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Solutions and Equilibrium

Solutions and Equilibrium. Solutions (homogeneous mixtures) are a part of our everyday life. Do you use mouthwash, toothpaste, cough syrup?

adrienne-conway
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Solutions and Equilibrium

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  1. Solutions and Equilibrium

  2. Solutions (homogeneous mixtures) are a part of our everyday life. Do you use mouthwash, toothpaste, cough syrup? • My swimming pool is one very large solution. The presence of algae makes my pool green and slimy. I then need to add an algicide which is another solution. Bromine is in solution as well as other chemicals (total alkalinity).

  3. Solubility is of fundamental importance in a large number of practical applications ranging from ore processing to the use of medicines and the transport of pollutants. • Solubility is often said to be one of the “characteristic properties of a substance”. • used to describe a substance • indicates its polarity • helps distinguish it from another substance • becomes a guide for the applications of that substance • You will often see in a compounds description: “is insoluble in water or alcohol but soluble in concentrated sulfuric acid”

  4. Solubility of a substance is useful when separating mixtures. The synthesis of chemical compounds, by the milligram in a lab or by the ton in industry makes use of the relative solubility of the end product. • Cooks, chemists, farmers, pharmacists and gardeners need to know which compounds are soluble and which are insoluble. • You will separate a mixture or five substances for first lab using the property of solubility as an important property.

  5. What properties do solutions have? • How do we describe the concentrations? • Why do some substances dissolve while others do not?

  6. Solution Definitions: • Solution – a homogeneous mixture that is uniform throughout. The same substances will be in the same relative amounts. Solutions are not pure substances because they can have variable composition (example salt water). • Solvent – any substance that has other substances dissolved in it. The substance that is present in the largest amount (by volume, mass or number of moles) is usually the solvent.

  7. Solute – a substance dissolved into the solvent. • Concentrated – a solution with a high level of solute ( > 1.0 M) • Dilute – a solution with a low level of solute (< 1.0 M) • A solution can be a gas, solid (alloy) or a liquid. • Aqueous – a liquid solution in which water is the liquid (aq) • Miscible – liquids which will dissolve in each other • Immiscible – liquids which do not dissolve in each other.

  8. Solubility and Saturation • The ability of a solvent to dissolve a solute depends on the forces of attraction between the particles. There is always some attraction between the solvent and the solute. • Solubility – the mass of the solute that dissolves in a given quantity of solvent, at a certain temperature. May be g/mL or molar solubility (moles of solute in 1 L of solution). • Saturated – no more solute will dissolve in a solution. Excess solute can be seen.

  9. Unsaturated – a solution which is not yet saturated. More solute can dissolve. • Sparingly soluble or slightly soluble – the solubility is between insoluble (less than 0.1g per 100 mL) and soluable (1g per 100 mL). • The solubility of gasses do not use the same terms as that for solids and liquids. An example is the amount of dissolved oxygen in water. 0.0009 g/100 mL is the solubility of oxygen at 20oC in fresh water. This is very soluble.

  10. Why Does Something Dissolve? • Substances of “like” polarity tend to be more soluble in each other than substances that differ in polarity. • Remember – polarity is a measure of how electrons are shared between bonding elements. It also is dependent on shapes of molecules determined by VESPR. “Like dissolves like.” Polar solvents dissolve other polar molecules and ionic compounds. Nonpolar molecules dissolve other nonpolar molecules. Alcohols which have characteristics of both tend to dissolve both polar and nonpolar but will not dissolve ionic solids.

  11. The Process of Dissolving at the Molecular Level • Step 1 – the forces between the particles in the solid must be broken. Either between the ions or between the molecules. This requires energy. • Step 2 – Some of the intermolecular forces between the particles in the liquid must be broken. This also requires energy. • Step 3 – There is an attraction between the particles of the solid and the particles of the liquid. This step gives off energy.

  12. Sugar (polar) in water (polar). • Iodine (nonpolar) in benzene (nonpolar) • Electronegativity:

  13. What type of bond? • Polar, nonpolar or ionic • N and H 3.0 – 2.1 • F and F 4.0 - 4.0 • Ca and O 3.5 -1.0 • Al and Cl 3.0 – 1.5 0.9 polar 0 nonpolar 2.5 ionic 1.5 polar

  14. What affects Solubility? • Not all substances dissolve in the same manner or amount.

  15. 1. The solubility of most substances increases as the temperature of the solvent increases. For some substances, the opposite is true. At 25oC NaCl 36.2 g/100g H2O At 100oC NaCl 39.2 g/100g H2O Sodium sulfate At 40oC 50g/100g H2O At 100oC 41g/100g H2O

  16. Gases have a higher solubility in cold water than in hot water. • …….So temperature affects the rate at which something will dissolve.

