Solution 3 : colligative properties continued

Colligative properties

Solution SBI 2008 19b1

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.

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 than before osmosis



o = sugar molecules0 = water molecules

0

00

0

0

00

0

0

0 0 0

000 0 0

00 0

o

o

o

oo

o o

oo o

o

o

o

0

Solution 1 Solution 2

Semipermeable membrane

0

00

0

0

00

0

0

0 0

0

000 0 0

00 0

o

o

o

oo

o o

oo o

o

o

o

0

Solution 1 Solution 2

Initial condition(before osmosis occurred)

After osmosis process occurred

Solution 1 is more concentrated than solution 2



Osmotic pressure was formulated by Jacobus Henricus van’t Hoff as:Л = M.R.T

where Л = osmotic pressure (atm)M = molar concentrationR = 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))



Example of Osmotic pressure17.1 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 MT = 25 + 273 = 298 K, then put the data into

the formula Л = M.R.T= 0.1 x 0.082 x 298 = 2.4436 atm

Question for Colligative properties


Question 1. The osmotic pressure of KOH in 5.0 l at 27°C is 9.56 atm. If = 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 = 0.95, at 25°C

Determine: the vapor pressure, boiling point, freezing point and osmotic pressure of this solution!!!!!

Definition of Acid, Base and Salt

Solution SBI 2008 20

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:

or Ka = [H

+][Cl

-]

[HCl] [H2O]

[H3O+][Cl

-]

Ka = [HCl]

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

Definition(Cont’d)


H2SO4 is also acid. Why?

Because it can also release H+, based on this reactionH2SO4(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

Weak acid: H3PO4, H2SO3, HCN, CH3COOH



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 are:

Kb = [Na

+][OH

-]

[NaOH] Kb =

[Ca2+

][OH-]2

[Ca(OH)2] and



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 other sign of their weakness is the ionization degree ().



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)

saltNaCl Na+ + Cl-

cation left over acid (from HCl)



If we have K2SO4, from which acid and base, is it formed?

K2SO4 ionize as the following:

K2SO4 2K+ + SO42-

cation 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 as neutralization reaction.



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.



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 1

N2 = normality (M2 x n2) of solution 2



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 11.8 mL, determine the concentration of Ca(OH)2?

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 molarThe relation between pH and pOH is= pH + pOH = pKwpKw is constant value of water, with a value 14 or if Kw is: 10-14

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

We can say that acid when the pH < 7 and base > 7.

0 7 141 2 3 4 5 6 8 9 10 11 12 13

NeutralThe acidity increased The basidity increased

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 1So pOH= -log [OH-] = -log 0.1 = 1The pH of NaOH can be calculated using pKw = pH +

pOH

So pH = pKw – pOH = 14 – 1

pH = 13.

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 3 : colligative properties continued

Documents

Transcript of Solution 3 : colligative properties continued