Solution 3 : Colligative Properties Continued

Colligative properties 4. Osmotic Pressure Osmosis is a process of permeating of a dilute solution to more concentrate solution through a semipermeable membrane. Semipermeable membrane: is thin layer which can only be passed through by solvent molecules but it cannot be passed by the solute molecules.

Solution SBI 2008 19b1

Colligative properties (cont’d) As an example of osmosis process can been in the following figure, where two sugar solutions with different concentration placed in one beaker glass separated by semipermeable membrane with concentration of solution 1 is more concentrated than solution 2. After the osmosis, the volume of solution 2 is decreased, as it was permeated through membrane into solution 1. While the sugar molecules remain the same. So solution 2 is now becomes more concentrated Solutionthan SBI 2008 19b2

Colligative properties (cont’d) Semipermeable membrane Solution 1 0 o 0 o0 o 0 0 0 0 o o o o 0 o 0o 0

Solution 2 0 o

0

0

o 0 0 00 o 0 o 0 0 0

Initial condition (before osmosis occurred)

Solution 1

Solution 2

0 0 0 o0 0 o 0 0o 0 0 0 o o o o 0 o 0o 0

o o 0 0 00 o 0 o 0 0 0

After osmosis process occurred

o = sugar molecules 0 = water molecules Solution 1 is more concentrated than solution 2

Solution SBI 2008 19b3

Colligative properties (cont’d) Osmotic pressure was formulated by Jacobus Henricus van’t Hoff as: Л = M.R.T where Л = osmotic pressure (atm) M = molar concentration R = universal gas constant (0.082 atm.l/mole.K) T = absolute temperature (K) = °C + 273 For electrolyte solution, the formula used is: Л = M.R.T.i with i = (1 + (n-1)α)

Solution SBI 2008 19b4

Colligative properties (cont’d) Example of Osmotic pressure 17.1 g of sugar (Mr 342) is dissolved to 500 mL of water. Calculate the osmotic pressure at 25°C. Answer: First calculate M of sugar: M=(mass/Mr)x(1000/v)= (17.1/342)x(1000/500) = 0.1 M T = 25 + 273 = 298 K, then put the data into the formula  Л = M.R.T= 0.1 x 0.082 x 298

Solution SBI 2008 19b4

Question for Colligative properties Question 1. The osmotic pressure of KOH in 5.0 l at 27°C is 9.56 atm. If α = 1 and Mr KOH 74, calculate the amount of KOH dissolve in this solution Question 2. If we are provided the following data: 10 % by mass of NaOH (Mr 40). The vapor pressure of water is 74.1 mmHg, Kb: 0.52 °C/mole, Kf: 1.86°C/mole, ρ: 1 g/mL and α = 0.95, at 25°C Determine: the vapor pressure, boiling point,

Solution SBI 2008 19b5

Definition of Acid, Base and Salt 1. Acid according to Arrhenius is a substance that produces hydrogen ions (H+) in solution , while according to Brosnted-Lowry is a proton donor. HCl(aq)  H+(aq) + Cl-(aq) or HCl(aq) + H2O(l) H3O+(aq) + Cl(aq)

The equilibrium constant value (Ka) of this acid is: +

Ka =

-

[H ][Cl ] [HCl]

+

or Ka =

-

[H3O ][Cl ]

[HCl] [H2O]

The concentration of water is constant (1) as it is as the solvent

Solution SBI 2008 20

Definition(Cont’d)

H2SO4 is also acid. Why? Because it can also release H+, based on this reaction H2SO4(aq)  2H+(aq) + SO42-(aq) Can you write the constant equilibrium value for this acid? There are two kinds of acid: strong and weak. The specific sign of weak acid is the Ka value always given (for strong acid, normally not written) Strong acid: HCl, H2SO4, HNO3, HBr, HI, HClO4

Solution SBI 2008 21

Definition(Cont’d)

2. Base according to Arrhenius is a substance that produces hydroxide ions (OH-) in solution , while according to Brosnted-Lowry is a proton acceptor. NaOH(aq)  Na+(aq) + OH-(aq) Ca(OH)2(aq)  Ca2+(aq) + 2OH-(aq) The equilibrium constant value (Kb) of these base + 2+ - 2 [Ca ][OH ] are:[Na ][OH ] Kb = Kb = and [NaOH] [Ca(OH)2] Solution SBI 2008 22

