Topic 8 Solutions PDF

 ❖ Organic Compounds - are compounds of carbon except the carbonates, cyanides, carbides, and oxides. ❖ Inorganic Compounds - include oxides, bases, acids, and salts. TOPIC 8. SOLUTIONS  Feature Organic Compounds...

 ❖ Organic Compounds - are compounds of carbon except the carbonates, cyanides, carbides, and oxides. ❖ Inorganic Compounds - include oxides, bases, acids, and salts. TOPIC 8. SOLUTIONS  Feature Organic Compounds Inorganic Compounds Carbon-based, with C-H Lacks C-H bonds; often Composition bonds. metal and non-metal. Living organisms Minerals and non-living Source (biological origin). sources. Soluble in organic Soluble in water (ionic Solubility solvents. compounds). Flammability Usually flammable. Non-flammable. Structure Complexity Complex (chains, rings). Simple, often crystalline. TOPIC 8. SOLUTIONS  ❖ Acids - are substances that taste sour, turns blue litmus paper red, and corrodes metals producing H2 gas. ❖ Bases -are substances that taste bitter, turns red litmus paper blue, and have a slippery feel. ❖ Salts - when acids and bases react, they neutralize each other producing salt along with water. TOPIC 8. SOLUTIONS TOPIC 8. SOLUTIONS  1. Arrhenius ▪ defined an acid as a substance that produces H+ ions and a base as substance that produces OH- ions in aqueous solution. ▪ free H+ ions are unlikely to exist in aqueous solution, they are strongly associated with surrounding H2O molecules resulting in H3O+ ions. ▪ assumes that all bases contain OH- ions, but some do not TOPIC 8. SOLUTIONS  Arrhenius Acid Arrhenius Base produces H+ in produces OH- in aqueous solution aqueous solution H2O HCl H+ + Cl- NaOH H2O Na+ + OH- TOPIC 8. SOLUTIONS  2. Bronsted and Lowry ▪ Independently proposed a theory that defines an acid as any substance that can donate H+ ion to another substance, and a base as any substance that can accept H+. ▪ conjugate acid of the base – the product that forms as a result of gaining an H+ ▪ conjugate base of the acid – the product that forms as a result of losing an H+ TOPIC 8. SOLUTIONS  Bronsted-Lowry Acid Bronsted-Lowry Base proton donor proton acceptor TOPIC 8. SOLUTIONS  Example: Identify the acid-base pairs of the following equations TOPIC 8. SOLUTIONS  3. Lewis ▪ Proposed a broader definition of acids and bases. In the Lewis definition, an acid is any species that acts as an electron pair acceptor and a base as an electron pair donor TOPIC 8. SOLUTIONS  Lewis Acid Lewis Base electron pair acceptor electron pair donor TOPIC 8. SOLUTIONS  Theory Acid Base produces H+ ions in produces OH- in Arrhenius aqueous solutions aqueous solution Bronsted-Lowry proton donor proton acceptor electron pair Lewis electron pair donor acceptor TOPIC 8. SOLUTIONS  Example: Consider the reaction : NH3 + HCl → NH4+ + Cl− According to the Arrhenius theory, which compound acts as an acid, and why? According to the Brønsted-Lowry theory, identify the acid-base pairs in this reaction. Using the Lewis theory, explain why NH3 acts as a base. TOPIC 8. SOLUTIONS  Electrolytes are substances that, when dissolved in water or other solvents, produce ions, enabling the solution to conduct electricity. They are essential in both chemistry and biology due to their role in conducting electrical signals and maintaining balance in various systems. TOPIC 8. SOLUTIONS TOPIC 8. SOLUTIONS  1. Strong Electrolytes ▪ Dissociate completely in solution, producing a high concentration of ions. ▪ Examples: ▪ Acids: Hydrochloric acid (HCl), sulfuric acid (H₂SO₄). ▪ Bases: Sodium hydroxide (NaOH), potassium hydroxide (KOH). ▪ Salts: Sodium chloride (NaCl), potassium nitrate (KNO₃). TOPIC 8. SOLUTIONS  2. Weak Electrolytes ▪ Partially dissociate in solution, producing fewer ions. ▪ Examples: ▪ Weak acids: Acetic acid (CH₃COOH). ▪ Weak bases: Ammonia (NH₃). 3. Non-Electrolytes ▪ Do not produce ions in solution; therefore, they do not conduct electricity. ▪ Examples: Sugars (glucose, sucrose) and alcohols (ethanol). TOPIC 8. SOLUTIONS TOPIC 8. SOLUTIONS  ❖ Solution - homogenous mixture of one or more solutes and a solvent ❖ Solute - is the substance being dissolved ❖ Solvent - is the dissolving agent and is usually the most abundant substance in the mixture TOPIC 8. SOLUTIONS  What gas is the solute in soft drinks? carbon dioxide What is another solute in soft drinks? sugar and flavorings TOPIC 8. SOLUTIONS  1. A homogeneous mixture of 2 or more components whose ratio can be varied. 