20 b week iv chapters 11
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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

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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


20 b week iv chapters 11

e

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


20 b week iv chapters 11

-2∂

First Solvation Shell of the

Solvent

-2∂

+∂

-2∂

+

-2∂

+∂

+∂

Solvated Na+

Fig. 10-6a, p. 450


20 b week iv chapters 11

Dissolution of K2SO4 in water

2nd solvation shell

Fig. 11-3, p. 480


20 b week iv chapters 11

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.


20 b week iv chapters 11

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

potential


20 b week iv chapters 11

The Hydrogen bonds between the water molecules of the solvents

makes the molecule more stable in solution than in the solid

H-bonds

Fructose C6H12O6

Hydrated Fructose C6H12O6


20 b week iv chapters 11

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


20 b week iv chapters 11

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


20 b week iv chapters 11

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

+


20 b week iv chapters 11

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)


20 b week iv chapters 11

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


20 b week iv chapters 11

Types of Acids

Table 11-1, p. 483


20 b week iv chapters 11

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+ +


20 b week iv chapters 11

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


20 b week iv chapters 11

Electrolyte Solutions can Carry current

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

Fig. 11-3, p. 480


20 b week iv chapters 11

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

Electrolytes.

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


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