Tài liệu Chapter 8B: Systematic Treatment of Equilibriu docx

A Systematic Method for Solving Multiple Equilibria Problems > Equilibrium-constant expressions » Mass concentration-balance equations >» A single charge-balance equation... Mass conc

Chapter 8B:

Systematic Treatment of Equilibrium

| Introduction

> Concentrations of individual species

> Determining which species are present

» Reactions

>» Equations

>» Steps in Figure 9-1

Cl-l Introduction

> Concentrations of individual species

> Determining which species are present

> Reactions

>» Equations

> Assumption

> Steps in Figure 9-1

ll A Systematic Method

for Solving Multiple Equilibria Problems

> Equilibrium-constant expressions

» Mass (concentration)-balance equations

>» A single charge-balance equation

Mass (concentration)-balance equations

> Equilibrium concentrations of various species

in a solution to one another and to the analytical concentration

> Two Steps:

> Reactions

> Equations

Example 9-1:

Write mass-balance expressions for a 0.0100 M solution of HCI that is in

equilibrium with an excess of solid BaSO,,(s)

> BaSO,(s) <=> Ba*t + SO,*

> HCI+H,O <=> H,O++CI:

> 2H;O <=> HẠ.O'T + OH

> SO,-+H,0t <=> HSO, + H,O

[Ba2+] = [SO,2- ] + [HSO, |

[H:Or ] = [CI-]+[OH-]-[HSO/r] = Cu + [OH'| - [HSO/]

=0.0100 +[OH'] - [HSO/]

[Cl-] = Cag

Example 9-2:

Write mass-balance expressions for the system formed

when a 0.010 M NH, solution is saturated with AgBr

A single charge-balance equation

> Solutions are neutral

> The Sum of molar concentration of positive charge = The sum of molar concentration of negative charge

> Concentration of charge by an ion =

molar concentration of that ion x its charge

>» Steps:

> List all ions in the solution and their charges and concentrations

> The Sum of molar concentration of positive charge =

The sum of molar concentration of negative charge

> SO/'+H.O*<=> HSO¿+ H,O

2[Ba2*] +[H,O+ ] = 2[SO,2- ~+[HSO, ]+[OH-'] + [CI-]

Example 9-2:

Write a charge-balance equation for the system formed

when a 0.010 M NH, solution is saturated with AgBr

Steps for solving problems involving multiple equilibria

>» Pertinent Equilibria

> The definition of unknown

> Equilibrium Constant Expressions

>» Mass balance equation

> Charge balance equation

>» Check No of unknown species concentrations and No of independent equations

> Approximations:

> Check of assumptions

>» Calculate unknown concentrations

lil Calculation of Solubility by the Systematic

Method

>» Metal Hydroxides (Examples 9-5)

>» The effect of pH on solubility

(1) pH 1s fixed (Example 9-6) (2) pH 1s variable or unknown

> Metal Hydroxides: Example 9-5:

Calculate the molar solubility of Fe(OH), in water

1 Pertinent Equilibria:

Fe(OH)s (s)< ===> Fe' + 3 OH

2 The definition of unknown:

Molar solubility Fe(OH)3 = [Fe* °]

3 Equilibrium Constant Expressions

Ksp = [Fe* *]J[OH]® = 2 x 10° [Eq 1]

K„ = [H:O' J[OH] = 1 x 10" [Eq 2]

4 Mass balance equation

[H3;0'] + 3[Fet?] = [OH] [Eq 3]

5 Charge balance equation

[H:O ]+ 3[Fe''] = [OH] [Eq 4]

6 3 chemical species ([Fe* *], [OH], [HsO* ])

3 Independent equations

Approximations:

First assume [H;0°*] << 3[Fe*?]

Then Eq 3: [HạO'| + 3[Fe”] = [OH] becomes

3 x 10° is not less than 3 x 10° so that

actually 3[Fe*’] << [H;0*] and Eq 3 becomes

In conclusion :

The effect of the self-ionization of water is

significant for relatively insoluble metal hydroxides This is an example of the common ion effect !!!

Any Questions?