  17. 2. Pressure The solubility of a gas increases as the pressure of the gas above the solution increases. Soft drinks are bottled under high pressure. Henry’s Law At a given temperature the solubility of a gas in a liquid (S) is directly proportional to the pressure of the gas above the liquid (P). S1= S2 P1 = P2 P then S

  18. If the solubility of a gas in water is 0.77 g/L at 3.5 atm of pressure, what is the solubility in g/L at 1.0 atm of pressure? The temperature is constant at 25oC. • If S1 = S2 then S2 = S1 x P2 ___ ____ _______ P1 P2 P1 = 0.77 g/lx 1.0 atm _________________ 3.5 atm = 0.22 g/L

  19. 3. Surface Area In order to dissolve, the solvent must come in contact with the solute. Smaller pieces increase the amount of surface of the solute interacting with the solvent.

  20. The formation of Insoluble compounds (Precipitation Reactions) • When a solution of NaOH is added to a solution of CuSO4 a blue solid precipitate forms. The mixture of solutions contains Na+, OH-, Cu 2+, and SO42- ions. • When we consult a solubility table we find that the Cu(OH)2 is insoluble, the precipitate formed must be Cu(OH)2. An equation for this reaction can be written as: 2NaOH (aq) + CuSO4 (aq) Na2SO4 (aq) + Cu(OH)2 (s)

  21. In solution these compounds exist as dissociated ions as shown below: 2Na+ (aq) + 2OH- (aq) + Cu2+ (aq) + SO42- (aq) Na+ (aq) + SO42- (aq) + Cu(OH)2 (s) • Notice that the Na+ (aq) and SO42- (aq) remain unchanged by the reaction. These ions are called spectator ions because they do not react. So if we show only the ions changed by the reaction, the equation becomes: Cu2+ (aq) + 2OH- (aq) Cu(OH)2 (s) This is called the net ionic equation.

  22. Precipitation Reactions and Solubility Rules A solid dissolves in water because the water molecules around the solid’s surface collide with the particles of the solid. The attraction between polar water and a partially charged particle or ion enables the water molecules to pull these particles away from the crystal, a process called dissociation. This is followed by solvation. Predicting Precipitates We can predict the formation of a precipitate if we know the solubility of the compounds. Patterns in solubility have lead to the creation of solubility rules. Solubility rules are based on what happens when salts are placed into water, they either dissolve or they do not dissolve.

  23. Precipitation occurs because two SOLUBLE substances containing solvated ions are mixed and a pair of ions form a substance that is INSOLUBLE. The insoluble substance is the precipitate (solid) that is formed. • When you mix two solutions containing ionic compounds you can predict the products based on DOUBLE REPLACEMENT reaction. • If one of the predicted products is not soluble, it will precipitate out and a reaction occurs. • If both predicted products are soluble, no reaction occurs (all ions remain solvated).

  24. Simple Rules for the Solubility of Ionic Compounds in Water 1. Most nitrate (NO3-) salts are soluble. 2. Most salts containing the alkali metals cations (Li+, Na+, K+, Rb+, Cs+) and the ammonium (NH4+) cation are soluble. 3. Most chloride, bromide, and iodide salts are soluble. Except those containing Ag+, Pb2+, and Hg+ 4. Most sulfate salts are soluble. Except those containing Ba2+, Pb2+, Hg2+, Sr2+ and Ca2+. 5. Most hydroxide salts are insoluble except those of the alkali metal cations and a few alkaline earth metals: Ba2+, Sr2+, and Ca2+. 6. Most sulfides (S2-), carbonate (CO32-), chromate (CrO42-), and phosphate (PO43-) salts are insoluble.

  25. Will a precipitate occur? Example: Will a precipitation reaction occur when Ba(NO3)2 (aq) and Na2SO4 (aq) are mixed? Possible products from a double displacement reaction: BaSO4 (s) and NaNO3 (s) • From the solubility table BaSO4 is not soluble so it would form a precipitate. Ba(NO3)2 (aq) + Na2SO4 (aq) BaSO4 (s) + 2 NaNO3 (aq) We can also write a complete (total) ionic equation: Ba2+ (aq) + 2NO3- (aq) + 2Na+ (aq) + SO42- (aq) 2Na+ (aq) + 2NO3- (aq) + BaSO4 (s) or finally we can write a net ionic equation which shows only the ions which precipitate from solution. Ba2+ (aq) + SO42- (aq) BaSO4 (s)

