9.01 Solutions. The Chemistry of Matter in Water. Dr. Fred Omega Garces. Chemistry 100, Miramar College. 1 Solutions. Aug 17

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1 9.01 Solutions The Chemistry of Matter in Water Dr. Fred Omega Garces Chemistry 100, Miramar College 1 Solutions

2 8.01 Solutions How water Dissolves Salts 2 Solutions

3 Components of Solution Homogeneous systems : Solutions Solution - Homogeneous mixture of two or more substances Components of solution Solute - Substance being dissolve Solvent - Substance in which solute is dissolved in. If solvent is water, then solution is considered Aqueous. 3 Solutions

4 9 Types of Solution (derived from 3 phases) Solute Solvent Solution Gas in a gas Air Gas in a liquid carbonated water Gas in a solid whipped cream, foam Liquid in a gas fog Liquid in a liquid wine. Solid in a solid 14 karat gold Solid in a liquid salt water. 4 Solutions

5 Properties of Solution Characteristics: Distribution of particles is uniform Components in solution do not separate upon standing Components cannot be separated by filtration. Solute / Solvent mixes in ratios - up to the solubility limit. Solution is almost always transparent. Compounds of solution may be separated by other methods i.e., distillation or chromatography. 5 Solutions

6 Properties of Solution Characteristics: Distribution of particles is uniform Components in solution do not separate upon standing Components cannot be separated by filtration. Solute / Solvent mixes in ratios - up to the solubility limit. Solution is almost always transparent. Compounds of solution may be separated by other methods i.e., distillation or chromatography. 6 Solutions

7 Suspended in Solution: Dissolution Solubility - The process in which substances dissolves at the molecular level. Solubility - The maximum amount of that solute that dissolves in a given amount of solvent in a given temperature. Unsaturated Saturated SuperSaturated A solution that has the capacity to dissolve solute A solution that contains all solute it can dissolve (There are no residue) A solution that contains more solute (in dissolved form) than normal Miscible When two liquids are soluble in each other Immiscible When two liquids are not soluble in each other. 7 Solutions

8 Suspended in Solution: Dissolution Supersaturated Solution. SuperSaturated A solution that contains more solute (in dissolved form) than normal 8 Solutions

9 Nature of Solute and Solvent Dissolving Process: Why does water not mix with oil? Yet water mixes with alcohol. An oil layer floating on water. For a Solubility Factor: Solute and Solvent characteristic: In aqueous solution, water as the solvent will dissolve only other polar molecules Oil is a nonpolar substance which will only dissolve other nonpolar substances such as organic solvents. The result is the immisciblity of the two liquid. Like Dissolves Like substance to dissolve, the water-water hydrogen bonds must be broken to make a hole for each solute particle. However, the water-water interactions will break only if they are replaced by similar strong interactions with the solute. 9 Solutions

10 Dissolution of Solid Solute What is the driving force which cause solutes to dissolve to form solutions? Ionic versus covalent solute 1. Ion solutes dissolve by dissociation into their ions. 2. Covalent solutes dissolve by H-bonding to water or by LDF Picture of NaCl and Ethanol dissolving 10 Solutions

11 Dissolving at the molecular Level Ionic substances NaCl, MgCl 2, AgCl The salt; NaCl and ionic compound (3D lattice) O H H O H H - H H H δ+ H H δ+ H δ δ+h H δ+ + δ δ δ H H δ+ δ+ H δ+ H δ+ The Na+ and Cl- are arranged in 3 D alternating lattice The positive Cl- attracts the (d+) of hydrogen in H 2 O while the negative Na+ attracts the (d -) of oxygen in H 2 O. This occurs because opposite attracts. The solvent (H 2 O) interaction to the ion is called Hydration. H 2 O literally pulls the lattice arrangement of the solid salt apart because of the attraction of the d+ or d- of water for the - or + ions. Competing factors are: 1) M + & X - (i.e., Na + and Cl - ) 2) M + for d - oxygen of H 2 O & X - for d + hydrogen of H 2 O. If (1) is favorable (lattice energy) then solute does not dissolve (insoluble) AgCl if (2) is favorable then hydration, solute does dissolve (soluble). NaCl 11 Solutions

12 Equilibrium: Dissolution = Crystallization Observe: After some time, no change in amount of solid precipitate at the bottom of the beaker. Concentration of the solution is maintained At the molecular level: Amount of salt dissolving into solution equals to amount of salt recrystallize out of solution. [Solute] 12 D [Solute] Solid Solution Solutions

