For Gases P Solubility P Solubility
Solution Composition Sodium Acetate crystallizing from a saturated solution
Solution Composition Sodium Acetate crystallizing from a saturated solution.
Solution Composition • Even for every soluble substance, there is a limit to how much solute can be dissolved in a given solvent. • For this reason, we categorize them as follows: • Saturated-describes a solution that contains as much solute as will dissolve at that temperature. • Unsaturated-describes a solution in which more solute can still be dissolved that is currently dissolved at that temperature. • Supersaturated-describes a solution that contains more solute than a saturated solution will hold, essentially more solute than can be dissolved by the given amount of solvent at that temperature.
Solubility Curves Saturated: Maximum dissolved Unsaturated: Less than maximum Supersaturated: More than maximum
SOLUBILITY CURVES - Crystals What is the solubility Of KNO3 at 40oC?
What is the solubility of KCl at 60oC? • What is the solubility of KNO3 at 20oC? • How many grams of KClO3 can dissolve in 200g of water at 70oC? • If 60 g of KCl is dissolved in 100g of water at 50oC, is this saturated, supersaturated or unsaturated?
Dilute Vs. Concentrated Solutions • Although a chemical compound always has the same composition, a solution is a mixture and the amounts of the solute present can vary in different solutions. • Sometimes qualitative terms are used to describe how much solute is in a solution. The solution has the same volume. • They are as follows: • Concentrated-describes a solution in which a relatively large amount of solute is dissolved in a solution. • Dilute-describes a solution in which a relatively small amount of solute is dissolved in a solution.
Solution Composition • One quantitative way of describing a solution’s composition is by calculating the percentage of solute contained in a given solution, otherwise known as mass percent. • Mass Percent expresses the mass of solute present in a a given mass of solution expressed in percent. • Formula:
Mass Percent • 1. A solution is prepared by mixing 1.00 g of ethanol, C2H5OH, with 100.0 g of water. Calculate the mass percent of ethanol in this solution. • 2. Although milk is not a true solution. It is really a suspension of tiny globules of fat, protein and other substances in water. It does contain a dissolved sugar called lactose. Cow’s milk typically contains 4.5% by mass of lactose, C12H22O11. Calculate the mass of lactose present in 175 g of milk.
Solution Composition Molarity • When a solution is described in mass percent, the solution is given in mass (grams). However it is more convenient to measure a solution in terms of volume. • Because of this, chemists describe solutions in terms of a specific concentration, amount of solute in mols per volume, in liters of solution. This is known as molarity. • Molarity-is the number of moles of solute per liter of solution. • HOW MUCH OF A SUBSTANCE IS DISSOLVED? • Formula:
Molarity Formula If the solute is given in grams, one must convent the solute to mols using molar mass. This formula is used for all aqueous solutions and not gases. Grams Moles = M x L Question: Adding 1.0 mol of solute to 1.0 L of water does not make a 1.0 M solution. Why not?
Solution Composition Molarity • 1. Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution. • 2. Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl into enough water to make 26.8 mL of solution. • It is important to note that a description of a solution’s composition may not reflect its true chemical nature. Solute concentrations are written in terms of the solute before it dissolves. • Example: 1.0 M NaCl means that the solution was prepared by dissolving 1.0 mol of solid NaCl units. But the solution contains 1.0 mol of Na+ ions and 1.0 mol of Cl- ions in the solution. It contains 1.0 M Na+ and 1.0 M Cl-.
Molarity Problems • Give the concentrations of all ions in each of the following solutions. 0.50 M Co(NO3)2 1 M FeCl3 Often we need to determine the mols of solute present in a given solution of a known molarity. When the molarity of a solution is multiplied by volume in Liters, we get the mols of solute present. Liters x Molarity = moles of solute L of soln x mols of solute/L of soln = moles of solute, where L of soln is crossed out
Molarity Problems • How many moles of Ag+ ions are present in 25 mL of a 0.75 M AgNO3 solution? • Calculate the number of mols of Cl- ions in 1.75 L of 1.0 x 10-3 M AlCl3? • Calculate the moles of solute needed to make a 1.00 L solution of an aqueous 0.200 M K2Cr2O7?