  26. Example: Will a precipitation reaction occur if 150 mL of 1.0M NaI (aq) is added to 200 mL of 0.5M AgNO3 (aq)? The solubility table shows that AgI is insoluble. AgNO3 (aq) + NaI (aq) AgI (s) + NaNO3 (aq) Net ionic equation Ag+ (aq) + I- (aq) AgI (s)

  27. How do we measure how much has dissolved? • Molarity (M) is the most common method for describing the concentration of a solution, and is defined as the number of moles of solute present per litre of solution. M = n/V (V in litres) Example: Calculate the molarity of 75.0 g of KBr (s) dissolved in 500 mL of water. M= mol solute/ V (in litres)

  28. Mol KBr = 75.0 g / 119 g/mol = 0.630 mol V solution = 500 mL / 1000 mL = 0.5 L M = 0.630mol/0.5 L = 1.26 M KBr (aq) • Example: What are the concentrations of Ca2+ (aq) and NO3- (aq) in 1.5 M Ca(NO3)2 (aq)? Ca(NO3)2 Ca2+ (aq) + 2NO31- (aq) 1.5 M 1.5 M + 3.0 M

  29. Standard Solutions • Solutions prepared by dissolving a precise mass of solute in a precise volume of solvent are known as standard solutions. Volumetric flasks are used to prepare standard solutions. • Dilutions It is often necessary to dilute a certain solution. Dilution is the process by which a solution is made less concentrated by the addition of more solvent. When a solution has been diluted its solute particles are spread out more. When solutions are diluted, the amount of solute particles remains the same. Since n=MV (concentration x volume) we can use the following equation: M1V1 = M2V2

  30. Example: 200.0 mL of 3.0 M NaOH (aq) is diluted to 400 mL. What is the concentration of the diluted solution? Initail # moles = final # moles M1V1 = M2V2 (3.0 M x 0.2 L) = ( M2 x 0.4L) M2 = 1.5 M NaOH (aq) • Example: What volume of a stock solution of 6.0 M HCl (aq) is needed to make 900 mL of a 1.3 M solution? M1V1 = M2V2 (6.0 M x V1) = (1.3 M x 0.9 L) V1 = 0.2 L or 200 mL

  31. Molality • Molality (m) is the number of moles of solute per kilogram of solvent. • M= n/kg • Example: A solution contains 15.5 g of NH2CONH2 in 74.3 g H2O. Calculate the molality. • Molality = mol solute/kg solvent

  32. Mol NH2CONH2 = 15.5 g / 60.0 g/mol (molar mass) = 0.258 mol • Kg H2O = 74.3 gx1 kg 1000 g = 0.0743 Kg • m = mol solute Kg solvent = 0.258 mol 0.0743 Kg = 3.48 m

  33. Other ways to measure Solubility • Mole fraction (no units) is the number of moles of one compound in a mixture divided by the total number of moles of all components in the mixture. Xa = na/ (na + nb) • Example: What are the mole fractions of an ethanol and water solution prepared by dissolving 3.45 g C2H5OH in 21.1 g H2O? Mol ethanol = 3.45 g / 46.0 g/mol = 0.075 mol Mol of water = 21.0 g / 18.0 g/mol = 1.18 mol Total # moles = mol of ethanol + mol of water = 0.075 mol + 1.18 mol = 1.255 mol

  34. Mol fraction of ethanol = 0.075 / 1.255 = 0.05976 Mol fraction of water = 1.18 / 1.255 = 0.9402 • Mol fraction ethanol + water = 1

  35. % by mass mass of solute x 100 (mass of solute + mass of solvent) • Example: A 5 % NaCl solution would consist of 5.0 g of NaCl (s) and 95 g of water. ( I chose 100 g because it was easier to take 5%). 1 mL of H2O = 1 g H2O. So 5.0 gNaCl (s) in 95 mL of H2O (l). • Example: 24 g of NaCl (s) is dissolved in 152 g H2O (l). Calculate the % by mass of this solution. Mass % NaCl = 24 gx 100 (24 g + 152g) = 24/176 x 100 = 14 %

  36. Colligative Properties • Properties of solutions which depend on the number of solute particles but not on their nature. • They are dependent on the number of solute particles dissolved in a given volume of solvent. They include: vapour pressure lowering, boiling point elevation, freezing point depression and osmotic pressure.

  37. Solution Phase Diagram

  38. Freezing Point Depression • The presence of a solute lowers the freezing point of a solution relative to that of the pure solvent. For example, pure water freezes at 0°C (32°F); if one dissolves 10 grams of sodium chloride (table salt) in 100 grams of water, the freezing point goes down to −5.9°C (21.4°F). If one uses sucrose (table sugar) instead of sodium chloride, 10 grams in 100 grams of water gives a solution with a freezing point of −0.56°C (31°F).

