The Nature and Properties of Solutions

1. The Nature and Properties of Solutions

2. Solutions are homogeneous (1 phase) mixtures where 1 of the components (solvent) is found in larger quantities than the rest. All other components are said to be dissolved in the solvent. These components are called solutes.

3. Both solutes and solvents can be liquids, solids, or gases. Create a chart with solutes along the side and solvents at the top which shows examples of: gas in gas, gas in solid, gas in liquid, liquid in gas, etc.

4. SOLVENT S O L U T E oxygen in air (nitrogen) oxygen in water air bubbles in ice water in air alcohol in water mercury in silver tin in copper (bronze) Sugar in water (syrup) Invisible dust in air

5. When solutes are dissolved in solvents the solutes formula is written followed by a bracketed subscript which follows. Examples: magnesium chloride is dissolved in water MgCl2(aq) iodine is dissolved in alcohol I2(al) Aqueous solutions have water as the solvent. They are always indicated by (aq) after the formula.

6. Properties of Solutions The ability to conduct electricity can be used to classify solutions. Electrolytes are substances which conduct electricity when dissolved in water. Ionic compounds are electrolytes and most molecular compounds are non-electrolytes. Solutions can also be categorized as acidic, basic or neutral. Litmus paper can be used in this determination.

7. H H Explaining Solutions Why does solid NaCl dissolve easily in water? H and O atoms in water molecules do not share electron pairs equally. O

8. H H O -ve H +ve +ve H -ve Explaining Solutions Why does solid NaCl dissolve easily in water? H and O atoms in water molecules do not share electron pairs equally. O Water molecules have oppositely charged ends. They are polar molecules

9. Moving water molecules collide with the ions of Na and Cl in solid NaCl crystals. H H H H H H H H O O O O Why does solid NaCl dissolve easily in water?

10. Here is a small crystal of NaCl Na1+ Cl1- Na1+ Cl1- Na1+ Cl1- Drop the crystal in a container of water Na1+ Cl1- Na1+

11. Here is a small crystal of NaCl Na1+ Cl1- Na1+ Cl1- Na1+ Cl1- Drop the crystal in a container of water Na1+ Cl1- Na1+

12. If the positive end of a water molecule strikes a chloride ion with enough energy it pulls it away. The same thing happens if the negative oxygen end of a water molecule strikes a Na1+ ion. H H O Na1+ Na1+ Cl1- Na1+ Na1+ Cl1- Cl1- Na1+

13. These types of interactions are called intermolecular and the NaCl crystal is dissociating. In reality each ion of Na and Cl become surrounding by a number of water molecules. H H O Na1+ Na1+ Cl1- Na1+ Na1+ Cl1- Cl1- Na1+

14. H H H H H H H H H H H H H H H H O O O O O O O O Na1+ These complexes are called hydrated ions. All ions in water become hydrated. Cl1- Na1+ Cl1- Na1+ Cl1- Na1+

15. Some substances do not easily dissolve in water.

16. When air is exhaled in water it does not easily dissolve. Why?

17. Air is made up mostly of nitrogen and oxygen. N2 and O2. Since they don't dissolve easily in water they must be non-polar. N N Since all three pairs of electrons are equally shared this molecule is non-polar. Both N2 and O2 molecules are non-polar so they are not strongly attracted by polar water molecules.

18. In general like dissolves like. Polar materials dissolve easily in polar solvents and non-polar materials dissolve easily in non-polar solvents. Water is often called the universal solvent because it dissolves so many different substances. This is due to the strong forces of attraction water molecules have on each other and on positive and negative particles in other substances.

19. The H end of one water molecule is strongly attracted to the O end of another water molecule. The special force of attraction is called a hydrogen bond and it occurs between molecules of substances with H and O, or H and N, or H and F. -ve -ve H H O O +ve +ve H H

20. H H O CH3OH +ve -ve Any molecular substance containing O atoms bonded to H atoms has polar regions which exert these attractive H bonds. For instance alcohols have OH groups. This allows them to easily mix with water. Attractive force

21. H H O CH3OH H bond Any molecular substance containing O atoms bonded to H atoms has polar regions which exert these attractive H bonds. For instance alcohols have OH groups. This allows them to easily mix with water.

22. H H O CH3OH H bond Alcohol will dissolve in water but this solution does not conduct electricity. Why? There are no mobile ions present.

