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Exam 1 Stats. Average 51.8 ± 22.1 High 92.5 (2 of you) Low 14. Chapter 8 Activity. The “true” nature of ionic species in solution. Ions are charged molecules As a result, they tend to attract polar solvent molecules (like water, for instance) and other ions Hydrated radius

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Exam 1 stats
Exam 1 Stats

  • Average 51.8 ± 22.1

  • High 92.5 (2 of you)

  • Low 14


Chapter 8 activity

Chapter 8Activity


The true nature of ionic species in solution
The “true” nature of ionic species in solution

  • Ions are charged molecules

  • As a result, they tend to attract polar solvent molecules (like water, for instance) and other ions

    • Hydrated radius

    • Ionic atmosphere

  • The thickness of the ionic atmosphere is a function of the ionic strength of the solution “the concentration of charge”


Hydrated radius
Hydrated radius

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+

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+

water

Chloride

Sodium ion


Ionic atmosphere and shielding

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+

+

+

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+

Ionic Atmosphere and Shielding

Low ionic strength

High ionic strength

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Calcium

ion

Chloride

ion

Sulfate

ion

Sodium

ion


Activity
Activity

  • All ionic species in any equilibrium expression are more accurately expressed as activities.

    A Ca2+ = [Ca2+]gCa2+

    or

    A + H2O ↔ A- + H3O+

    Ka = A A-A H3O+ / [A] = [A-]gA-[H3O+]gH3O+ / [A]

    where g is the activity coefficient and is a function of the ionic radius of the ion and the ionic strength of the solution.


Activity continued
Activity - continued

  • The higher the ionic strength of the solution, the larger the smaller the activity coefficient.

  • Rationalization: At higher ionic strengths the ion cloud around any ion is thicker, which weakens the attractive forces between the ion and its counterpart, inhibiting recombination.


Ionic strength
Ionic strength

  • A measure of the concentration of ions in solution

    m = ½ ∑ cizi2

    0.010 M Ca(NO3)2

    m = ½ ([NO3-](-1)2 + [Ca2+](2+)2) =

    ½ {(0.02)(-1)2 + (0.01)(2)2 } = 0.03 M

    gCa2+@m=0.03 = ?

    Use extended Debye-Huckle eq or Table 8.1 and extrapolation


Take home message
Take home message

  • At high ionic strengths, solubility increases slightly (by a factor of 2-10).

  • pH is influenced by ionic strength

  • Weak acid and base dissociation is influenced a little by ionic strength

  • Significant figures?


Example 8 12
Example 8-12

  • Solubility of Hg2Br2

    • in pure water

    • in 0.00100 M KNO3

    • in 0.0100 M KNO3

    • in 0.100 M KNO3


In pure water
In pure water

Hg2Br2 Hg22+ + 2Br-

Ksp = [Hg22+]gHg22+ [Br-]2gBr-2

2[Hg22+] = [Br-] and let [Hg22+] = x

Ionic strength is very low.

gHg22+ and gBr- are close to 1.00

so,

Ksp = 4 [x]3 = 5.6E-23

x = 2.4E-8 M



In 0 00100 m kno3
in 0.00100 M KNO3 -1

m = ½ ((.001)(+1)2 + (0.001)(-1)2 = 0.001 M

gHg22+@ m = 0.001 = 0.867

gBr-2 @ m = 0.001 = 0.964

Ksp = 4 gHg22+gBr-2 [x]3 = 5.6E-23

x = 2.6E-8M


In 0 0100 m kno3
in 0.0100 M KNO3 -1

m = ½ ((.01)(+1)2 + (0.01)(-1)2 = 0.01 M

gHg22+@ m = 0.001 = 0.660

gBr-2 @ m = 0.001 = 0.898

Ksp = 4 gHg22+gBr-2 [x]3 = 5.6E-23

x = 3.0E-8M


In 0 100 m kno3
in 0.100 M KNO3 -1

m = ½ ((0.1)(+1)2 + (0.1)(-1)2 = 0.1 M

gHg22+@ m = 0.001 = 0.335

gBr-2 @ m = 0.001 = 0.75

Ksp = 4 gHg22+gBr-2 [x]3 = 5.6E-23

x = 4.2E-8M


Ph and ionic strength
pH and ionic strength -1

  • True definition of pH

  • pH = -logAH+ = -log {[H+]gH+}

  • pH of a 0.00100 M HCl solution

    • Ionic strength, m, = 0.001; gH+ = 0.967

    • pH = -log(.001*.967) = 3.01

  • pH of a 0.100 M HCl solution

    • Ionic strength, m, = 0.1; gH+ = 0.83

    • pH = -log(0.1*0.83) = 1.08

  • pH of a 0.00100 M HCl/0.100 M NaCl solution

    • Ionic strength, m, = 0.1; gH+ = 0.83

    • pH = -log(0.001*0.83) = 3.08


What is the concentration h an oh in a 0 100 m nacl solution
What is the concentration H -1+ an OH- in a 0.100 M NaCl solution?

Kw = [OH-]gOH-[H+]gH+ = 1.01E-14

At m = 0.100, gOH- = 0.76 and gH+ = 0.83

H2O is the only source of H+ and OH- so let x = [H+] = [OH-]

Kw = (0.76)(0.83) x2

x = 1.27E-7 M


Edta prob 12 31
EDTA Prob 12-31 -1

  • A 50 mL sample containing Ni2+ is treated with 25.00 mL of 0.0500 M EDTA to complex all of the Ni2+. The excess EDTA is back-titrated, requiring 5.00 mL of 0.0500 M Zn2+. What is the concentration of Ni2+ IN THE ORIGINAL SAMPLE?