  39. The reason that the salt solution has a lower freezing point than the sugar solution is that there are more particles in 10 grams of sodium chloride than in 10 grams of sucrose. Since sucrose, C 12 H 22 O 11 has a molecular weight of 342.3 grams per mole and sodium chloride has a molecular weight of 58.44 grams per mole, 1 gram of sodium chloride has almost six times as many sodium chloride units as there are sucrose units in a gram of sucrose. In addition, each sodium chloride unit comes apart into two ions (a sodium cation and a chloride anion ). The freezing point depression of a solution containing a dissolved substance, such as salt dissolved in water, is a colligative property.

  40. One can calculate the change in freezing point (ΔT f) relative to the pure solvent using the equation: Δ T f= K fm where K fis the freezing point depression constant for the solvent (1.86°C·kg/mol for water) (can be looked up on a table) ,m is molality (the number of moles of solute in solution per kilogram) of solvent for ALL particles. Because the presence of a solute lowers the freezing point, the Department of Highways puts salt on our roads after a snowfall, to keep the melted snow from refreezing. Also, the antifreeze used in automobile heating and cooling systems is a solution of water and ethylene glycol (or propylene glycol); this solution has a lower freezing point than either pure water or pure ethylene glycol.

  41. ΔTf = Kfxm What is the freezing point of these solutions? 1.4 mol of Na2SO4 in 1750 of H2O m = 1.4 mol 1.75 kg = 0.8 m 3 particles are formed when Na2SO4 ionizes 3 x 0.8m = 2.4 m ΔTf = Kfxm = 1.86 oC/mx 2.4 m = 4.46 oC 0 oC – 4.5 oC = - 4.46 oC

  42. What is the freezing point of 0.60 mol of MgSO4 in 1300 g of H2O? m = 0.60 mol 1.3 kg = 0.46 m 2 particles x 0.46 m = 0.92 m ΔTf = 1.86 oCx 0.92 m = 1.71 oC 0oC – 1.71 oC = - 1.71 oC

  43. Boiling Point Elevation • The boiling point of a solution is higher than that of the pure solvent. The actual boiling point elevation is the difference in temperature between the boiling points of a solution and the pure substance. Accordingly, the use of a solution, rather than a pure liquid, in antifreeze serves to keep the mixture from boiling in a hot automobile engine. As with freezing point depression, the effect depends on the number of solute particles present in a given amount of solvent, but not the identity of those particles. If 10 grams of sodium chloride are dissolved in 100 grams of water, the boiling point of the solution is 101.7°C (215.1°F; which is 1.7°C (3.1°F) higher than the boiling point of pure water).

  44. The formula used to calculate the change in boiling point (ΔT b) relative to the pure solvent is similar to that used for freezing point depression: Δ T b= K bm , where K bis the boiling point elevation constant for the solvent (0.52°C·kg/mol for water), and m has the same meaning as in the freezing point depression formula. Note that Δ T brepresents an increase in the boiling point, whereas Δ T frepresents a decrease in the freezing point.

  45. Boiling Point Δ Tb = Kb xm What is the boiling point of a solution that contains 1.2 mol of sodium chloride in 800 g of water? m= 1.2 mol NaCl 0.8 kg = 1.5 m Total molality = 1.5 mx 2 = 3m Δ Tb = Kbxm = 0.512 oC/mx 3 m = 1.54 oC 100 oC + 1.54 oC = 101.54 oC

  46. What is the boiling point of a solution that contains 1.25 mol of CaCl2 in 1400 g of water? m= 1.25 mol 1.4 kg = 0.89 m CaCl2 = 3 particles when it ionizes 3 x 0.89 m = 2.67 m Δ Tb = Kbxm = 0.512 oC/mx 2.67 m = 1.37oC 100 oC + 1.37 oC = 101.37 oC

  47. Calculating Molecular Mass • A solution of 7.50 g of a nonvolatile compound in 22.60 g of water boils at 100.78 oC . What is the molecular mass of the solute? • Molality of the solution: 100.78 oC – 100 oC = 0.78 oC ( ΔTb) If ΔTb = Kbxm then m = Δ Tb Kb m = 0.78 oC 0.512 oC/m m = 1.5 m

  48. Now calculate moles in solution…. • 1.5 mx 0.0226 kg = 0.034 mol Molecular mass of solute = mass of solute moles of solute = 7.5 g 0.0344 mol = 218 g/mol

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