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• Chapter 10, 11( except 11.4 and and 11.6 -7) • Intermolecular potentials. ion-dipole, ion-ion • Solutions. Interactions in solution, Boiling Point Elevation, Freezing Point Depression and Osmotic Pressure, Electrolyte solutions

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20 b week iv chapters 11

• Chapter 10, 11( except 11.4 and and 11.6 -7)

• Intermolecular potentials.

ion-dipole, ion-ion

• Solutions. Interactions in solution, Boiling Point Elevation, Freezing Point Depression and Osmotic Pressure, Electrolyte solutions

• Dissolution reactions(rxns) and Arrhenius type Acid/Base rxns

Midterm Friday:

Chaps 9(no 9.7), 10, 11.-11.3

One side of 1 page notes(must be hand written), closed book

Review Session Today @ 2-3 pm, in FRANZ 1178

20 B Week IV Chapters 11


In Solutions, for example when NaCl(s) is dissolved in H2O(l).

+ H2O

Dissolution of a polar solid by a polar solid by a polar liquid

A non-polar liquid e.g., benzene, would not dissolve NaCl?

NaCl(s) + H2O(l) Na+(aq) +Cl-(aq)

(aq) means an aqueous solution, where water is the solvent,

major component.

The solute is NaCl, which is dissolved, minor component

Water molecules solvates the ions the

Cation (Na+) and the Anion (Cl-). The forces at play here are

Ion dipole forces

Fig. 10-6, p. 450



First Solvation Shell of the









Solvated Na+

Fig. 10-6a, p. 450


Dissolution of K2SO4 in water

2nd solvation shell

Fig. 11-3, p. 480


In a solution of solvent A and solute B, the important thermodynamic variables

are V, T, molar Concentration of [A] and [B] and the relative composition X

The Molarity, moles of A or B/ Liters of Solution

[A]=nA/Vsol and [B]=nB/Vsol in units of mols L-1

The composition in Mole Fraction: XA= nA/(nA + nB) and XB= nB/(nA + nB)

XA + XB =1 and XA=1 -XB

Because Volume depends on the Temperature, the Molarity is not always the

Best variable, the Molality

{B} = Moles of Solute/ kg of Solvent mol kg-1

{B}= nB//kg of solvent, since the density of water is defined as kg/L,

Therefore, for ideal aqueous solutions the Molality is, in effect, the

moles of solute/Liters of solvent.


The dissolution reaction with water as a solvent: A(s)A(aq) A is a molecular solid and the molecular units, or monomers, do not dissociate in solution. Like fructose with lots of OH ( hydroxyl groups) for forming H-bonds and which makes the monomer more stable inaqsoln

Electro Static



The Hydrogen bonds between the water molecules of the solvents

makes the molecule more stable in solution than in the solid


Fructose C6H12O6

Hydrated Fructose C6H12O6


The dissolution of salts such as NaCl(s), NaOH(s)(basic in soln, sodium hydroxide) and NH4Cl(acidic in soln, Ammonium Chloride )

AB(s)  A+(aq) + B+(aq)

In some case like NaCl no reaction occurs with the solvent just hydration

+ H2O


Lone pair


NH4Cl(s)  NH+4(aq) + Cl-(aq)

NH+4(aq) + H2O(l)H3O+((aq) + NH3((aq)

Proton Transfer

Notice that the Number of Elements and the Amount of Charge are balanced

On both sides reaction the rxn

Such a reaction has the correct Stoichiometry and is said to be balanced


NH+4(aq) + H2O(l)  H3O+(aq) + NH3(aq)

By definition, the Ammonium ion NH+4 is an Arrhenius acid

(Svante Arrhenius) since it increased the concentration(molarity) of the

H3O+(aq), the hydronium ion in solution by reacting with the solvent,

H2O in the case



In the general an Acid AH, the solute, in the Arrhenius sense will react

with the solvent H2O, in this case, it could have been

AH(aq) + H2O(l)  H3O+(aq) + A-(aq)

A strong acid such as HCl will completely dissociate in solution

HCl(g) + H2O(l)  H3O+(aq) + Cl-(aq)

The reaction goes to completion: all the HCl molecules produce H3O+(aq)


For a relatively weak Acids such as HF, the reaction does not go to completion

HF(g) + H2O(l)  H3O+(aq) + F-(aq)

And there is still lots of HF in solution solvated by H2O molecules, the rxn therefore goes to equilibrium

HF forms a strong H-bonded network


Types of Acids

Table 11-1, p. 483


The Neutralization reaction

H2O(l) + H2O(l)  H3O+(aq) + OH-(aq)

where H2O(l) amphoteric acts both as an acid as well as a base ths reaction goes to equilibrium

 H+ +


NaOH(s)  Na+(aq) + OH-(aq)

Where OH-, the hydroxide ion, is a strong base

In acid base reactions for example:

HCl(g) + H2O(l)  H3O+(aq) + Cl-(aq)

HCl is the strong Acid and H2O is the Base

HCl + NaOH  H2O + NaCl  Na+(aq) + Cl-(aq)

Which is just salt water


Electrolyte Solutions can Carry current

Dissolution of K2SO4(s)  2K+(aq) + SO-4(aq)

Fig. 11-3, p. 480


The Solubility limit of Potassium Sulfate in aqueous soln

is 120 gL-1 at 25 °C

Molecules that dissolve in solution to produce ions are called


Some compounds have rather limited solubility in water

e.g., example of BaSO4(s) can dissolve to 2.5 mgL-1

BaSO4(s)  Ba++(aq) + SO2-4(aq)

The equilibrium of this reaction is therefore on the side of

Ba++(aq) + SO2-4(aq)  BaSO4(s)

And the solid would precipitate out of soln