Solution 2 : Colligative Properties

COLLIGATIVE PROPERTIES

1. The Vapor Pressure of Solutions Depression = Penurunan Tekanan Uap Jenuh Larutan Vapor pressure is pressure when a liquid and its vapor is in the reversible state. Francois M. Raoult (1830-1901) did the research on this field, and he stated:”The decrease of vapor pressure equals to the ratio of the number mole of solute and total number of moles in the solution” Solution SBI 2008 8

po - ps po

=

n

Colligative Properties(Cont’d)

n+N

po = vapor pressure of pure liquid ps = vapor pressure of solution n = number of moles of solute N = number of moles of solvent This formula can also be expressed as: p = po . X p = vapor pressure decrease (po - ps) X = mole fraction of solute Solution SBI 2008 9

Colligative Properties(Cont’d)

The Raoult law can not be used for electrolyte solution. For electrolyte solution we must consider the factor i which is the ionization factor. The value of i is: i = (1 + (n-1)α); i = van’t Hoff factor n = the number of ions produced α = degree of ionization For example: HCl ionized as HCl  H+ + Cl-, number of n is 2, if α is 1 then; i = (1 + (2-1) 1)=2 α For strong electrolyte close to 1 Solution (~1), but SBIweak 2008 10

Colligative Properties(Cont’d) - There are three type of electrolytes: 1. Strong electrolyte: it can be dissolved perfectly in water. Examples of this are salts, strong base and acid; such as: NaCl, KCl, MgCl2, ZnSO4, HCl, CuSO4 , H2SO4, NaOH, KOH, Mg(OH)2, HCl etc 2. Weak electrolyte: only part of it dissolves in water. Examples of this are weak base and weak acid, such as: acetic acid (CH3COOH), ammonium hydroxide (NH4OH), cyanide acid (HCN) 3. Non electrolyte: the substance that can not conduct an electric current. Example of this is many of organic compound; such as sugar (C12H22O11), glucose (C6H12O6), benzene (C6H6) etc. Solution SBI 2008 11

Colligative Properties(Cont’d)

Example: Determine the vapor pressure of sugar solution 34.2% at 20°C (vapor pressure of water is 17.512 mmHg) ! (Mr sugar 342, water=18) Answer: 1. We must find the mole fraction of sugar solution Assume the mass of solution is 100 g, so the mass of sugar based on the mass percent is: Sugar mass = (34.2%/100%) x 100 g = 34.2 g Mass of water is = 100 g – mass of sugar = 100 g – 34.2 g = 65.8 g Solution SBI 2008 12

Colligative Properties(Cont’d)

Calculate mole of each component Mole of sugar= (34.2 g/342 g/mole) x 1 mole = 0.1 mole Mole of water= (65.8 g/18 g/mole) x 1 mole = 3.7 moles Mole fraction (X) sugar= 0.1/(0.1 + 3.7) = 0.0263 So, the decrease of vapor pressure is ∆p = po x Xsugar = 17.512 x 0.0263 = 0.4606 mmHg So the pressure of solution is: po - ∆p = 17.512 – 0.4606

Solution SBI 2008 13

Colligative Properties(Cont’d)

Question 1. Calculate the vapor pressure of solution of 19.6 g H2SO4 (Mr 98) and 90 g of water! The pressure of water in this condition is 74.1 mmHg and α H2SO4 is 0.89 Answer: Calculate mole of each components Mole H2SO4 = 19.6/98 = 0.2 Mole H2O = 90/18 = 5 X H2SO4 = 0.2/5.2 = 0.038 H2SO4  2H+ + SO42-

Solution SBI 2008 14

Colligative Properties(Cont’d)

the number of ions = 3; I = (1 + (3-1).0.89) = 2.78 ∆p = po x X x i = 74.1 x 0.038 x 2.78 = 7.923 The pressure of the solution = 74.1 – 7.923 = 66.177

Solution SBI 2008 14

Colligative Properties(Cont’d)

2. Boiling Point Elevation (Kenaikan titik didih) Boiling point of a liquid (Tb) is temperature when the vapor pressure of the liquid equal to surround pressure. Raoult formulated this phenomena as: ∆Tb = Kb . m ∆Tb = Boiling Point Elevation Change Kb

= molal boiling point constant (°C/mole)

m = molality For electrolyte solution: ∆Tb = Kb . m . i The value of Kb depends on the solvent used. Solution SBI 2008 14

Colligative Properties(Cont’d)

To calculate the boiling point of the solution (T) is then= The boiling point of the solvent + ∆Tb Example 1: Calculate the boiling point of solution made of 12.4 g of ethylenglicol (C2H6O2) and 400 g of water! Kb of water 0.52 °C/mole Answer: calculate the m of the solution m = (12.4/62) x (1000/400) = 0.5 Raoult formulated this phenomena as: ∆Tb = Kb . m = 0.52 °C/mole x 0.5 = 0.26°C So the solution will boil at = 100 + 0.26° =

Solution SBI 2008 15

Colligative Properties(Cont’d)

Question 2: A chemical substance wants to be determined its molecular relative mass. 10 g of this substance is dissolved in 500 g of water. If Kb of water is 0.52 °C/mole, then the solution boils at 100.1733°C. Question 3. 10.125 g of HBr (Mr 81 g/mole) is dissolved in 250 g of water. The solution boils at 100.4628°C. If Kb of water is 0.52 °C/mole, determine the α of HBr! Solution SBI 2008 16

Colligative Properties(Cont’d)

3. Freezing Point Depression (Penurunan titik beku) Freezing point (Tf) is temperature when liquid phase and solid phase is in equilibrium. Raoult formulated this phenomena as: ∆Tf = Kf . m ∆Tf = Freezing Point Depression Change Kf

= molal feezing point constant (°C/mole)

m = molality For electrolyte solution: ∆Tf = Kf . m . i The value of Kf depends on the solvent used. Solution SBI 2008 17

Colligative Properties(Cont’d)

To calculate the freezing point of the solution (T) is then= The feezing point of the solvent + ∆Tf Example 1. Determine a freezing point of solution composed of 45 g of glucose (Mr 180) in 2000 g of water (Kf water is 1.86 °C/mole) Answer: calculate m of the glucose solution m = (45/180) x (1000/2000) = 0.125 Wit Raoult formula : ∆Tf = Kf . M = 1.86 x 0.125 = 0.2325 °C The freezing point is then = 0 – 0.2325 = -.2325°C

Solution SBI 2008 18

Colligative Properties(Cont’d)

Question 4 5 g of non electrolyte soultion in 1000 g of water (Kf water is 1.86 °C/mole) freeze at -0.202°C. Calculate the Mr of this non electrolyte. Question 5. 29.4 g of H3PO4 (assume α as 0.78) dissolves in 600 g of water. In what temperature this solution will freeze?

Solution SBI 2008 19