Unit 4: Introduction to Solutions

Unit 4: Introduction to Solutions
Pre-AP Chemistry This chapter we will discuss the major types of chemical reactions that we will encounter in this class. But before we get into the reactions themselves, we have to discuss solutions, their properties, and how compounds react in them first.

Solutions Solution – homogeneous mixture
Solute – substance being dissolved Solvent – present in greater amount 2

Commonly Used Solutions
Make-up Laundry detergent Motor oil Gasoline Food preservatives Deodorant Lawn fertilizers & weed killers Shampoo Air fresheners Furniture polish Toothpaste and mouthwash Oven cleaner Glass cleaner etc…

Solution Vocabulary Alloy – a solid solution of metals
Bronze = Cu + Sn Brass = Cu + Zn Soluble – “will dissolve in” Insoluble – “will not dissolve in” Miscible – refers to two gases or two liquids that form a solution More specific than “soluble” e.g. food coloring and water Immiscible – refers to two gases or liquids that will not form a solution Suspension – appears uniform while being stirred, but settles over time

Types of Solutions Solute Solvent Solution Gaseous Solutions
Liquid air (nitrogen, oxygen, argon gases) humid air (water vapor in air) Liquid Solutions liquid solid carbonated drinks (CO2 in water) vinegar (CH3COOH in water) salt water (NaCl in water) Solid Solutions dental amalgam (Hg in Ag) sterling silver (Cu in Ag) Solutions are not limited to gases and liquids; solid solutions also exist. • Amalgams, which are usually solids, are solutions of metals in liquid mercury. • Network solids are insoluble in all solvents with which they do not react chemically; covalent bonds that hold the network together are too strong to be broken and are much stronger than any combination of intermolecular interactions that might occur in solution. • Most metals are insoluble in all solvents but do react with solutions such as aqueous acid or base to produce a solution; in these cases the metal undergoes a chemical transformation that cannot be reversed by removing the solvent. 5

Solution Composition Solute – a substance in a smaller amount dissolved in a larger amount of another substance (the solvent) Concentration – the number of moles present in a certain volume of solution Expressed as the amount of solute dissolved in a given amount of solution. An intensive quantity Molarity (M) expresses the concentration in units of moles of solute per liter of solution Can be used as a conversion factor between volume of solution and amount (mol) of solute Most chemical process (almost all biochemical processes) occur in solution. Some vocabulary for parts of a solution are listed here. First, we have solute. That is the substance that occurs in smaller amounts and is dissolved in a larger amount of another substance, which is called the solvent. For instance, in salt water, salt is the solute and water is the solvent. Concentration refers to a measure of the number of moles present in a given volume of solution. Concentration is an intensive property like density, meaning the amount of the solution does not affect the concentration. Concentration is given as an amount of solute dissolved in an amount of solution. Molarity is something you will become very familiar with. This is a measure of concentration that is expressed in moles of solute per liter of solution. If you are given molarity, symbolized by the big M, you have to know the units that it is given in – always mol/L. This value can be used as a conversion factor between volume and mol just like molar mass converts between mass and mol.

Expressing Concentration “The amount of solute in a solution”
A. mass % = mass of solute mass of solution B. parts per million (ppm) (also, ppb and ppt) commonly used for minerals or contaminants in water supplies C. molarity (M) = moles of solute L of solution used most often in this class D. molality (m) = moles of solute kg of solvent MOLARITY - Most common unit of concentration Most useful for calculations involving the stoichiometry of reactions in solution Molarity of a solution is the number of moles of solute present in exactly 1 L of solution: moles of solute molarity = liters of solution Units of molarity — moles per liter of solution (mol/L), abbreviated as M Relationship among volume, molarity, and moles is expressed as VL M Mol/L = L (mol) = moles (L) There are several different ways to quantitatively describe the concentration of a solution, which is the amount of solute in a given quantity of solution. 1. Molarity – Useful way to describe solution concentrations for reactions that are carried out in solution or for titrations – Molarity is the number of moles of solute divided by the olume of the solution Molarity = moles of solute = mol/L liter of solution – Volume of a solution depends on its density, which is a function of temperature 2. Molality – Concentration of a solution can also be described by its molality (m), the number of moles of solute per kilogram of solvent – Molality = moles of solute kilogram solvent – Depends on the masses of the solute and solvent, which are independent of temperature – Used in determining how colligative properties vary with solute concentrations 3. Mole fraction – Used to describe gas concentrations and to determine the vapor pressures of mixtures of similar liquids – Mole fraction () = moles of component total moles in the solution – Depends on only the masses of the solute and solvent and is temperature independent 4. Mass percentage (%) – The ratio of the mass of the solute to the total mass of the solution – Result can be expressed as mass percentage, parts per million (ppm), or parts per billion (ppb) mass percentage = mass of solute  100% mass of solution parts per million (ppm) = mass of solute  106 parts per billion (ppb) = mass of solute  109 – Parts per million (ppm) and parts per billion (ppb) are used to describe concentrations of highly dilute solutions, and these measurements correspond to milligrams (mg) and micrograms (g) of solute per kilogram of solution, respectively – Mass percentage and parts per million or billion can express the concentrations of substances even if their molecular mass is unknown because these are simply different ways of expressing the ratios of the mass of a solute to the mass of the solution 7

