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Exam 1 Stats Average 51.8 ± 22.1 High 92.5 (2 of you) Low 14 A 100-83 4 B 82-64 11 C 63-41 15 D 40-22 8 F 21-0 6

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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 - PowerPoint PPT Presentation

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Page 1: Exam 1 Stats

Exam 1 Stats

• Average 51.8 ± 22.1• High 92.5 (2 of you)• Low 14

A 100-83 4

B 82-64 11

C 63-41 15

D 40-22 8

F 21-0 6

Page 2: Exam 1 Stats

Chapter 8Activity

Page 3: Exam 1 Stats

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”

Page 4: Exam 1 Stats

_

+

Hydrated radius

_ +

water Chloride Sodium ion

Page 5: Exam 1 Stats

+

Ionic Atmosphere and Shielding

_

Sulfateion

Sodium ion

-+ +

++

+++

+

++ +

+

++-

+ +++

+++

+

+

+

+- -

--

---

+- -

--

---

--- -

----

- --

--

-

-+

Calcium ion

Chloride ion

Low ionic strength High ionic strength

Page 6: Exam 1 Stats

Activity

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

A Ca2+ = [Ca2+]Ca2+

or

A + H2O ↔ A- + H3O+

Ka = A A- A H3O+ / [A] = [A-]A-[H3O+]H3O+ / [A]where is the activity coefficient and is a

function of the ionic radius of the ion and the ionic strength of the solution.

Page 7: Exam 1 Stats

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.

Page 8: Exam 1 Stats

Ionic strength• A measure of the concentration of ions in

solution

= ½ ∑ cizi2

0.010 M Ca(NO3)2

= ½ ([NO3-](-1)2 + [Ca2+](2+)2) =

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

Ca2+@=0.03 = ?

Use extended Debye-Huckle eq or Table 8.1 and extrapolation

Page 9: Exam 1 Stats

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?

Page 10: Exam 1 Stats

Example 8-12

• Solubility of Hg2Br2

– in pure water

– in 0.00100 M KNO3

– in 0.0100 M KNO3

– in 0.100 M KNO3

Page 11: Exam 1 Stats

In pure water

Hg2Br2 Hg22+ + 2Br-

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

2

2[Hg22+] = [Br-] and let [Hg2

2+] = x Ionic strength is very low.

Hg22+ and Br- are close to 1.00so,

Ksp = 4 [x]3 = 5.6E-23x = 2.4E-8 M

Page 12: Exam 1 Stats

Activity coefficient as a function of ionic strength Table 8 -1

= 0.001 = 0.01 = 0.1

Hg22+ 0.867 0.660 0.335

Br- 0.964 0.898 0.75

Page 13: Exam 1 Stats

in 0.00100 M KNO3

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

Hg22+ @ = 0.001 = 0.867

Br-2

@ = 0.001 = 0.964

Ksp = 4 Hg22+ Br-2 [x]3 = 5.6E-23

x = 2.6E-8M

Page 14: Exam 1 Stats

in 0.0100 M KNO3

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

Hg22+ @ = 0.001 = 0.660

Br-2

@ = 0.001 = 0.898

Ksp = 4 Hg22+ Br-2 [x]3 = 5.6E-23

x = 3.0E-8M

Page 15: Exam 1 Stats

in 0.100 M KNO3

= ½ ((0.1)(+1)2 + (0.1)(-1)2 = 0.1 M

Hg22+ @ = 0.001 = 0.335

Br-2

@ = 0.001 = 0.75

Ksp = 4 Hg22+ Br-2 [x]3 = 5.6E-23

x = 4.2E-8M

Page 16: Exam 1 Stats

solubility vs log()

2.2E-08

2.7E-08

3.2E-08

3.7E-08

4.2E-08

4.7E-08

-8 -6 -4 -2 0

log()

solu

bili

ty o

f H

g 2B

r 2

Page 17: Exam 1 Stats

pH and ionic strength

• True definition of pH• pH = -logAH+ = -log {[H+]H+}• pH of a 0.00100 M HCl solution

– Ionic strength, , = 0.001; H+ = 0.967– pH = -log(.001*.967) = 3.01

• pH of a 0.100 M HCl solution– Ionic strength, , = 0.1; H+ = 0.83– pH = -log(0.1*0.83) = 1.08

• pH of a 0.00100 M HCl/0.100 M NaCl solution – Ionic strength, , = 0.1; H+ = 0.83– pH = -log(0.001*0.83) = 3.08

Page 18: Exam 1 Stats

What is the concentration H+ an OH- in a 0.100 M NaCl solution?

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

At = 0.100, OH- = 0.76 and H+ = 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

Page 19: Exam 1 Stats

EDTA Prob 12-31

• 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?