2. The dissolved solute is molecular or ionic in size (less than 1 nm). 3. Liquid or gaseous solutions can be colored or colorless and are usually transparent. 4. The solute will not settle out of the solution. 5. The solute can be separated from the solvent by physical means. TOPIC 8. SOLUTIONS  Example: ▪ Sweet tea is prepared by dissolving an instant tea packet in water. Which substance is the solvent? ▪ A solution of alcohol and water is prepared by adding 25 mL of water to 75 mL methyl alcohol. Which substance is the solute? TOPIC 8. SOLUTIONS  Solubility describes the amount of a substance that will dissolve in a specified amount of solvent at a particular temperature. For example: 36 g NaCl/100 g H2O at 20°C Miscible is the term used if 2 liquids will dissolve in each other. Immiscible is used if the liquids will not dissolve in each other. TOPIC 8. SOLUTIONS  NaCl soluble AgNO3 soluble AgCl insoluble AgOH insoluble TOPIC 8. SOLUTIONS  Example: ▪ Use the ionic compound solubility rules to predict the solubility of a. Barium sulfate b. Ammonium carbonate TOPIC 8. SOLUTIONS  “Like dissolves like” ▪ Polar compounds dissolve in polar solvents, like water and alcohol (CH3CH2OH) Acetone [(CH3)2CO] dissolves in water because it has a net dipole on the O to C bond, making it polar. ▪ Nonpolar compounds dissolve in nonpolar solvents, like petroleum ether and CCl4 Hexane [CH3(CH2)4CH3] dissolves in petroleum ether because they are both nonpolar. TOPIC 8. SOLUTIONS ▪ Many ionic compounds dissolve in water because they form ion to dipole forces with water (a strong intermolecular force). The ions become surrounded by water (become hydrated). ▪ The cation is attracted to the partially negative O in water ▪ The anion is attracted to the partially positive H in water. TOPIC 8. SOLUTIONS  ▪ Most solids’ solubility increases with increasing temperature. (See red lines.) ▪ All gases solubility decreases with increasing temperature. (See blue lines.) TOPIC 8. SOLUTIONS ▪ Pressure does not affect the solubility of solids or liquids, but there is a large effect with gases. ▪ The solubility of gas in a liquid is proportional to the pressure of the gas over the liquid. ▪ Sodas are canned under high pressure. When you open a can, the pressure decreases and bubbles form, releasing the excess gases. TOPIC 8. SOLUTIONS  Example: ▪ Under what conditions are gases most soluble in liquids? a. high temperature, high pressure b. high temperature, low pressure c. low temperature, high pressure d. low temperature, low pressure TOPIC 8. SOLUTIONS ▪ Saturated solutions contain as much dissolved solute as the solvent will hold at a given temperature. Saturated solutions are always in equilibrium with undissolved solute. undissolved solute → → dissolved solute ▪ Unsaturated solutions contain less solute than the amount needed to saturate the solution. TOPIC 8. SOLUTIONS ▪ Supersaturated solutions contain more solute than the amount needed to saturate the solution at a particular temperature. ▪ Supersaturated solutions are unstable – stirring, adding a crystal of solute – will cause the excess solute to come out of solution. TOPIC 8. SOLUTIONS  Example: ▪ What mass of this compound will dissolve at 30°C? ▪ 6.0 g of solute is dissolved in 100 g of water at 60°C. The solution is allowed to cool to 25°C. No crystals form. The solution is: TOPIC 8. SOLUTIONS  1. Particle Size ▪ A solid can dissolve only at the surface that is in contact with the solvent. ▪ Smaller crystals have a larger surface to volume ratio than large crystals. ▪ Smaller crystals dissolve faster than larger crystals. TOPIC 8. SOLUTIONS TOPIC 8. SOLUTIONS  2. Temperature ▪ Increasing the temperature increases the rate at which most compounds dissolve. This occurs because solvent molecules strike the surface of the solid more frequently, causing the solid to dissolve more rapidly. ▪ The dissolved solute particles are also carried away from the solid by the higher kinetic energy solvent molecules, allowing more solvent to hit the surface. TOPIC 8. SOLUTIONS  3. Concentration of Solution TOPIC 8. SOLUTIONS  4. Agitation or Stirring ▪ Stirring