Concentration = # of moles volume (L) V = 1000 mL V = 1000 mL
n = 8 moles [ ] = 32 molar V = 1000 mL V = 1000 mL V = 5000 mL n = 2 moles n = 4 moles n = 20 moles [ ] = 2 molar [ ] = 4 molar [ ] = 4 molar 8

MOLES LITERS OF GAS AT STP (22.4 L/mol) MASS IN GRAMS NUMBER OF
Molar Volume (22.4 L/mol) MASS IN GRAMS MOLES NUMBER OF PARTICLES Molar Mass (g/mol) 6.02  1023 particles/mol Looking back at this conversion diagram, we will now know all four branches as we add on the Liters of Solution box using Molarity as a conversion factor. Remember, all our conversion factors so far have been per 1 mole. Molarity is a conversion factor that is mol/L. However you use the conversion factor, moles will be on either top or bottom and liters will be on the flip side. Molarity (mol/L) LITERS OF SOLUTION 9

Molarity Calculations
Hydrobromic acid (HBr) is a solution of hydrogen bromide gas in water. Calculate the molarity of hydrobromic acid solution if L contains 1.80 mol of hydrogen bromide. How many moles of KI are in 84 mL of 0.50 M KI? We are going to perform some calculations with molarity involved. Question 1 asks us for molarity after giving us moles and volume. Be careful though. Make sure you convert your volume to L before giving a molarity value. Question 2 asks us to go backward by giving us molarity and volume (again in mL) and asking us for moles. Use this as a conversion factor. Line up those units! Question 3 asks us about mass of a solute. Well we know in order to get to mass from volume and molarity, we need to get to moles so we can use the molar mass. Finally, we are given a mass and molarity and are asked about volume. What do we have to do to that mass to use it? Convert it to moles!!

How many grams of solute are in 1.75 L of M sodium hydrogen phosphate (Na2HPO4)? How many liters of 3.30 M sucrose (C12H22O11) contain 135 g of solute? We are going to perform some calculations with molarity involved. Question 1 asks us for molarity after giving us moles and volume. Be careful though. Make sure you convert your volume to L before giving a molarity value. Question 2 asks us to go backward by giving us molarity and volume (again in mL) and asking us for moles. Use this as a conversion factor. Line up those units! Question 3 asks us about mass of a solute. Well we know in order to get to mass from volume and molarity, we need to get to moles so we can use the molar mass. Finally, we are given a mass and molarity and are asked about volume. What do we have to do to that mass to use it? Convert it to moles!!

You have 10.8 g potassium nitrate. How many mL of solution will make this a 0.14 M solution? We are going to perform some calculations with molarity involved. Question 1 asks us for molarity after giving us moles and volume. Be careful though. Make sure you convert your volume to L before giving a molarity value. Question 2 asks us to go backward by giving us molarity and volume (again in mL) and asking us for moles. Use this as a conversion factor. Line up those units! Question 3 asks us about mass of a solute. Well we know in order to get to mass from volume and molarity, we need to get to moles so we can use the molar mass. Finally, we are given a mass and molarity and are asked about volume. What do we have to do to that mass to use it? Convert it to moles!!

Solution Stoichiometry Examples
Given the reaction Pb(NO3)2(aq) + KI (aq)  PbI2(s) + KNO3(aq), what volume of 4.0 M KI solution is required to yield 89 g PbI2? How many mL of a M CuSO4 solution will react w/excess Al to produce 11.0 g Cu? Aluminum sulfate is also produced.

Given the unbalanced reaction Cu AgNO3  Ag + Cu(NO3)2 , how many grams of Cu are required to react with 1.5 L of 0.10M AgNO3? 79.1 g of zinc react with 0.90 L of 2.5M HCl to form zinc chloride and hydrogen gas. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP?

5. A common antacid contains magnesium hydroxide, which reacts with HCl to form water and magnesium chloride solution. As a government chemist testing commercial antacids, you use 0.10 M HCl to simulate the acid concentration in the human stomach. How many liters of stomach acid react with a tablet containing 0.10 g of magnesium hydroxide? 6. Mercury (II) nitrate reacts with sodium sulfide solution to produce solid mercury (II) sulfide and sodium nitrate solution. In a laboratory simulation, L of M mercury (II) nitrate reacts with L of 0.10 M sodium sulfide. How many grams of mercury (II) sulfide form?

