Chapter 6 Problems

Chapter 6 Problems

6.19, 6.21, 6.24 6-29, 6-31, 6-39, 6.41 6-42, 6-48,

6-19. A solution contains 0.0500 M Ca2+ and 0.0300 M Ag+.

Can 99% of Ca2+ be precipitated by sulfate without precipitating Ag+? What will be the concentration of Ca2+ when Ag2SO4 begins to precipitate?



MM 000500.001.0*0500.0 Ca2+ at equilibrium

CaSO4 Ca2+ + SO42- Ksp

= 2.4 x 10-5 = [Ca2+][SO42-]CaSO4 Ca2+ + SO4

2- Ksp= 2.4 x 10-5 = [0.000500][SO4

2-]

[SO42-] = 0.048 M

Ag2SO4 2Ag+ + SO42- Ksp = 1.5 x 10-5

Q = [Ag+]2[SO42-] Q = [0.03]2[0.000500] Q = 4.3 x 10-5

Q>K

Sep.

is N

OT fe

asib

le



Ag2SO4 2Ag+ + SO42- Ksp = 1.5 x 10-5

K/[Ag2+]2 = [SO42-] 1.5 x 10-5/[0.0300]2 = [SO4

2-]

1.67 x 10-2= [SO42-]

Find Ca2+

CaSO4 Ca2+ + SO42- Ksp

= 2.4 x 10-5

[Ca2+] = 0.0014 M

= [Ca2+][1.67 x 10-2]

About 2.8 % remains in solution

6.21 If a solution containing 0.10 M Cl-, Br-, I- and Cr2O4

2- is treated with Ag+, in what order will the anions precipitate?

AgCl Ag+ + Cl- Ksp = 1.8 x 10-10=[Ag][Cl]

AgBr Ag+ + Br- Ksp = 5.0 x 10-13=[Ag][Br]

AgI Ag+ + I- Ksp = 8.3 x 10-17=[Ag][I]

Ag2CrO4 2Ag+ + CrO4

- Ksp = 1.2 x 10-12=[Ag]2[Cl]

AgCl Ag+ + Cl- Ksp = 1.8 x 10-10=[Ag][0.1]

AgBr Ag+ + Br- Ksp = 5.0 x 10-13=[Ag][0.1]

AgI Ag+ + I- Ksp = 8.3 x 10-17=[Ag][0.1]

Ag2CrO4 2Ag+ + CrO4

- Ksp = 1.2 x 10-12=[Ag]2[0.1]

SOLVE for Ag+ required at equilibrium

1.8 x 10-9=[Ag]5.0 x 10-12=[Ag]8.3 x 10-16=[Ag]3.5 x 10-6=[Ag]

6-24. The cumulative formation constant for SnCl2(aq) in

1.0 M NaNO3 is 2=12. Find the concentration of SnCl2 for a solution in which the concentration of Sn2+ and Cl- are both somehow fixed at 0.20 M.

SnCl2(aq)

Sn2+ (aq) + 2Cl- (aq) SnCl2 (aq) 2=12

2=12 222

]][[)]([

ClSnaqSnCl

22

]20.0][20.0[)]([ aqSnCl

MaqSnCl 096.0)]([ 2

Complex Formation

complex ions (also called coordination ions)

Lewis Acids and Basesacid => electron pair acceptor (metal)base => electron pair donor (ligand)

Effects of Complex Ion Formation on Solubility

Consider the addition of I- to a solution of Pb+2 ions

Pb2+ + I- PbI+

PbI+ + I- PbI2 K2 = 1.4 x 101

PbI2 + I- PbI3- K3 =5.9

PbI3+ I- PbI42- K4 = 3.6

221 100.1

]][[

][x

IPb

PbIK



Pb2+ + I- <=> PbI+

PbI+ + I- <=> PbI2 K2 = 1.4 x 101

Pb2+ + 2I- <=> PbI2 K’ =?