13 Dynamic Equilibrium Equilibrium Situation in which changes occur at equal rates so no net change is apparent. LeChatelier Principle A change (stress) on a system at equilibrium will cause the system to self adjust to reduce the stress until a new equilibrium is reestablish. Example: Traffic at a toll bridge Vapor Pressure Sugar dissolving 13 Solutions

14 Ionic Vs Covalent Compounds: Electrolyte Vs. Nonelectrolyte Substance when dissolve can break-up to ions (NaCl) or stay intact (sugar). Type: % ionization: Solubility Electrolyte: conducts electricity. Strong electrolyte 100 % ionization very soluble weak electrolyte less 100% ionization slightly soluble zero ionization insoluble ionic or covalent. Nonelectrolyte: no conduction 14 Solutions

15 Factors Affecting Solubility Nature of Solute / Solvent. - Like dissolves like (IMF) Temperature Factor - i) Solids/Liquids- Solubility increases with Temperature Increase K.E. increases motion and collision between solute / solvent. ii) gas - Solubility decreases with Temperature Increase K.E. result in gas escaping to atmosphere. Pressure Factor - i) Solids/Liquids - Very little effect Solids and Liquids are already close together, extra pressure will not increase solubility. ii) gas - Solubility increases with Pressure. Increase pressure squeezes gas solute into solvent. 15 Solutions

16 Principles of Solubility Homogeneous systems: Solutions Solution - Homogeneous mixture of two or more substances Components of solution Solute - Substance being dissolve Solvent - Substance in which solute is dissolved in. If solvent is water, then solution is considered aqueous. 16 Solutions

17 Solution Composition Homogeneous systems: Solutions Solution - Homogeneous mixture of two or more substances Components of solution Solute- Substance being dissolve Solvent-Substance which dissolves solute. If solvent is water, then solution is considered aqueous. 17 Solutions

18 Solution concentration Concentration- Amount of solute in given amount of solvent. Most common type: Molarity Molarity - moles solute / Liter solution How many moles in 16.0 g CuSO 4 (MW = g/mol)? If this amount of CuSO 4 is dissolved in 0.100L of water what is the Molarity of the solution (mol / L)? To make a 1.00-M solution ofcuso 4, 16.0 g, or mol, of CuSO 4 (the blue crystalline solid) is placed in a 100-mL volumetric flask. Exactly 100 ml of water is measured out and slowly added to the volumetric flask. When enough water had been added so that the solution volume is exactly 100 ml. The emphasizes here is that molar concentrations is defined as moles per liter of solution and not per liter of water or other solvent. 18 Solutions

19 Solution and Concentration 5 ways of expressing concentration- Molarity(M) - moles solute / Liter solution Mass percent (% m)- (grams solute/total grams of solution)*100 Molality * (m) - moles solute / Kg solvent Normality(N) - Number of equivalent / Liter solution mole fraction( c A) - moles solute / Total moles solution But learn the main two * Note that molality is the only concentration unit in which denominator contains only solvent information rather than solution. 19 Solutions

20 Concentration Relationship Molecular Weight Moles mass } Solute Molc Wt Density Moles mass Volume } Solvent c m* %m M * Volume used must be volume of solvent N 20 Solutions

21 % Concentration % Concentration w/w = Wt Solute 100 g 100 g % (pph) Wt Soln g w/v = Wt Solute 100 g 100 g % (pph) Vol Soln ml v/v = Vol Solute 100 ml 100 g % Vol Soln ml ppm & ppb (For dilute solid in liquid solution) m/v = Wt Solute 1,000,000 g 10 6 g ppm (ppm) Vol Soln ml m/v = Wt Solute 1,000,000,000 g 10 9 g ppb (ppb) Vol Soln ml 21 Solutions

22 % Concentration: % Mass Example Example#1 3.5 g of CuSO4 is dissolved in 100mL solution. Assume the density of the solution is 1.0 g/ml, what is concentration of the solution in % mass? %m = 3.5 g CuSO4 100g Solution = 3.5% (m/m) 22 Solutions Example#2: What is the v/v (%) & (ppm) concentration if one drop of alcohol (1/20 ml = 0.05 ml) is added to 1.00L? Anws: %(v/v) = (0.05mL/1000mL) 100 = % (pph) ppm(v/v)= (0.05mL/1000mL) 106 = 50 ppm Example#3: The legal BAC in California cannot exceed.080% of a person s blood by volume. How much alcohol (oz & ml) if a 160lb person has 4.7 L blood and has 0.12% BAC. 1 L = 1000ml, ml = 1.0 oz Anws: 5.64 ml or oz