Molality mass of solvent only 16

Molality Find the molality of a solution containing 75 g of MgCl2 in kg of water. How many grams of NaCl are required to make a 1.54m (molal) solution using kg of water? 17

Preparing and Diluting Molar Solutions
The volume of a solution is made up of solute and solvent. To prepare a solution of a specific molarity: Weigh the solid (solute) needed. Transfer the solid to a volumetric flask that contains about half the final volume of solvent. Dissolve the solid thoroughly by swirling. Add solvent until the solution reaches its final volume. To dilute the solution to a lesser molarity, add only solvent to increase the volume. Often in lab, we are asked to prepare solutions of a specific molarity. Remember that we need moles of solute and volume of solution. The volume of the solution is composed of both the solute and the solvent so you have to account for the volume of the solvent you are adding. Some steps to help you with this are listed here. First, we have to weigh the solute needed. How would we know how much solute we needed in grams if we are only given molarity and volume? Convert to moles of solute and use molar mass to convert to mass. Then transfer this solute into a volumetric flask containing about half the final volume of solvent. This is to allow us to account for the volume of solute without overflowing. Next, dissolve the solute thoroughly. Once this is done, add enough solvent to reach the desired volume. Many times, chemists (and students in lab) use the same solution over and over. To simplify the process of creating solutions of a specific molarity, labs usually have a stock solution of the desired solution which is a large quantity of the solution at a high concentration. This is so that those who use the solution can dilute it to any particular molarity by calculations. To dilute a stock solution, you must add more solvent, not solute. This increases the volume of the solution without increasing the moles of solute. We are going to do some calculations with this so that you may understand this better.

How to Mix a Standard Solution
Use a VOLUMETRIC FLASK to make a standard solution of known concentration Step 1> add the weighed amount of solute in the volumetric flask Step 2> add distilled water (about half of final volume) Step 3> cap volumetric flask, and shake to dissolve solute completely Step 4> add distilled water to volume marker (calibration mark) The solution process may be exothermic (release heat). This may cause the liquid to show a larger volume than is real. Allow the solution to return to ambient (room) temperature and check volume again. 19

Making Solutions Assume that 1 L of 3 M NaOH solution is needed for a class lab. Determine the mass of sodium hydroxide required to create the solution. What mass of salt is required to prepare 500 mL of M NaCl solution? To prepare a solution that contains a specified concentration of a substance, it is necessary to dissolve the desired number of moles of solute in enough solvent to give the desired final volume of solution. Solute occupies space in the solution so the volume of the solvent that is needed is less than the desired volume of solution. To prepare a particular volume of a solution that contains a specified concentration of a solute, calculate the number of moles of solute in the desired volume of solution and then covert the number of moles of solute to the corresponding mass of solute needed. 20

Diluting Solutions Concentration – a measure of solute to solvent ratio Concentrated – lots of solute Dilute – not much solute (“watery”) To dilute solutions, add more solvent (often water) to solution. Moles of solute remain the same. Concentration of a solution describes the quantity of a solute that is contained in a particular quantity of solvent or solution Knowing the concentration of solutes is important in controlling the stoichiometry of reactant for reactions that occur in solution A concentrated solution contains a large amount of solute in a given amount of solution. A 10 mol/L solution would be called concentrated. A dilute solution contains a small amount of solute in a given amount of solution. A 0.01 mol/L solution would be called dilute. 21

Diluting Stock Solutions
A solution of a desired concentration can be prepared by diluting a small volume of a more-concentrated solution, a stock solution, with additional solvent. – Calculate the number of moles of solute desired in the final volume of the more-dilute solution and then calculate the volume of the stock solution that contains the amount of solute. – Diluting a given quantity of stock solution with solvent does not change the number of moles of solute present. – The relationship between the volume and concentration of the stock solution and the volume and concentration of the desired diluted solution is (Vs) (M s) = moles of solute = (Vd) (M d). 22

Diluting Stock Solutions (2)
Dilution of solutions  Acids (and sometimes bases) are purchased in concentrated form (concentrate) and are easily diluted to any desired concentration. Safety Tip: When diluting, add acid or base to water (A&W!) Dilution Equation: MCVC=MDVD C  “concentrate” D  “dilute” 23

Diluting Molar Solutions Examples
Concentrated H3PO4 is 14.8 M. What volume of concentrate is required to make L of M H3PO4? What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution?

Isotonic saline is a 0.15 M aqueous solution of NaCl that simulates the total concentration of ions found in many cellular fluids. Its uses range from a cleansing rinse for contact lenses to a washing medium for red blood cells. How would you prepare 0.80 L of isotonic saline from a 6.0 M stock solution? Often in lab, we are asked to prepare solutions of a specific molarity. Remember that we need moles of solute and volume of solution. The volume of the solution is composed of both the solute and the solvent so you have to account for the volume of the solvent you are adding. Some steps to help you with this are listed here. First, we have to weigh the solute needed. How would we know how much solute we needed in grams if we are only given molarity and volume? Convert to moles of solute and use molar mass to convert to mass. Then transfer this solute into a volumetric flask containing about half the final volume of solvent. This is to allow us to account for the volume of solute without overflowing. Next, dissolve the solute thoroughly. Once this is done, add enough solvent to reach the desired volume. Many times, chemists (and students in lab) use the same solution over and over. To simplify the process of creating solutions of a specific molarity, labs usually have a stock solution of the desired solution which is a large quantity of the solution at a high concentration. This is so that those who use the solution can dilute it to any particular molarity by calculations. To dilute a stock solution, you must add more solvent, not solute. This increases the volume of the solution without increasing the moles of solute. We are going to do some calculations with this so that you may understand this better.