221 100.1

]][[

][x

IPb

PbIK

Overall constants are designated with

This one is



Pb2+ + I- PbI+

PbI+ + I- PbI2 K2 = 1.4 x 101

PbI2 + I- PbI3- K3 =5.9

PbI3+ I- PbI42- K4 = 3.6

221 100.1

]][[

][x

IPb

PbIK

Acids and Bases & EquilibriumEquilibrium

Section 6-7

Strong Bronsted-Lowry Acid

A strong Bronsted-Lowry Acid is one that donates all of its acidic protons to water molecules in aqueous solution. (Water is base – electron donor or the proton acceptor). HCl as example

Strong Bronsted-Lowry Base

Accepts protons from water molecules to form an amount of hydroxide ion, OH-, equivalent to the amount of base added.

Example: NH2- (the amide ion)

Question

Can you think of a salt that when dissolved in water is not an acid nor a base?

Can you think of a salt that when dissolved in water IS an acid or base?

Weak Bronsted-Lowry acid

One that DOES not donate all of its acidic protons to water molecules in aqueous solution.

Example? Use of double arrows! Said to reach

equilibrium.

Weak Bronsted-Lowry base

Does NOT accept an amount of protons equivalent to the amount of base added, so the hydroxide ion in a weak base solution is not equivalent to the concentration of base added.

example: NH3

Common Classes of Weak Acids and Bases

Weak Acids carboxylic acids ammonium ions

Weak Bases amines carboxylate anion

Question: Question: Calculate the Concentration of H+ and OH- in Pure

water at 250C.

Equilibrium and Equilibrium and WaterWater

EXAMPLE: Calculate the Concentration of H+ and OH- in Pure water at 250C.

H2O H+ + OH-

Initial liquid - -

Change -x +x +x

Equilibrium Liquid-x +x +x

KW=(X)(X) = ?

Kw = [H+][OH-] = ?

EXAMPLE: Calculate the Concentration of H+ and OH- in Pure water at 250C.

H2O H+ + OH-

Initial liquid - -

Change -x +x +x

Equilibrium Liquid-x +x +x

KW=(X)(X) = 1.01 X 10-14

Kw = [H+][OH-] = 1.01 X 10-14

(X) = 1.00 X 10-7

Example

What is the concentration of OH- in a solution of water that is 1.0 x 10-3 M in [H+] (@ 25 oC)?

Kw = [H+][OH-]

1 x 10-14 = [1 x 10-3][OH-]

1 x 10-11 = [OH-]

“From now on, assume the temperature to be 25oC unless otherwise stated.”

pH

~ -3 -----> ~ +16pH + pOH = - log Kw = pKw = 14.00

Is there such a thing as Pure Water? In most labs the answer is NO Why?

A century ago, Kohlrausch and his students found it required to 42 consecutive distillations to reduce the conductivity to a limiting value.

CO2 + H2O HCO3- + H+

Weak Acids and Bases

HA H+ + A-

HA + H2O(l) H3O+ + A-

Ka

][

]][[

HA

AHKa

][

]][[ 3

HA

AOHKa

Ka’s ARE THE SAME

Weak Acids and Bases

B + H2O BH+ + OH-

][

]][[

B

OHBHKb

Kb

Relation Between KRelation Between Kaa and and KKbb

Relation between Ka and Kb

Consider Ammonia and its conjugate acid.

][

]][[

4

33

NH

OHNHKa

][

]][[

3

4

NH

OHNHKb

NH3 + H2O NH4+ + OH-

Kb

NH4+ + H2O NH3 + H3O+

Ka

H2O + H2O OH- + H3O+

][

]][[

][

]][[

3

4

4

33

NH

OHNH

NH

OHNHK

][][ 3 OHOHKw

Example

The Ka for acetic acid is 1.75 x 10-5. Find Kb for its conjugate base.