Standard Solution • Standard Solution- is a solution in which the concentration is accurately known. • A standard solution is made by weighing out the appropriate amount of pure solute and transferring it to a volumetirc flask (a flask of accurately know volume) and then slowly filling the flask with solvent and gently swirling it until the solution reaches to the one Liter mark.
Dilutions • To save time and money, laboratory solutions are made and prepared in concentrated form called stock solutions. If less concentrated solutions are needed, water or another solvent is added to achieve the desired molarity. This process is called diluting. • Dilution- is the process of adding solvent to an already made solution to in effect lower its concentration of solute. • A typical dilution involves determining how much solvent to add to a given amount of stock solution to achieve the desired concentration. They key thing to remember that only the solvent, usually water is the only thing added. The amount of solute in the more dilute solution will be the same as in the concentrated stock solution. • Mols solute after dilution = Mols of solute before dilution • In other words, adding a solvent like water does not alter the number of mols of solute present in the new solution.
Calculating Dilutions DILUTION FORMULA EXAMPLE
Dilution Problems • How much stock solution is needed to prepare 500 mL of 1.00 M acetic acid, HC2H3O2, from a 17.5 M stock solution of acetic acid. What volume of the stock solution is required? • What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H2SO4 solution?
Parts Per Million 4 g of salt is mixed with 1,000 g of water, What is the concentration in ppm? A sugar solution has a concentration of 12%,What is the concentration in ppm?
Stoichiometry of Solution Reactions • Because so many important reactions occur in solution, it is important to be able to do stoichiometric calculations for solution reactions. • There are 5 KEY STEPS for Solving Solution Problems • Write and balance the equation for the reaction. If the reaction involves ions, it is best to write the net ion equation. • Calculate the moles of reactants. • Determine the limiting reactant. • Calculate the moles of other reactants or products. • Convert to grams or other units, if required.
Stoichiometric Problems • Calculate the mass of solid NaCl that must be added to 1.50 L of a 0.100 M AgNO3 solution to precipitate all of the Ag+ ions in the form AgCl. Calculate the mass of AgCl formed.
STEPS for Serial Dilutions STEP 1 Fill a test tube A with 10 milliliters (mL) of a solution. STEP 2 Fill a second test tube B with 9 mL of a buffer. This buffer will serve to dilute the original solution. The buffer is frequently distilled water, but this actually depends on the composition of the solution in test tube A.
STEPS Continued STEP 3 Dilute the solution. Draw 1 mL of solution from test tube A with a pipette and transfer it to test tube B and mix thoroughly. The solution that originally had a volume of 10 mL now has a volume of 1 mL in test tube B. The solution, therefore, has been diluted by a factor of 10. STEP 4 Fill a test tube C with 9 mL of buffer. Move 1 mL of the solution in test tube B to test tube C using the technique described previously and thoroughly mix the contents of test tube C. The solution in test tube C has been diluted by a factor of 100.
STEPS Continued STEP 5 Examine the effects of serial dilution. The solution in test tube B has 1/10 the concentration of the solution in test tube A and the solution in test tube C has 1/10 the concentration of the solution in test tube B. The solution in test tube C, therefore, has 1/100 (1/10 x 1/10 = 1/100) the concentration of the solution in test tube A. STEP 6 Extend this procedure to perform longer serial dilutions. This process may be repeated as many times as necessary to achieve the desired solution. In an experiment involving concentration curves, use serial dilution to create a series of solutions with dilutions of 1, 1/10, 1/100, 1/1,000. Then perform an experiment on each dilution to measure how the behavior of a particular solution changes with concentration.