If 25.0 mL of 7.50 M sulfuric acid are diluted to exactly 500. mL, what is the mass of sulfuric acid per milliliter? Often in lab, we are asked to prepare solutions of a specific molarity. Remember that we need moles of solute and volume of solution. The volume of the solution is composed of both the solute and the solvent so you have to account for the volume of the solvent you are adding. Some steps to help you with this are listed here. First, we have to weigh the solute needed. How would we know how much solute we needed in grams if we are only given molarity and volume? Convert to moles of solute and use molar mass to convert to mass. Then transfer this solute into a volumetric flask containing about half the final volume of solvent. This is to allow us to account for the volume of solute without overflowing. Next, dissolve the solute thoroughly. Once this is done, add enough solvent to reach the desired volume. Many times, chemists (and students in lab) use the same solution over and over. To simplify the process of creating solutions of a specific molarity, labs usually have a stock solution of the desired solution which is a large quantity of the solution at a high concentration. This is so that those who use the solution can dilute it to any particular molarity by calculations. To dilute a stock solution, you must add more solvent, not solute. This increases the volume of the solution without increasing the moles of solute. We are going to do some calculations with this so that you may understand this better.

Beer’s Law Beer’s Law relates the absorption of light to the properties of the material through which the light is traveling. There is a dependence between the absorbance of light by a substance and the product of the molar absorptivity coefficient, a, the distance the light travels through the material, b, and the molar concentration, c. Also called the Beer-Lambert Law, the more concentrated a solution is, the more light it will absorb and the darker it will appear.

Using Beer’s Law Beer’s law is typically used in a process called spectroscopy. In this process, a beam of light is sent through a small sample of a solution. A detector on the other side of the solution records the amount of light that passed through the solution.

Beer’s Law Examples A red light is passed through a blue solution as shown. The molar absorptivity constant of the solution is 1.3x106 M-1 cm-1. The absorbance reading is Determine the concentration of the solution. If water is added to the solution in question #1, what do you think will happen to the concentration and the absorbance reading? 0.52 1 cm

Beer’s Law Examples A green light is passed through a purple solution as shown. The molar absorptivity constant of the solution is 1.8x104 M-1 cm-1. The absorbance reading is Determine the concentration of the solution. List at least two laboratory procedures you could do to increase the absorbance reading for the purple solution in #3. 0.18 1 cm

Calibration Curve Experiments with Beer’s Law typically create a “calibration curve” by recording absorbance values for solutions with known concentrations. Then, the absorbance for a solution of unknown concentration can be measured and the concentration may be calculated using the equation from the calibration curve. An example of a calibration curve is shown below.

Polar Nature of Water Covalent bonds between H and O have unequal sharing. Electrons spend more time closer to O, creating a slightly negative pole. Water molecule has bent shape; atoms form angle. The distribution of its bonding electrons and its overall shape makes water an ionizing solvent. Now we are going to talk about water and why it is important to our reactions. Water is a unique substance, and one that makes up a majority of our planet. The properties of the water molecule give make it important. We know a water molecule is composed of one Oxygen atom and two Hydrogen atoms. From chapter two, we should know that these are both nonmetals and they bond covalently, which means they share electrons. But these bonds are not an equal partnership. We will discuss all the reasons for this in later chapters, but what you really need to know here is that the electrons are more attracted to the O atom and spend more time closer to the oxygen atom. This makes the oxygen very slightly negative and the hydrogen atoms slightly positive. Also, water is not a linear molecule (meaning atoms are in a straight line). If you look on the wall, you can see the water molecule is bent and it forms an angle. These properties – electron distribution and shape make water an ionizing solvent. Now we are going to discuss what an ionizing solvent is.

Ionic Compounds in H2O To be soluble, the attraction between each ion and the water molecules must outweigh the attraction of the oppositely charged ions. If the electrostatic attraction among ions in the compound is greater than the attraction between ions and water molecules, the substance is insoluble. Next up we are going to discuss the differences between ionic and covalent compounds when they are dissolved in water. First, ionic compounds. We discussed that they separate into ions, but why do they do this. Well, when we talked about the polar nature of water, we learned that water has opposite poles with the oxygen being partially negative and the hydrogen being partially positive. It is these charges that attract the ions and pull them away from each other. The negative O end pulls at the positive ion while the positive H end pulls at the negative ion. When compounds are soluble (meaning they do dissociate) the attraction that the water has on the ions is greater than the attraction the ions have for each other. If this doesn’t occur and the ions are more attracted to each other than they are to the water, the substance does not dissociate and it is called insoluble.