23 Shark Sense Example#3 A shark can smell blood in water from several miles away. What is the concentration of 1 drop blood in 1mi 3 volume? Anws: Vol = 1mi 3 (5280 ft) 3 (12 in) 3 (2.54 cm) 3 1ml (1 mi) 3 (1 ft ) 3 (1 in ) 3 (1 cm) 3 Vol = ml ppt= (0.05mL/ ml) 10 9 = ppb = ppt = 12 ppquintlllion* *Quintillion = Solutions

24 Concentration: Molarity Example Suppose g of KMnO 4 is dissolved in enough water to give 250. ml of solution. (i) How many moles of KMnO 4 was used? (ii) What is the molar concentration of KMnO 4? In most stoichiometry problems, the first step is to convert the mass of material to moles. Molar Mass KMnO 4 = g/mol 250 ml 24 Solutions

25 Concentration: Molarity Example (1 st ) Suppose g of KMnO 4 is dissolved in enough water to give 250. ml of solution. (i) How many moles of KMnO 4 was used? (ii) What is the molar concentration of KMnO 4? In most stoichiometry problems, the first step is to convert the mass of material to moles. Molar Mass KMnO 4 = g/mol i) g KMnO 4 1 mol KMnO 4 = mole KMnO g KMnO 4 Now that the number of moles of substance is known, this can be combined with the volume of solution which must be in liters to give the molarity ml is equivalent to L. ii) Molarity KMnO 4 = mol KMnO L solution = M 25 Solutions

26 Molarity: second example i) What is the molar concentration of 1.00 g potassium dichromate (K 2 Cr 2 O 7, MW = g/mol) in 250mL solution? ii) What is the concentration as % w/v? Molarity = moles Solute mol g Molar (M) (M) L solution L 26 Solutions

27 Molarity: Second example i) What is the molar concentration of 1.00 g potassium dichromate (K 2 Cr 2 O 7, MW = g/mol) in 250mL solution? Molarity = moles Solute mol g Molar (M) (M) L solution L Calculate the concentration of 1.00 g K 2 Cr 2 O 7 in 250 ml solution? given goal Mass(g) M (g/l) Vol (L) Molar mass K 2 Cr 2 O 7 : g/mol Answer: C g D g F Molarity: 1.00 g mol 1 = mol = M g 0.250L L ii) What is the concentration as % w/v? Assume % w/v = 1.00g K 2 Cr 2 O 7 x 100 = % w/v (or w/w) density = 1.0 g/cc 250 ml solution 27 Solutions

28 Solution Preparation (Molarity as a conversion factor) When a targeted concentration is sought: What mass of K 2 Cr 2 O 7 is needed to prepare 500mL of M solution given Conc M Vol 0.500L Molar mass K 2 Cr 2 O 7 MW =294.2 g/mol goal mass (g) 28 Solutions

29 Solution Preparation (Molarity as a conversion factor) When trying to prepare a specific concentration: What mass of K 2 Cr 2 O 7 is needed to prepare 500.mL of M solution given Conc M Vol 0.500L Molar mass K 2 Cr 2 O 7 goal mass (g) MWt = g/mol Strategy: F g D g C L mol g = g L mol L 0.20 mol g = 29.4 g L mol Weigh 29.4 g of K 2 Cr 2 O 7 placed in 500mL Vol flask and dilute to mark 29 Solutions

30 Dilution Process Dilution: When the number of Solute:Solvent ratio decreases because the Volume of solvent is increased. 1Can OJ concentrated is diluted to prepare 4-can volume of OJ 30 Solutions

31 Dilute Solutions HNO 3 dilution: Suppose 10 ml of the M HNO 3 solution is diluted to 50.0mL. What is the new concentration? 10 ml M HNO 3 in 50 ml Volume. What is Conc? 100 ml of M HNO 3 Given: C1 = C2 = V1 = V2 = Goal or Methodology M 1 V 1 = M 2 V 2 C 1 V 1 = C 2 V 2 31 Solutions

32 Dilute Solutions Dilution Analysis: Suppose 10 ml of the M HNO 3 solution is diluted to 50.0mL. What is the new concentration? 100 ml of M HNO 3 10 ml M HNO 3 in 50 ml Volume. What is Conc? C 1 V 1 = C 2 V 2 Before Dilution: 10 ml Aliquot: 10ml of M Moles in aliquot- Moles = mol L L = mol After Dilution: 50 ml Volume: 50 ml vol. contains mol Moles of new solution mol = M new Vol new = M new L M = M new 32 Solutions