rapidly distributes the dissolved solute throughout the solution, eliminating the saturated solution that forms at the surface of the solid. ▪ Moving dissolved solute away from the surface increases the contact between water molecules and the solid and increases the rate of dissolving. TOPIC 8. SOLUTIONS  Example: ▪ Which would most likely increase the solubility of a solid in water? a. Stirring b. Grind the solid to increase its surface area c. Increase the pressure d. Increase the temperature e. All of the above TOPIC 8. SOLUTIONS  Qualitative expressions of concentration: ▪ A dilute solution contains a relatively small amount of dissolved solute. ▪ A concentrated solution contains a relatively large amount of solute. Hydrochloric acid is sold as a concentrated 12 M (moles/ L) solution. A dilute 0.1 M solution is commonly found in labs. TOPIC 8. SOLUTIONS  Quantitative expressions of concentration: TOPIC 8. SOLUTIONS  Example: ▪ Calculate the mass % NaCl in a solution prepared by dissolving 50. g NaCl in 150. g H2O. ▪ Calculate the mass of Na2CO3 and water needed to make 350. g of a 12.3% solution. TOPIC 8. SOLUTIONS  Example: ▪ Normal saline is a 0.90 m/v % NaCl solution. What mass of sodium chloride is needed to make 50. mL of normal saline? TOPIC 8. SOLUTIONS  Example: ▪ What volume of beer that is 6.0 % by volume alcohol contains 200. ml CH3CH2OH (ethyl alcohol)? TOPIC 8. SOLUTIONS  ▪ A 20.0 % solution of KCl has a mass of 400. g. What mass of KCl is contained in this solution? ▪ A solution is prepared by mixing 20.0 mL of propanol with enough water to produce 400.0 mL of solution. What is the volume percent of propanol in this solution? TOPIC 8. SOLUTIONS A 1.0 M KCl solution is prepared by dissolving 1.0 moles KCl in enough water to make 1.0 L of solution. TOPIC 8. SOLUTIONS  Example: ▪ Calculate the molarity of a solution prepared by dissolving 9.35 g KCl in enough H2O to make 250. mL solution. TOPIC 8. SOLUTIONS  Example: ▪ Electrolyte in automobile lead storage batteries is a 3.75 M sulfuric acid solution that has a density of 1.230 g/mL. Calculate the mass percent and molality of the sulfuric acid. TOPIC 8. SOLUTIONS  Solution: TOPIC 8. SOLUTIONS  ▪ A solution is prepared by mixing 1.00 g ethanol (C2H5OH) with 100.0 g water to give a final volume of 101 mL. Calculate the molarity, mass percent, mole fraction, ppm, and molality of ethanol in this solution. TOPIC 8. SOLUTIONS  Solution: TOPIC 8. SOLUTIONS  ▪ How many milliliters of 0.175 M Hg(NO3)2 is needed to completely precipitate 2.50 g KI? Hg(NO3)2 (aq) + 2 KI(aq) → 2KNO3(aq) + HgI2(s) g KI → mol KI → mol Hg(NO3)2→ mL soln TOPIC 8. SOLUTIONS  ▪ Dilution: Adding solvent to a concentrated solution to make a more dilute solution. ▪ When you dilute a concentrated solution, only the volume of solution changes. The quantity of solute remains the same. ▪ Volume (V) × Molarity (M) = moles of solute V 1 × M1 = V 2 × M2 TOPIC 8. SOLUTIONS  Example: ▪ How many milliliters of 12 M HCl are needed to make 500. mL of 0.10 M HCl? TOPIC 8. SOLUTIONS  ▪ What is the molarity of a solution in which 5.85 g of NaCl is dissolved in 200. mL of solution? ▪ What is the molarity of the resulting solution when 300. mL of a 0.400 M solution is diluted to 800. mL? TOPIC 8. SOLUTIONS  ▪ A colligative property is any property of a solution that depends on the number of solute particles, and not on the nature of the particles. Solutions will have ▪ Lower vapor pressures than the pure solvent ▪ Higher boiling points than the pure solvent ▪ Lower freezing points than the pure solvent ▪ Osmosis and osmotic pressure are also colligative properties of solutions. TOPIC 8. SOLUTIONS  ▪ Because of solute–solvent intermolecular attraction, higher concentrations of nonvolatile solutes make it harder for solvent to escape to the vapor phase. Therefore, the vapor pressure of a solution is lower than that of the pure solvent. TOPIC 8. SOLUTIONS  ▪ The vapor pressure of a volatile solvent over the solution is the product of the mole fraction of the solvent times the vapor pressure of the pure solvent. Psoln = Observed vapor pressure of solution χsolv = Mole fraction of solvent P°solv = Vapor pressure of pure solvent ▪ In ideal solutions, it is assumed that each substance will follow Raoult’s Law. TOPIC 