SOLUTIONS. Engr. Yvonne Ligaya F. Musico

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1 SOLUTIONS

2 SOLUTION A homogeneous mixture of two or more substances, the relative proportion of which may vary within certain limits.

3 COMPONENTS OF SOLUTION SOLUTE component which is in small quantity SOLVENT component that is in the greatest abundance and typically determines the physical state of the solution.

4 EXAMPLE State of Solution Gas State of Solvent State of Solute Example Gas Gas Air Liquid Liquid Gas Oxygen in water Liquid Liquid Liquid Alcohol in water Liquid Liquid Solid Salt in water Solid Solid Gas Hydrogen in palladium Solid Solid Liquid Mercury in silver Solid Solid Solid Silver in gold

5 SOLUBILITY the amount of solute needed to form a saturated solution in a given quantity of solvent. SATURATED solution that can hold no more of the solute at a particular temperature. UNSATURATED solution which contains less amount of solute that is required to saturate it at that temperature.

6 FACTORS AFFECTING SOLUBILITY Solute-solvent interactions Pressure effect Temperature effect

7 Solute-Solvent Interaction The stronger the attraction between solute and solvent molecules, the greater the solubility. Polar liquids tend to dissolve readily in polar solvents. Non-polar liquids tend to be insoluble in polar liquids. Substance with similar intermolecular attractive forces tend to be soluble in one another LIKE DISSOLVES LIKE

8 MISCIBLE VS IMMISCIBLE MISCIBLE liquids that mix in all proportions. IMMISCIBLE liquids that do not dissolve significantly in one another.

9 Pressure Effect The solubility of a gas in any solvent is increased as the pressure of the gas over the solvent increases. By contrast, the solubility of solids and liquids are not appreciably affected by pressure. The solubility of the gas increases in direct proportion to its partial pressure above the solution.

10 Pressure Effect The relationship between pressure and solubility of gas is expressed by HENRY S LAW: C kp g g Where: Cg = concentration of the gas k = proportionality constant or Henry s law constant P g = partial pressure of the gas over the solution.

11 Sample Problem Calculate the concentration of CO in a softdrink that is bottled with a partial pressure of CO at 4.0 atm over the liquid at 5 o C. The Henry s law constant for CO in water at this temperature is 3.1 x 10 - mol/l - atm.

12 Solution C kp g g = (3.1 x 10 - mol/l-atm)(4.0 atm) = 0.1 mol/l or 0.1 M

13 Practice Exercise Calculate the concentration of CO in the softdrink after the bottle is opened and equilibrates at 5 o C under CO partial pressure of 3.0 x 10-4 atm. Ans.: 9.3 x 10-6 M

14 Temperature Effect The solubility of most solid solutes in water increase as the temperature of the solution. In contrast to gas solutes, the solubility of gases in water decreases with increasing temperature.

15 Concentration It refers to the weight or volume of the solute present in a specified amount of the solvent or solution.

16 QUALITATIVELY WAYS OF EXPRESSING CONCENTRATIONS QUANTITATIVELY

17 QUALITATIVELY DILUTE a solution with a relatively small concentration of solute. CONCENTRATED a solution with a large concentration.

18 QUANTITATIVELY Percent by weight Percent by volume Mole fraction and Mole Percent Molality (m) Molarity (M) Normality

19 1. Percent by Weight weight of component % by weight x100 weight of solution

20 Problem 1 A solution is made by dissolving 13.5 g of glucose, C 6 H 1 O 6, in a 0.1 kg water. What is the weight percentage of solute in a solution?

21 Solution % by weight C weight H1O6 x weight of of C H O solution 100 But, weight of solution = weight of solute + weight of solvent = 13.5 g g = g % by weight g C6H1O x % 113.5g

22 Problem A commercial bleaching solution contain 3.6% weight of sodium hypochlorite, NaOCl. What is the weight of NaOCl in a bottle containing 500 g of bleaching solution.

23 Solution % by weight NaOCl weight NaOCl weight of solution x100 weight weight NaOCl 3.6 x g (3.6)(500g) NaOCl 90.5g NaOCl 100

24 . Percent by Volume volume of solute % by volume x100 volume of solution

25 Sample Problem What is % by volume of 1 ml of alcohol in a 100 ml of wine?

26 Solution % by volume volume of volume of alcohol solution x100 % by volume 1 ml 100 ml x100 1%

27 3. Mole Fraction and Mole Percent Mole fraction of component moles of component moles of solution or moles of component Mole% moles of solution Mole% mole fraction x 100 x100

28 NOTE: The sum of the mole fractions of solute and solvent must be equal to one (1)

29 Sample Problem Calculate the mole fractions and mole % of ethyl alcohol, C H 5 OH, and water, H O, in a solution by dissolving 13.8 g of alcohol in 7 g of water.

30 Solution Mole C Mole H H O 5 OH weight CH5OH MW C H OH weight H O MW H O g 0.30 mole 46 g / mole 7 g 1.5 mole 18 g / mole Moles of solution mole of solute mole of solvent moles

31 Solution 83.3% (0.833)(100) 100 % 16.7% (0.167)(100) 100 % O x H of fraction Mole O H Mole x OH H C of fraction Mole OH H C Mole OH H C of fraction Mole O H of fraction Mole or moles moles O H of fraction Mole moles moles OH H C of fraction Mole solution of moles OH H C of moles OH H C of fraction Mole

32 4. Molality (m) Molality ( m) moles of solute kilogram of solvent

33 Problem Calculate the molality of a solution made by dissolving 6 g of ethylene glycol, C H 6 O, in 8000 g of water.

34 Solution m kg mole m mole mole g g O H MW C O H C g O H C of moles O H kilogram of O H C of moles m Molality / 6 6 ) (

35 5. Molarity (M) Molarity ( M ) moles of solute Liter of solution

36 Problem Calculate the molarity of a solution made by dissolving 4.0 g of calcium bromide, CaBr, in enough water to give 00 ml of solution.