23. Quantitative Aspects of Solutions

24. Solutions are homogeneous mixtures. Solutions involve 2 components. The substance doing the dissolving (solvent) and the substance being dissolved (solute). Typically the amount of solute dissolved is measured and compared to the total volume of solution. This quantity is known as the concentration of solution. A 710 mL bottle of coke has 30 g of sugar. What is the concentration in g/L (M/V)? 30 g / 0.710 L = 42 g / L

25. What mass of sugar is there in a 355 mL can of coke? 42 g / L x 0.355 L = 15 g of sugar Some solutions, like alcohol mixtures, list the quantity of alcohol as a percentage by volume since this number is bigger than the percentage by mass for solutes with a density smaller than water. A can of regular beer is 5% (V/V) alcohol by volume. What volume of alcohol is their in a 355 mL can of beer? 5/100 x 355 mL = 18 mL of alcohol

26. Which has more alcohol 45 mL of 40% (V/V) rye whiskey (typical shot) 310 mL of 7% (V/V) vodka cooler 341 mL of 4% (V/V) Coors light bottle of beer 180 mL of 12% (V/V) glass of red wine 45 mL x 40/100 = 18 mL in shot of rye 310 mL x 7/100 = 21.7 mL in the cooler 341 mL x 4/100 = 13.6 mL in the beer 180 mL x 12/100 = 21.6 mL in the wine

27. Concentration Parts per Million ppm

28. A 1.0 L sample of water is found to have 0.0012 g of lead. The molar concentration works out to be a very small number. To avoid using really small numbers for concentrations of dilute solutions another more practical scale is used. This scale is called parts per million. What is the ppm of lead for the example above? 1000 L would have 1.2 g of lead so it is 1.2 g in 1000 L or 1200 mg / 1000 L or 1.2 mg / 1.0 L

29. ppm can be expressed in a variety of ways 1 ppm = 1 g/1000 L or 1 ppm = 1 g / 1000 000 mL or 1 ppm = 1 g / 106 mL or 1 ppm = 1000 mg / 1000 L or 1 ppm = 1 mg / L Calculating ppm In a chemical analysis 3.4 mg of lead was found in 100 mL of tap water. Find the ppm of lead. ppm = 1 mg/L = 3.4 mg / 0.1 L = 34 ppm

30. What fraction of a part per million (ppm) is a part per billion (ppb)? 1/1000 So 1 ppm = ? ppb 1 ppm = 1000 ppb An even smaller concentration unit is a part per trillion (ppt) 1000 ppb = 1 ppt 1 mg in 1.0 L is 1 ppm 1 mg in 1000 L is a ppb 1 mg in 1 000 000 L is a ppt

31. Measuring Quantities of Solutes in Solutions mol mol kmol M = g dm3 m3 L L The quantity of solute can be measured in grams or moles. The total volume of the solution is measured in L. The amount of solute in a given volume of solution is measured using these units: or = = or molL-1 kmolm-3 moldm-3

32. n C = V This leads to the development of the following equation: # of moles of solute Concentration of a solution = Volume, in L, of solution

33. Preparing Solutions From Solid Reagents Sample Problem Describe how to prepare 500 mL of a 0.035 M solution of sodium thiosulfate. Given: V = 500 mL = 0.500 L C = 0.035 M Asked to Find: Mass of Na2S2O3

34. Mass Mole Use n= m/MM Concentration Use C = n/V

35. Preparing Solutions From Solutions Determining Concentrations of Concentrated Reagents Concentrations of solutions, in molL-1, can be determined from density and percentage composition.

36. Sample Problem A solution of concentrated (conc.) HCl (hydrochloric acid) has a density of 1.25 g/mL and it is 35% HCl by mass. Find the concentration of the HCl.

37. Given: density = 1.25 g/mL, 35% HCl Change density into units of mass and volume m = 1.25 g, V = 1 mL = 0.00100 L Mass Mole Concentration

38. Step 1 - Find mass of HCl 35% of 1.25 g = 0.4375 g Step 2 - Find nHCl = m/mm = 0.4375 g/ 36.45 g/mol = 0.01199 mol Step 3 - Find C = n/V = 0.01199 mol/ 0.00100 L = 12 M

39. Describe how to prepare 1.5 L of 0.75 M HCl from this concentrated reagent. Solution:Find the volume of the concentrated reagent needed to prepare the solution. Given: Cd = 0.75 M , Vd = 1.5 L Cc = 12 M, Vc = ?

40. # of moles Concentrated Reagent # of moles Diluted Reagent = CcVc = CdVd Cc Cc Vc = 0.75 M x 1.5 L 12 M = 0.094 L = 94 mL

41. 1. Get a 1.5 L Volumetric Flask 2. Measure 94 mL of concentrated HCl using gloves, apron, shield 3. Half fill the 1.5 L Volumetric flask with distilled water 4. Add the 94 mL of conc. HCl 5. Top up with distilled water to the calibration mark. AW not WA

42. Describe how to prepare 2.0 L of a 1.5 M solution of ammonium hydroxide from a concentrated reagent which is 14.5 M.

43. # of moles Concentrated Reagent # of moles Diluted Reagent = CcVc = CdVd Cc Cc Vc = 1.5 M x 2.0 L 14.5 M = 0.207 L = 210 mL

44. 1. Get a 2.0 L Volumetric Flask 2. Measure 210 mL of concentrated NH4OH using gloves, apron, shield 3. Half fill the 2.0 L Volumetric flask with distilled water 4. Add the 210 mL of conc. NH4OH. Top up with distilled water to the calibration mark.