Solubility of Ionic Compounds
Water interacts strongly with dissolved reactants and can affect their bonds. Electrolyte – a substance that conducts a current when dissolved in water Soluble ionic compounds dissociate completely into ions and create a large current; called strong electrolytes. Solvated ions are surrounded by solvent molecules. Oppositely charged ions separate when dissolved in water, become surrounded by water molecules, and spread randomly throughout the solution. The H2O above the arrow indicates that water is required but is not a reactant in the usual sense. Solubility refers to how well a compound will stay in solution (dissolved). We are going to go through some definitions first to better understand solutions. We just discussed some properties of water and we learned that water is an ionizing solvent. What this means is that water can affect the bonds of the substances that are dissolved in it. An electrolyte is a substance that conducts current when dissolved in water. The current comes from having positive and negative entities present. Ionic compounds that are called “soluble” are those that break apart or dissociate completely into ions. For instance, salt (NaCl) is an electrolyte. When you dissolve it in water, salt separates into Na+ and Cl- ions. Ionic compounds that dissociate completely are called strong electrolytes, while those that just dissociate a little are called weak electrolytes. The current conducted by a solution of these ions is coincides with how much they ionize. When a compound ionizes, the ions are surrounded by molecules of the solvent (in this case water) and the ions are considered to be “solvated” – meaning surrounded by solvent. In a water solution, the differently charged ions become separated and solvated by the water. They then spread throughout the solution. Here is an example of an ionization reaction. The potassium bromide is dissolved in water (shown above the arrow) and it separates into positive potassium ions and negative bromine ions. Notice that the compound was a solid, but when dissolved in the water, the ions are now in aqueous solution. The water is placed above the arrow because it does not really participate in the reaction, but it is necessary for the ionization to occur. (The water is not changing).

Dissolving of Salt in Water
Na+ ions Water molecules Cl- ions When sodium chloride crystals are dissolved in water, the polar water molecules exert attracting forces which weaken the ionic bonds. The process of solution occurs the ions of sodium and chloride become hydrated. NaCl(s) + H2O  Na+(aq) + Cl-(aq) 35

Dissociation of NaCl Animation
The attraction of water dipoles for ions pulls ions out of a crystalline lattice and into aqueous solution. The ion-dipole forces exist in the solution as well, lessening the tendency for ions to return to the crystalline state. The combination of an ion in solution and the neighboring water dipoles to which it is attracted is called a hydrated ion. Animation 36

Covalent Compounds in H2O
Many covalent compounds dissolve in water. Do not dissociate into ions Remain intact molecules Do not conduct an electric current; called nonelectrolytes. Acids are H-containing covalent compounds that do dissociate into ions in water. Covalent compounds are not made up of ions, but many of them do dissolve in water. They don’t dissociate into ions, but remain intact molecules. Dissolving these compounds just means that the molecules are apart from each other. These compounds do not conduct electric current because they do not have the charged particles – called non-electrolytes. There is a special class of covalent compounds that we will discuss again in further detail later in this chapter that do dissociate into ions. These compounds contain hydrogen and are called acids.

The H+ Ion Water interacts most strongly with the hydrogen cation (H+)
H+ is just a proton H+ attracts the negative pole of surrounding water molecules so strongly that it forms a covalent bond to one of them H3O+ (hydronium ion) We just mentioned that acids are hydrogen containing covalent compounds that do dissociate. When they dissociate, the hydrogen cation separates from the anion. Water interacts very strongly with this cation, which as we have discussed before, is just a proton because the hydrogen atom has lost its electron. The hydrogen cation is so attracted to the negative O pole of water molecules that it even forms a covalent bond and forms the hydronium ion. When acids dissolve in water and ionize, this hydronium ion is formed. In many cases, it might still be indicated in the reaction equation as H+ and H2O.

Formation of Hydronium Ions
1+ 1+ 1+ + H+ H2O H3O+ hydrogen ion water hydronium ion (a proton) 39

H+ Ions Examples Nitric acid is a major chemical in the fertilizer and explosives industries. In aqueous solution, each molecule dissociates and the H becomes a solvated H+ ion. What is the molarity of H+ (aq) in 1.4 M nitric acid? How many moles of H+(aq) are present in 451 mL of M hydrobromic acid? We are going to work a few examples now concerning the H+ ion. We are going to determine the amount and/or concentration of H+ ions when acids are dissolved in solution.

H+ Ion Examples How many moles of H+(aq) are present in 585 mL of 3.50 M H3PO4? What is the molarity of the H+ ion in 1.6 M sulfuric acid?