Kw = Ka x Kb

a

wb K

KK

105

14

107.51075.1

100.1

bK

Example Calculate the hydroxide ion concentration in

a 0.0100 M sodium hypochlorite solution.OCl- + H2O HOCl + OH-

The acid dissociation constant = 3.0 x 10-8

][

]][[

OCl

OHHOClKb

1st Insurance Problem

Challenge on page 120Challenge on page 120

Chapter 8

ActivityActivity

Write out the equilibrium constant for the following expression

Fe3+ + SCN- Fe(SCN)2+

Q: What happens to K when we add, say KNO3 ?

A: Nothing should happen based on our K, our K is independent of K+ & NO3

-

]][[

])([3

2

SCNFe

SCNFeK

K decreases when an inert salt is added!!! Why?

8-1 Effect of Ionic Strength 8-1 Effect of Ionic Strength on Solubility of Saltson Solubility of Salts

Consider a saturated solution of Hg2(IO3)2 in ‘pure water’. Calculate the concentration of mercurous ions.

Hg2(IO3)2(s) Hg22+ + 2IO3

- Ksp=1.3x10-18

A seemingly strange effectseemingly strange effect is observed when a salt such as KNO3 is added. As more KNO3 is added to the solution, more

solid dissolves until [Hg22+] increases to 1.0 x 10-6 M. Why?

1823

22 103.1]][[ IOHgK sp

IICCEE

somesome -- - --x-x +x+x +2x+2xsome-xsome-x +x+x +2x+2x

182 103.1]2][[ xxK sp7109.6][ x

Increased solubility

Why? Complex Ion?

No Hg2

2+ and IO3- do not form complexes

with K+ or NO3-.

How else?

The Explanation

Consider Hg22+ and the IO3

-

-2+

Electrostatic attraction

The Explanation


-

2+


-

Hg2(IO3)2(s) The Precipitate!!The Precipitate!!

The Explanation


-

-2+


Add KNO3

K+ K+

K+K+

K+K+

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

The Explanation


-

-2+

K+ K+

K+K+

K+K+

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

Hg22+ and IO3

- can’t getCLOSE ENOUGH to form Crystal latticeOr at least it is a lot “Harder” to form crystal lattice

The potassium hydrogen tartrate example

K+-O

O

OH

OH

O

OH

potassium hydrogen tartrate

Alright, what do we mean by Ionic strength?

Ionic strength is dependent on the number of ions in solution and their charge.

Ionic strength () = ½ (c1z12+ c2z2

2 + …)

Or Ionic strength (m) = ½ cizi2

Examples Calculate the ionic strength of (a) 0.1 M

solution of KNO3 and (b) a 0.1 M solution of

Na2SO4 (c) a mixture containing 0.1 M KNO3

and 0.1 M Na2SO4.() = ½ (c1z1

2+ c2z22 + …)

Alright, that’s great but how does it affect the equilibrium constant?

Activity = Ac = [C]c

AND

bB

baA

a

dD

dcC

c

bB

aA

dD

cC

BA

DC

AA

AAK

][][

][][

Relationship between activity and ionic strength

x

x

xz

3.31

51.0log

2

Debye-Huckel Equation

2 comments

= ionic strength of solution = activity coefficientZ = Charge on the species x = effective diameter of ion (nm)

(1)What happens to when approaches zero?(2)Most singly charged ions have an effective radius of about 0.3 nm

Anyway … we generally don’t need to calculate – can get it from a table

Activity coefficients are Activity coefficients are related to the hydrated related to the hydrated radius of atoms in radius of atoms in moleculesmolecules

Relationship between and

Back to our original problem

223

22

2

3223

22

][][ IOHgIOHgsp IOHgAAK


Hg2(IO3)2(s) Hg22+ + 2IO3

- Ksp=1.3x10-18

At low ionic strengths -> 1


223

22

2

3223

22



Hg2(IO3)2(s) Hg22+ + 2IO3

- Ksp=1.3x10-18

In 0.1 M KNO3 - how much Hg22+ will be dissolved?


223

22

2

3223

22



Hg2(IO3)2(s) Hg22+ + 2IO3

- Ksp=1.3x10-18

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Chapter 6 Problems