33 Dilution Example Example (100 RS): There is a bottle of M sucrose stock solution in the lab. Give precise directions to your assistant to prepare ml of a M sucrose solution. 240mL ml of M sucrose Suppose you have M sucrose stock solution. How do you prepare 250 ml of M sucrose solution? Concentration M Sucrose C 1 V 1 = C 2 V 2 C 1 =? V 1 =? C 2 =? V 2 =? Try to solve this Problem 33 Solutions

34 Dilution Example in Analytical Chemistry 10.0 g MnSO4 H2O is placed in a 1-L volumetric flask. What volume is necessary to prepare 250mL of 0.050% solution? Assume d soln 1 g/ml 10.0g MnSO 4 H 2 O 1 L 1 L 1000 ml 100 = 1.00% solution C 1 V 1 = C 2 V 2 V 1 = C 2 V 2 C 1 C 1 = 1.00 % V 1 =? V 1 = C 2 V 2 C 1 = 0.05 % 250 ml 1.00 % C 2 = % V 2 = 250 ml V 1 = 12.5 ml of 1.00 % Solution Aliquot 12.5 ml of 1.00% in to 250mL flask Dilute to the 250mL mark with water. 34 Solutions

35 Solution Stoichiometry How is concentration used in stoichiometry problems? 35 Solutions

36 Solution Stoichiometry: Titration From Previous problem: Suppose ml of the M HNO 3 is titrated with NaOH. What volume of M NaOH is necessary to neutralize the acid solution. 36 Solutions

37 Solution Stoichiometry: Titration From Previous problem: Suppose ml of the M HNO 3 is titrated with NaOH. What volume of M NaOH is necessary to neutralize the acid solution. Reaction: HNO 3 (aq) + NaOH (aq) H 2 O (l) + NaNO 3 (aq) 6.8 e -3 M 1.0 e -3 M F g D ml V? Moles HNO 3 = 0.100L x (6.8e -3 mol / 1 L) = 6.8 e -4 mol HNO 3 D g 4 At neutralization, moles HNO 3 = moles NaOH = 6.8 e -4 mol NaOH 4 g 6 Volume NaOH = 6.8 e -4 mol NaOH x (1 L / 1.00e -3 mol NaOH) = L Answer: Vol NaOH = ml 37 Solutions

38 Solution Stoichiometry: Titration End of Chapter: Practice problems Solutions

39 Solution Stoichiometry: Titration End of Chapter: Practice problems Solutions

40 Solution Stoichiometry: Titration End of Chapter: Practice problems Solutions

41 Solution Stoichiometry: Titration End of Chapter: Practice problems a) AA: 750ml 4g 100mL = 30 g: 1day 60 g 25g Cb: 750ml = g: 1day 375 g 100mL Lp: 500ml 10g 100mL = 50 g: 1day 100 g b) Total Kcal = 60g ( 4Kcal 4Kcal 9Kcal ) + 375g( ) + 100( ) = 2640 Kcal g g g 41 Solutions

42 Solution Stoichiometry: Titration End of Chapter: Practice problems a) AA: 750ml 4g 100mL = 30 g: 1day 60 g Cb: 750ml 25g 100mL = g: 1day 375 g Lp: 500ml 10g 100mL = 50 g: 1day 100 g b) Total Kcal = 60g ( 4Kcal 4Kcal 9Kcal ) + 375g( ) + 100( ) = 2640 Kcal g g g 42 Solutions

43 Solution Stoichiometry: Titration End of Chapter: Practice problems KCl solution volume = 15mL = 0.015L i) Evap Dish = g ii) Evap Dish + KCl solution = g iii) Evap Dish + KCl residue = g Mass Residue (iii - i) = g = 4.18 g Mass solution (ii - i) = g = g a) m:m KCl = 4.18 g g b) Molarity KCl = 4.18 gkcl 100 = 24.0 % mol g L = 3.74 M c) C 1 V 1 = C 2 V 2 : C 2 = C 1 V 1 V 2 = 3.74 M 10mL 60 ml = M 43 Solutions

44 Solution at a Glance Solutions can be describe by the following: Solvent The component of a solution present in the greatest quantity Solute The component of solution present in the lesser quantity Solution A homogeneous mixture of two or more substances in which each substance retains its chemical identity Concentration of a Solution The amount of solute in a specific amount of solution. Molarity (M) moles of solute Liters of solution 44 Solutions