Solvation Terminology
Dissociation: separation of an ionic solid into aqueous ions e.g. NaCl(s)  Na+(aq) + Cl-(aq) Ionization: breaking apart of polar molecules into aqueous ions e.g. HNO3(aq) + H2O(l)  H3O+(aq) + NO3–(aq) Molecular Solvation: molecules (covalent) dissolve, but remain intact e.g. C6H12O6(s)  C6H12O6(aq)

Ionization/Dissociation Examples
Predict the ionization or dissociation for the following chemical species: Sodium hydroxide Hydrochloric acid Sulfuric Acid Acetic Acid Potassium fluoride Calcium nitrate

Electrolyte Review Strong Electrolytes exhibit nearly 100% dissociation e.g. Weak electrolytes exhibit little dissociation “Strong” or “Weak” is a property of the substance. One cannot be changed into the other. When written in a chemical equation, a strong electrolyte is shown as individual ions while a weak electrolyte is shown in molecular form. NaCl Na Cl– NOT in water: In aq. solution: CH3COOH CH3COO – H+ NOT in water: In aq. solution: 44

Electrolytes Electrolytes - solutions that carry an electric current
Electrolyte — any compound that can form ions when it dissolves in water – When strong electrolytes dissolve, constituent ions dissociate completely, producing aqueous solutions that conduct electricity very well. – When weak electrolytes dissolve, they produce relatively few ions in solution, and aqueous solutions, of weak electrolytes do not conduct electricity as well as solutions of strong electrolytes. – Nonelectrolytes dissolve in water as neutral molecules and have no effect on conductivity. strong electrolyte weak electrolyte nonelectrolyte NaCl(aq) Na+ + Cl- HF(aq) H+ + F- 45

Factors Affecting the Rate of Dissolution
Temperature: as T increases, rate increases Particle Size: as particle size decreases, rate increases Mixing: more mixing/stirring, rate increases Nature of solvent or solute: identity determines whether a substance will dissolve and to what extent Formation of a solution from a solute and a solvent is a physical process, not a chemical one. Both solute and solvent can be recovered in chemically unchanged form using appropriate separation methods. Dissolution of a solute in a solvent to form a solution does not involve a chemical transformation. Substances that form a single homogeneous phase in all proportions are said to be completely miscible in one another. If two substances are essentially insoluble in each other, they are immiscible. 46

SUPERSATURATED SOLUTION
Solution Saturation UNSATURATED SOLUTION more solute dissolves SATURATED SOLUTION no more solute dissolves SUPERSATURATED SOLUTION becomes unstable, crystals form Maximum amount of a solute that can dissolve in a solvent at a specified temperature and pressure is its solubility. – Solubility is expressed as the mass of solute per volume (g/L) or mass of solute per mass of solvent (g/g) or as the moles of solute per volume (mol/L). – Solubility of a substance depends on energetic factors and on the temperature and, for gases, the pressure. • A solution that contains the maximum possible amount of solute is saturated. • If a solution contains less than the maximum amount of solute, it is unsaturated. When a solution is saturated and excess solute is present, the rate of dissolution is equal to the rate of crystallization. • Solubility increases with increasing temperature — a saturated solution that was prepared at a higher temperature contains more dissolved solute than it would contain at a lower temperature, when the solution is cooled, it can become supersaturated. increasing concentration 47

Solubility and Temperature
Solids dissolved in liquids Gases dissolved in liquids To Sol. As To , solubility To Sol. As To , solubility 48

Solubility Tables Solubility: how much solute dissolves in a given amount of solvent at a given temperature Unsaturated: Solution can hold more solute; below line Saturated: Solution has “just the right” amount of solute; on line Supersaturated: Solution has “too much” solute dissolved in it; above the line Temp. (oC) Solubility (g/100 g H2O) KNO3 (s) KCl (s) HCl (g) 49

Solubility Table Solubility vs. Temperature
Solubility vs. Temperature Solubility (grams of solute/100 g H2O) KI KCl 20 10 30 40 50 60 70 80 90 110 120 130 140 100 NaNO3 KNO3 HCl NH4Cl NH3 NaCl KClO3 SO2 gases solids Temperature °C Solubility Table “Solubility Curves for Selected Solutes” Description: This slide is a graph of solubility curves for 10 solutes. It shows the number of grams of solute that will dissolve in 100 grams of water over a temperature range of 0cC to 10 cC. Basic Concepts The maximum amount of solute that will dissolve at a given temperature in 100 grams of water is given by the solubility curve for that substance. When the temperature of a saturated solution decreases, a precipitate forms. Most solids become more soluble in water as temperature increases, whereas gases become less soluble as temperature increases. Teaching Suggestions Use this slide to teach students how to use solubility curves to determine the solubilities of various substances at different temperatures. Direct their attention to the dashed lines; these can be used to find the solubility of KClO3 at 50 cC (about 21 g per 100 g of H2O). Make sure students understand that a point on a solubility curve represents the maximum quantity of a particular solute that can be dissolved in a specified quantity of solvent or solution at a particular temperature. Point out that the solubility curve for a particular solute does not depend on whether other solutes also are present in the solution (unless there is a common-ion effect; this subject usually is covered at a later stage in a chemistry course). Questions Determine the solubilities (in water) of the following substance at the indicated temperatures: NH3 at 50 oC; KCl at 90 oC; and NaNO3 at 0 oC. Which of the substances shown on the graph is most soluble in water at 20 oC? Which is lease soluble at that temperature? For which substance is the solubility lease affected by changes in temperature? Why do you think solubilities are only shown between 0 oC and 100 oC? In a flask, you heat a mixture of 120 grams of KClO3 and 300 grams of water until all of the KClO3 has just been dissolved. At what temperature does this occur? You then allow the flask to cool. When you examine it later, the temperature is 64 oC and you notice a white powder in the solution. What has happened? What is the mass of the white powder? Compare the solubility curves for the gases HCl, NH3, and SO2) with the solubility curves for the solid solutes. What generalizations(s) can you make about the relationship between solubility and temperature? According to an article in an engineering journal, there is a salt whose solubility in water increases as the water temperature increases from 0 oC to 65 oC. The salt’s solubility then decreases at temperatures above 65 oC, the article states. In your opinion, is such a salt likely to exist? Explain your answer. What could you do to verify the claims of the article? 50