8. SOLUTIONS  Example: ▪ Calculate the expected vapor pressure at 25°C for a solution prepared by dissolving 158.0 g common table sugar (sucrose, molar mass = 342.3 g/mol) in 643.5 cm3 of water. At 25°C, the density of water is 0.9971 g/cm3 and the vapor pressure is 23.76 torr. ▪ Predict the vapor pressure of a solution prepared by mixing 35.0 g solid Na2SO4 (molar mass = 142.05 g/mol) with 175 g water at 25°C. Vapor pressure of pure water at 25°C is 23.76 torr. TOPIC 8. SOLUTIONS  Solution: TOPIC 8. SOLUTIONS  Solution: TOPIC 8. SOLUTIONS  ▪ For liquid–liquid solutions where both components are volatile, a modified Raoult’s law applies ▪ PTOTAL = total vapor pressure of a solution containing A and B ▪ χA and χB = mole fractions of A and B ▪ PA° and PB° = vapor pressures of pure A and pure B ▪ PA and PB = partial pressures of A and of B in the vapor above the solution TOPIC 8. SOLUTIONS TOPIC 8. SOLUTIONS  Example: ▪ A solution is prepared by mixing 5.81 g acetone (C3H6O, molar mass = 58.1 g/mol) and 11.9 g chloroform (HCCl3, molar mass = 119.4 g/mol). At 35°C, this solution has a total vapor pressure of 260 torr. Determine if this is an ideal solution. Vapor pressures of pure acetone and pure chloroform at 35°C are 345 and 293 torr, respectively. TOPIC 8. SOLUTIONS  Solution: TOPIC 8. SOLUTIONS ▪ Since vapor pressures are lowered for solutions, it requires a higher temperature to reach atmospheric pressure. Hence, boiling point is raised. tb = mK b TOPIC 8. SOLUTIONS  Example: ▪ What is the boiling point of a solution prepared by dissolving 0.10 mol sugar in 0.50 kg water? The normal boiling point of water is 100.0°C and the boiling point constant for water is 0.512 °C/m. TOPIC 8. SOLUTIONS ❖ The construction of the phase diagram for a solution demonstrates that the freezing point is lowered while the boiling point is raised. tb = mK b TOPIC 8. SOLUTIONS  Example: ▪ What is the freezing point of a solution prepared by dissolving 0.10 mol sugar in 0.50 kg water? The normal freezing point of water is 0.0°C and the freezing point constant for water is 1.86 °C/m. TOPIC 8. SOLUTIONS ▪ The change in temperature is directly proportional to molality (using the van’t Hoff factor). TOPIC 8. SOLUTIONS ▪ It takes into account dissociation of solutes in solution. ▪ Theoretically, we get 2 particles when NaCl dissociates. So, i = 2. ▪ In fact, the amount that particles remain together is dependent on the concentration of the solution. TOPIC 8. SOLUTIONS  ▪ A solution prepared by dissolving 18.00 g glucose in 150 g water has a boiling point of 100.34°C. Calculate the molar mass of glucose. Glucose is a molecular solid that is present as individual molecules in solution. TOPIC 8. SOLUTIONS  ▪ Solution: TOPIC 8. SOLUTIONS  ▪ Determine the mass of ethylene glycol (C2H6O2, molar mass = 62.1 g/mol), the main component of antifreeze, to be added to 10.0 L of water to produce a solution for use in a car’s radiator that freezes at –10.0°F (–23.3°C). Assume the density of water is exactly 1 g/mL. TOPIC 8. SOLUTIONS  ▪ Solution: TOPIC 8. SOLUTIONS  ▪ A chemist is trying to identify a human hormone that controls metabolism by determining its molar mass. A sample weighing 0.546 g was dissolved in 15.0 g benzene, and the freezing- point depression was determined to be 0.240°C. Calculate the molar mass of the hormone. TOPIC 8. SOLUTIONS  ▪ Solution: TOPIC 8. SOLUTIONS ▪ Some substances form semipermeable membranes, allowing some smaller particles to pass through, but blocking larger particles. ▪ The net movement of solvent molecules from solution of low to high concentration across a semipermeable membrane is osmosis. The applied pressure to stop it is osmotic pressure. TOPIC 8. SOLUTIONS ▪ Osmotic pressure is a colligative property. ▪ ∏ = Osmotic pressure (atm) ▪ M = Molarity of the solution ▪ R = Ideal gas constant ▪ T = Temperature (Kelvin) ▪ If two solutions separated by a semipermeable membrane have the same osmotic pressure, no osmosis will occur. TOPIC 8. SOLUTIONS Example: To determine the molar mass of a certain protein, 1.00x10–3 g of it was dissolved in enough water to make 1.00 mL of solution. Osmotic pressure of the solution was found to be 1.12 torr at 25.0°C. Calculate the molar mass of the protein. TOPIC 8. SOLUTIONS  Solution: TOPIC 8. SOLUTIONS