> The effect of pH on solubility: (a) pH is fixed (known)

Example 9-6: Calculate the molar solubility of PobCO, in a solution

4 Mass Balance Equation

[Pb*?] = [COs “] + [HCO;] + [H;CO]

5 Charge balance equation cannot be determined

because the buffer species have not been given,

6 Unknowns are [Pb"'], [CO; “], [HCO; ], [H;CO;]

Four independent equations and four unknowns

= 469x10"'

7 Approximations: not necessary because we can obtain an exact

equation based upon [H:O:]= 1.0 X 107

[Pb* 2] = [CO32] + 2.13 X 103[COs2] + 4.73 X 102[COs?]

[CO37] = [Pb* 2]/(2.60 X 10°) and substituting this into the

solubility product equation:

Solve solubility product equation:

Any Questions?

> The effect of pH on solubility: (b) pH is unknown

Example 9-7: Calculate the solubility of CuS in water

2 Definition of unknown: Solubility = [Cu*]

3 Mass Balance Equation:

7 Assumption:

First, we can see that the salt is fairly insoluble (K,, = 8 X 10°”)

so we'll use assume [H;0"] = 10”,

Also, due to low solubility, assume 2IS] + [HS]<< [OH]

since we assume [H*] = 10”, [OH] = 10°

9 Equilibrium Expressions becomes:

Any Questions?

Review of Textbook for Chem115/116: Chapters

17 & 23: Complex (coordinate compound)

> Coordinate covalent bonds: a bond formed when

both electrons of the bond are donated by one atom

Ag* + 2(:NH;) © [H,N: Ag :NH,]*

Electron configuration of Ag [Kr]4d!°5s!5P°

Agt [Kr]4d!°5s° 5 P®

Sp hybrid orbitals: accommodate 2 pairs of electrons

> Complex ion: Lewis base

> Complex (Coordinate compound): a compound

consisting either of complex ions and other ions of

opposite charge or of neutral complex species

> Formation of Complex Ion: Example 9-8

The solubility product of Cul is 1.0x10-!* The formation constant K, for the reaction of Cul with I to give Cul, ~is 7.9x10*+ Calculate the

molar solubility of Cul in a 1.0x10* M solution of KI

Step 1 Pertinet Equilibria Step 4 Mass-balance Expression

Cul (s) <==> Cut + [|

Cul (S) + Ï <==> Cul,” [L] — Cay + [Cu | [Cul,’]

KI—~- K+

Step 2 Define of Unknown

Step 5 Charge-balance Equation

[Cu?] + [KT] = [I] + [CuL] [Cut] + 1.0x10 = [I] + [CuL-] Step 3 Equilibrium-Constant Expression

Step 6 Number of Independent Equations and unknown

Three unknown: [Cu*], [I-], and [Cul, ]

Three independent equations

Step 7 Assumptions

[Cut] << [I] because Ksp of Cul Is very small

[Cut] - [Cul] << 1.0x10-4

Hence: Charge-Balance Equation is simplified as

[Cu] + [KT] = [IL] + [CuL] [Cut] + 1.0x10-4 = [I-] + [Cul,"]

[I-] = 1.0x1074 + [Cut] - [CuL,-] = 1.0x10* M

Step 8 Solve Equations

[Cur] = 1.0x10®M

[CuL] = 7.9x10° M

Solubility of Cul (s) = [Cu*] + [Cul] = 1.0x10° + 7.9x103

= 8.9x10° M

Step 9 Check Assumptions

[Cut] << [I] = 1.0x10-4 (True)

[Cu"] - [CuL] << 1.0x10 (True)

Both assumptions are correct Therefore, our solubility calculated based upon these two assumptions is valid!

Mission accomplished !!!

Any Questions?

VỊ The Separation of lons by Control of the Concentra

of the Precipitating Reagent

A Separation of metal hydroxides by controlling [OH]

Example 9-9: Can Cu* be quantitatively separated from a solution that is 0.100 M in Mg* by controlling pH?

pH = -log [H30°] = 8.9 which is fairly basic

Answer: Yes, we can quantitatively separate Cu** from

Mg** by keeping solution 6.3 <pH < 8.9

Any Questions?

Y Mass (concentration)-balance equations

V A single charge-balance equation

Y Calculation of Solubility by the Systematic Method

Metal Hydroxides

Formation of Complex Ion Separation of ions

Homework

> $-A, B, C, E, F, G, 2, 4, 11, 12, 18, 20, 24,

26

Before working on Homework,

Practice with all examples that we discussed in the class and examples in the textbook!!