Understanding Saturation
Solubility (grams of solute/100 g H2O) KI KCl 20 10 30 40 50 60 70 80 90 110 120 130 140 100 NaNO3 KNO3 HCl NH4Cl NH3 NaCl KClO3 SO2 gases solids Temperature °C Understanding Saturation Describe each situation below according to saturation and appearance using the given solubility table: Per 100 g H2O,100 g 60°C. Per 100 g H2O, 20 g 80°C. Cool solution (B) very slowly to 30°C Quench solution (B) in an ice bath to 30°C Per 100 g H2O, 70 g 50°C 51

Gas In a gas, the energy of attraction is small relative to the energy of motion. On average, the particles are far apart. Large interparticle distance has several macroscopic consequences: A gas moves randomly throughout its container and fills it. Gases are highly compressible (can be squeezed and shrunk). Gases flow and diffuse through one another easily.

Liquid In a liquid, the attractions are stronger because the particles are in contact. But their kinetic energy allows them to tumble randomly over and around each other. Therefore, a liquid conforms to the shape of its container but has a surface. With very little free space between the particles, liquids compress only very slightly. Liquids flow and diffuse but much more slowly than gases.

Solid In a solid, the attractions dominate the motion to such an extent that the particles remain in position relative to one another. With the particles very close together and positions fixed, a solid has a specific shape. Solids compress even less than liquids. Solids do not flow significantly.

Phase Changes Phase changes are determined by the interplay between kinetic energy and intermolecular forces. As the temperature increases, the average kinetic energy increases as well so the faster moving particles can overcome the attractions more easily. Lower temperatures allow the forces to draw the slower moving particles together. The process by which a gas changes into a liquid is called condensation. A liquid changing to a gas is called vaporization. The process by which a liquid changes into a solid is called freezing. A solid changing to a liquid is called melting or fusion. The process by which a solid becomes a gas (without becoming a liquid first) is called sublimation. A gas changing to a solid is called deposition.

Phase Diagrams Phase diagrams are used to depict the phase changes of a substance at various conditions of temperature and pressure. A phase diagram has these four features: Regions of the diagram: Each region corresponds to one phase of the substance. A particular phase is stable for any combination of pressure and temperature within its region. If any other phase is placed under those conditions, it will change to the stable phase. Lines between regions: The lines separating the regions represent the phase transition curves. Any point along a line shows the pressure and temperature at which the two phases exist in equilibrium. The critical point: The liquid-gas line ends at the critical point. Beyond the critical temperature, a supercritical fluid exists rather than separate liquid and gaseous phases. The triple point: The three phase transition curves meet at the triple point: the pressure and temperature at which three phases are in equilibrium.

Phase Diagram

Water – the Exception! Most substances have a solid phase that is more dense than the liquid phase, giving a positive slope to the solid-liquid line. (Because the liquid occupies slightly more space than the solid, an increase in pressure favors the solid phase, in most cases.) Unlike almost every other substance, solid water is less dense than liquid water (due to hydrogen bonds causing the solid phase to have a greater volume because of large spaces between molecules). An increase in pressure favors the phase that occupies less space, which for water is the liquid phase. Therefore, the solid-liquid line for water has a negative slope.

Phase Diagram of CO2 vs. H2O

Phase Diagram Examples
Using the phase diagram for water, determine what phase water is in at the following conditions: 2 atm, 55°C 1 atm, 75°C 1 atm, 200°C 5 atm, -100°C

Colligative Properties
Colligative properties refer to the physical properties that are changed by the presence of any solute particles. The chemical identity of the solute is not important. Only the concentration of the solute particles is significant in altering physical properties. The physical properties that change when a solute is present include vapor pressure, boiling point, melting point, and osmotic pressure.

Boiling Point Elevation / Freezing Point Depression
The lowering of vapor pressure caused by a nonvolatile nonelectrolyte solute leads to an increase in the boiling point and a decrease in the freezing point of the solvent. Higher temperatures are needed to make the vapor pressure of the system equal atmospheric pressure (boiling point). Lower temperatures are needed to overcome the interference with the crystallization process caused by the attractive forces between the solute and solvent. The changes in the boiling and melting points of a solution depend on the solvent and on the molal concentration of the solute.