37 Solution Molarity ( M ) moles of CaBr Liter of solution M moles of mole L CaBr 0.1 M weight of CaBr MW CaBr 4.0 g 00 g / mole 0.0 mole

38 6. Normality (N) Normality ( N) equivalent of solute Liter of solution equivalent of solute weight of solute equivalent weight of solute

39 Equivalent Weight weight of substances that are equivalent to one another in chemical reaction. The equivalent of an oxidizing agent or reducing agent is the weight of the substance required to gain or lose 1 mole of electron.

40 Equivalent Weight Equivalent Weight = molecular weight (MW) f (factor)

41 Reacting Capacity of Solute ACID BASE SALT f Total number of H + replaceable Total number of replaceable OH - Total (+/-) charge

42 Sample Problem 1 Calculate the equivalents of: (a) 0 g of H SO 4 (b) 50 g of Mg(OH) (c) 30 g of NaOH

43 Sample Problem Calculate the normality of a Na CO 3 solution containing grams in 15 ml solution.

44 Solution N = equivalent of solute Liter of solution N = (0.59) (0.15) N =.075 N equiv. of Na CO 3 = wt./equiv. wts. = g. (106 g/mol)/( equiv/mol) = 0.59 equiv/mol

45 RELATIONSHIP BETWEEN NORMALITY AND MOLARITY N = n solute x f liter of solution N = molarity (M) x f

46 Sample Problem 5 Find: (a) The molarity (M) of 0.1 N H 3 PO 4 (b) The normality (N) of 0.5 M NaOH (c) The normality (N) of.50 M HCl

47 Dilution It refers to the reduction of concentration of a solution. The most important thing to remember concerning dilution is that you are only adding solvent. You are not adding solute when you dilute.

48 Dilution Therefore: moles of solute before dilution = moles of solute after dilution or Number of millimoles of solute before = millimoles of solute after dilution or Number of grams of solute before dilution = number of grams of solute after dilution Since the definition for Molarity is: Molarity = moles of solute / volume of solution in liters Solving for moles of solute gives: moles of solute = M x V of soln in liters or millimoles of solute = M x V of solution in ml

49 Dilution If Moles of solute before dilution = moles of solute after dilution Then M x V in liters before dilution = M x V in liters after dilution or M 1 V 1 = M V where M 1 = Molarity before dilution V 1 = volume of solution before dilution M = Molarity of solution after dilution V = Volume of solution after dilution

50 Dilution Actually one can use Molarity or mass % as the concentration term so you could have the following alternative where mass % is used: mass % x grams solution before dilution = mass % x grams of solution after dilution

51 Sample Problem 1 How would you prepare 500 ml of 3 M HCl using 6 M HCl from the stockroom.

52 Solution 1. Determine the volume of 6M of HCl to use applying the dilution equation M V 1 6M V 3M 500 ml V V M V 3M 500 ml 6M 50 ml of 6M HCl

53 Solution. Determine the amount of water to be added to 6M HCl Since the total volume after dilution is 500 ml and the volume of 6M to use is 50 ml then: Volume of water = = 50 ml of water

54 Sample Problem A chemist starts with 50.0 ml of a 0.40 M NaCl solution and dilutes it to ml. What is the concentration of NaCl in the new solution?

55 Solution M V M 50 ml M 1000 ml M M V.4M 50ml 1000 M 0.0M

56 Sample Problem 3 How would you prepare 800 grams of a 3% Hydrogen Peroxide solution using 10% H O solution

57 Solution 1. Determine the mass of 10% H O to be used using the dilution equation mass of H O before dilution = mass of H O after dilution mass % x mass of solution before dilution = mass % x mass of solution after dilution 10 ( mass of solution before) = (3) (800) mass of solution before = (3) (800) 10 mass of solution before = 40 grams

58 Solution. Determine the mass of water to be added Total mass of 3% = mass of 10% + mass of water added mass of water to be added = mass of water to be added = 560 g

59 Sample Problem 4 A chemist wants to make 500. ml of M HCl by diluting a 6.0 M HCl solution. How much of that solution should be used?

60 Solution M V 1 6.0M V 0.050M 500 ml V V M V 0.05M 500ml 6.0M 4.17 ml of 6.0M HCl

61 Practice Exercise 1. How much.0 M NaCl solution would you need to make 50 ml of 0.15 M NaCl solution?. How would you prepare 1000 ml of a 5% glucose solution using a 0% glucose solution. How much 0% glucose and how much water to be mixed? 3. What would be the concentration of a solution made by diluting 45.0 ml of 4. M KOH to 50 ml? 4. What would be the concentration of a solution made by adding 50 ml of water to 45.0 ml of 4. M KOH? 5. How much 0.0 M glucose solution can be made from 50. ml of 0.50 M glucose solution?

62 Thank You for Listening

SOLUBILITY AS AN EQUILIBRIUM PHENOMENA

SOLUBILITY AS AN EQUILIBRIUM PHENOMENA Equilibrium in Solution solute (undissolved) solute (dissolved) Solubility A saturated solution contains the maximum amount of solute that will dissolve in a given