Definition(Cont’d)

There are also two kinds of base: Strong and weak. The specific sign of weak base is the Kb value always given (for strong base, normally not written) Strong base: NaOH, KOH, Ca(OH)2, Mg(OH)2, Ba(OH)2 Weak base: NH4OH Besides Ka for weak acid and Kb for weak base, the

Solution SBI 2008 23

Definition(Cont’d)

3. Salt is an ionic compound, composed of left over of an acid and cation. Salt is normally formed as a result of reaction between acid and base. Example: HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) salt NaCl  Na+ + Clcation left over acid (from HCl) Solution SBI 2008 24

Definition(Cont’d)

If we have K2SO4, from which acid and base, is it formed? K2SO4 ionize as the following: K2SO4  2K+ + SO42cation left over acid (from H2SO4) So K2SO4 is from reaction of 2KOH + H2SO4  K2SO4 + 2H2O Water is always produced in the reaction of acid and base, so sometimes we call this reaction Solution SBI 2008 25 as neutralization reaction.

Definition(Cont’d)

When we react acid and base, we can find concentration of one compound if other is known. The process of the determination of acid/base by base/acid is known as titration. In laboratory, normally base is put in the burette, while the acid is in Erlenmeyer. To know when the reaction is finished (the concentration of acid and base equal), we used an indicator to give a certain color as a sign where the reaction has completed. Indicator is a chemical that changes color and used to mark the end point of titration. Solution SBI 2008 26

Definition(Cont’d)

The formula used to calculate the unknown sample is: V1.M1.n1 = V2.M2.n2 or we can write as = V1.N1 = V2.N2 Where V1 = volume of solution 1 M1 = molar concentration of solution 1 n1 = valency of solution 1 V2 = volume of solution 2 M2 = molar concentration of solution 2 n1 = valency of solution 2 N1 = normality (M1 x n1) of solution Solution 1SBI 2008 27

Definition(Cont’d) Example 1: 10 mL 0.12 M HCl is titrated with NaOH to determine the concentration NaOH, the volume of NaOH required is 9.0 mL, determine the concentration of NaOH?

Answer: The valency of HCl and NaOH are 1. Why? So we just put all the data to the known equation V1.M1.n1 = V2.M2.n2 10 mL x 0.12 M x 1 = 9.0 mL x M NaOH x 1 M NaOH = 0.133 Question 1: 25 mL 0.15 M HCl is titrated with Ca(OH)2 to determine the concentration Ca(OH)2, the volume of Ca(OH)2 required is

Solution SBI 2008 28

pH pH scale is a log scale based on 10 and equal to –log [H+]. pH is a convenient way to represent solution acidity. So pH = –log [H+] The smaller the pH value, the stronger the acid will be (the base is less) and vice versa means the higher the pH value, the stronger the base will be (the acid is less) For a base, we calculate as pOH = -log [OH-] [H+] and [OH-] are concentration in molar The relation between pH and pOH is= pH + pOH = pKw Solution SBI 2008 29

pH(Cont’d) For acid or base which the valency more than 1, the molar concentration is times the valency. The above formulas are for strong acid and base. Below is the representative value of the degree of acidity 0

1

2

3

4

The acidity increased

5

6

7

8

9

10 11 12

13

14

Neutral

The basidity increased

We can say that acid when the pHSolution < 7 and SBIbase 2008> 30

pH(Cont’d) Example 2: determine the pH solution of 0.01 M HCl and 0.1 M NaOH! Answer: In HCl, the [H+] is= 0.01 M x 1 (because the valency is 1) So pH = -log [H+] = -log 10-2 = 2 For 0.1 NaOH, the [OH-] in NaOH is= 0.1 M x 1 So pOH= -log [OH-] = -log 0.1 = 1 The pH of NaOH can be calculated using pKw = pH + pOH

So pH = pKw – pOH = 14 – 1

Solution SBI 2008 31

pH(Cont’d) Question 2: calculate the pH solution when 4.9 g of H2SO4 is dissolved in 1 L water (Mr 98)! Question 3: If we dissolve 1.85 g Ca(OH)2 in 2000 mL water, calculate the pH (Mr 74)

Solution SBI 2008 32