Boiling Point Elevation / Freezing Point Depression
The increase in the boiling point is given by m is the molality of the solution kb is the boiling point elevation constant for the solvent i represents the number of particles produced by each part of the solute The decrease in the melting (freezing) point is given by The negative sign indicates a decrease in temperature kf is the freezing point depression constant

Uses of BP Elevation/ MP Depression
Chemists use the freezing point depression to make cooling baths below 0°C by dissolving large quantities of salt in water and then adding ice. Antifreeze in a car’s engine lowers the freezing point of water by adding a high concentration of ethylene glycol. Antifreeze also increases the boiling point of water to allow engines to run above 100°C without boiling over. A common cooking use is adding salt to water to increase the boiling point, allowing the food to cook at a higher temperature than otherwise possible. The most important use is the determination of molar masses of solutes.

Colligative Properties of Electrolyte Solutions
Example #1: Which of the following, when added to kg H2O, is expected to give the greatest increase in the boiling point of water? (kb = 0.052°C/m) 1.25 mol sucrose (C12H22O11) 0.25 mol iron (III) nitrate 0.50 mol ammonium chloride 0.60 mol calcium sulfate 1.00 mol acetic acid Discuss the similarity between molarity and molality of water when the solution is very dilute. dH2O = 1.00kg/L

Equations for Aqueous Ionic Reactions
Three types of equations: Molecular Total Ionic Net Ionic Atoms and charges must balance in ionic equations. Molecular equations show all reactants and products as if they were intact, undissociated compounds Total ionic equations show all the soluble ionic substances dissociated into ions. Spectator ions appear in the same form on both sides of the equation and are not involved in the actual chemical change. Net ionic equations eliminate the spectator ions and show the actual chemical change taking place Now we are going to talk some chemistry terms that are used to show reactions. We’ve already talked about chemical equations, but now we’re going to specialize a little more with respect to these equations. There are three types of equations for aqueous ionic reactions (that is reactions that occur with ionic species in a solution). These are molecular, total ionic, and net ionic. In all three of these equations, the atoms of each element and the charges ( something we haven’t really worked with up til now) must balance. First we have molecular equations. These look like all the chemical equations we’ve seen before. They show all the reactants and products completely as if they were undissociated compounds. Next, and this is the first real new one, we have the total ionic equation. These equations show the ionic compounds that will dissociate (soluble) in ion form. There are also spectator ions included. Spectator ions are those that remain the same on both side of the equation and are not really involved in the change going on. Finally, the net ionic equation gets rid of those spectator ions and shows the change that occurs. We are going to look at an example of each of these to help you understand.

Molecular Equation 2NaCl(aq) + Pb(NO3)2(aq) PbCl2(s)+ 2NaNO3(aq) Total Ionic Equation 2Na+(aq)+2Cl-(aq) + Pb2+(aq) + 2NO3-(aq)PbCl2(s) + 2Na+(aq)+2NO3-(aq) Net Ionic Equation 2Cl-(aq) + Pb2+(aq)  PbCl2(s) Spectator Ions

Molecular Equation Total Ionic Equation Net Ionic Equation Ok, we are going to examine an aqueous ionic reaction using the three equations we discussed on the previous slide. First, the molecular equation. Looking at this, it appears to be just like all the equations we’ve seen so far. It lists two reactants and two products with their respective physical states (notice that three of them are aqueous in solution and one is a solid) and it is balanced. Next, is the total ionic equation. For this one, we have to take the molecular equation and show all the substances that will dissociate in solution as the ions they form. So silver nitrate becomes silver cation and nitrate anion. Sodium chromate becomes sodium cation and chromate anion. Sodium nitrate on the product side becomes sodium cation and nitrate anion. Notice that silver chromate is not shown as dissociated ions. The reason for this we will discuss in just a bit, but pay attention to its state. The compound is listed as a solid, which means it is no longer in the solution so it will not dissociate. Ok, finally, the net ionic equation. Remember from the definition of net ionic, we get rid of spectator ions, those that are the same on both sides of the equation. Looking at the total ionic equation, which ions seem to be exactly the same on both sides? Nitrate and sodium. Therefore, we should get rid of those species for the net ionic. If we cancel them out, we arrive at the net ionic equation that shows the actual chemical change going on: silver cation and chromate anion combining to make a solid silver chromate.

Writing Net Ionic Equations
Write the net ionic equations for the following molecular equations and identify the spectator ions: Li2SO4(aq)+Ca(OH)2(aq)  CaSO4(s)+2LiOH(aq) 2NaI(aq) + Pb(NO3)2(aq)  2NaNO3(aq) + PbI2(s)

Write the net ionic equations for the following molecular equations and identify the spectator ions: AgNO3(aq) + KBr(aq)  AgBr(s) + KNO3(aq) H2SO4(aq) + NaOH(aq)  Na2SO4(aq) + H2O (l)

Write the net ionic equations for the following molecular equations and identify the spectator ions: Al2(SO4)3(aq)+6NH4OH(aq)3(NH4)2SO4(aq) +2Al(OH)3(s)

Unit 4: Introduction to Solutions

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