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Unit 4: Equilibrium. Chapter 8.2 Weak acids and Bases. Acid Ionization Constants 01. Acid Ionization Constant: the equilibrium constant for the ionization of an acid. HA( aq ) + H 2 O( l ) æ H 3 O + ( aq ) + A – ( aq ) Or simply: HA( aq ) æ H + ( aq ) + A – ( aq ).

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unit 4 equilibrium

Unit 4: Equilibrium

Chapter 8.2

Weak acids and Bases

acid ionization constants 01
Acid Ionization Constants 01
  • Acid Ionization Constant: the equilibrium constant for the ionization of an acid. HA(aq) + H2O(l) æ H3O+(aq) + A–(aq)
  • Or simply: HA(aq) æ H+(aq) + A–(aq)

Chapter 15

slide3

Acid Ionization Constants,

Ka and Kb

ACIDKaCONJ. BASE Kb

HF

HNO2

C9H8O4 (aspirin)

HCO2H (formic)

C6H8O6 (ascorbic)

C6H5CO2H (benzoic)

CH3CO2H (acetic)

HCN

C6H5OH (phenol)

7.1 x 10 –4

4.5 x 10 –4

3.0 x 10 –4

1.7 x 10 –4

8.0 x 10 –5

6.5 x 10 –5

1.8 x 10 –5

4.9 x 10 –10

1.3 x 10 –10

F–

NO2 –

C9H7O4 –

HCO2 –

C6H7O6 –

C6H5CO2 –

CH3CO2 –

CN –

C6H5O –

1.4 x 10 –11

2.2 x 10 –11

3.3 x 10 –11

5.9 x 10 –11

1.3 x 10 –10

1.5 x 10 –10

5.6 x 10 –10

2.0 x 10 –5

7.7 x 10 –5

Chapter 15

slide4

Strength of Acids and Bases

(a) Arrange the three acids in order of increasing value of Ka.

(b) Which acid, if any, is a strong acid?

(c) Which solution has the highest pH, and which has the lowest?

Chapter 15

slide5

Equilibria Involving Weak Acids and Bases

  • Consider acetic acid, HC2H3O2 (HOAc)
  • HC2H3O2 + H2O  H3O+ + C2H3O2-
  • Acid Conj. base
  • (K is designated Ka for ACID)
  • K gives the ratio of ions (split up) to molecules (don’t split up)
  • The value of Ka is used to calculate pH of weak acids.
slide6

Ionization Constants for Acids/Bases

Conjugate

Bases

Acids

Increase strength

Increase strength

equilibrium constants for weak acids
Equilibrium Constants for Weak Acids

Weak acid has Ka < 1

Leads to small [H3O+] and a pH of 2 - 7

equilibrium constants for weak bases
Equilibrium Constants for Weak Bases

Weak base has Kb < 1

Leads to small [OH-] and a pH of 12 - 7

equilibria involving a weak acid
Equilibria Involving A Weak Acid

You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH.

Step 1. Define equilibrium concs. in ICE table.

[HOAc] [H3O+] [OAc-]

Initial

Change

Equilib.

1.00 0 0

-x +x +x

1.00-xx x

equilibria involving a weak acid12
Equilibria Involving A Weak Acid

You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH.

Step 2. Write Ka expression

This is a quadratic. Solve using quadratic formula.

or you can make an approximation if X is very small!

equilibria involving a weak acid13
Equilibria Involving A Weak Acid

You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH.

Step 3. Solve Ka expression

First assume x is very small because Ka is so small.

Now we can more easily solve this approximate expression.

equilibria involving a weak acid14
Equilibria Involving A Weak Acid

You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH.

Step 3. Solve Ka approximate expression

x = [H3O+] = [OAc-] = 4.2 x 10-3 M

pH = - log [H3O+] = -log (4.2 x 10-3) = 2.37

slide15

Acid Ionization Constants

  • Initial Change Equilibrium Table: Determine the pH of 0.50M HA solution at 25°C. Ka = 7.1 x 10–4.

-

+

H

+

A

æ

HA

(aq)

(aq)

(aq)

Initial

(

M

)

:

0.50

0.00

0.00

Change

(M):

x

+

x

+

x

Equilib

0.50

x

x

x

(M):

Chapter 15

slide16

Acid Ionization Constants

  • Calculate the pH of a 0.036 M nitrous acid (HNO2) solution.
  • What is the pH of a 0.122 M monoprotic acid whose Ka is 5.7 x 10–4?
  • The pH of a 0.060 M weak monoprotic acid is 3.44. Calculate the Ka of the acid.

Chapter 15

slide17
In a solution of acetic acid, the equilibrium concentrations are found to be [CH3COOH] = 1.000; [CH3COO-] = 0.0042. Evaluate the pH of this solution and the equilibrium constant of ionization of acetic acid. (1.78x10-5)

From the ionization of acetic acid,

CH3COOH = CH3COO- + H+0.100             0.0042     0.0042

p k a log k a
pKa = - log Ka
  • The pKa of acetic acid is 4.75. Find the pH of acetic acid solutions of labeled concentration of 1.0 M (x=0.042; pH=2.38)
hc 2 h 4 no 2 h c 2 h 4 no 2 c 0 01m calculate ph and ka

R

I

C

E

[H+][C2H4NO2–]

[x]2

[x]2

Ka =

=

[HF]

[0.010-x]

[0.010]

HC2H4NO2 H+ + C2H4NO2– c=0.01M Calculate pH and Ka

HC2H4NO2

H+

C2H4NO2–

1

1

1

0.010

0

0

-x

+x

+x

0.010 - x

x

x

= 1.4x10–5

Since x is small 0.010 – x = 0.010

x= 3.74 x 10–5 M, pH = 3.43

=1.4 x 10 – 5

hbu h bu c hbu 0 01m and ph 3 40 calculate ka

R

I

C

E

[H+][Bu–]

[0.0004]2

Ka =

=

[HBu]

[.0096]

HBu  H+ + Bu– c(HBu = 0.01M and pH = 3.40 Calculate Ka

HBu

H+

Bu–

1

1

1

0.0100

0

0

-0.0004

+0.0004

+0.0004

0.0096

0.0004

0.0004

[H+] = 10–pH = 10–3.40 = 3.98 x 10–4

= 1.67x10–5

equilibria involving a weak base
Equilibria Involving A Weak Base

You have 0.010 M NH3. Calc. the pH.

NH3 + H2O  NH4+ + OH-

Kb = 1.8 x 10-5

Step 1. Define equilibrium concs. in ICE table

[NH3] [NH4+] [OH-]

initial

change

equilib

0.010 0 0

-x +x +x

0.010 - x xx

slide24

Acid Ionization Constants

  • Percent Dissociation: A measure of the strength of an acid.
  • Stronger acids have higher percent dissociation.
  • Percent dissociation of a weak acid decreases as its concentration increases.
m h 2 o mh oh c 0 01m ph 10 10 kb

R

I

C

E

[MH+] [OH–]

Kb =

=

[M]

M + H2O  MH+ + OH– c=0.01M, pH =10.10 Kb-?

M

MH+

OH–

1

1

1

0.010

0

0

-0.00013

+0.00013

+0.00013

0.00987

0.00013

0.00013

pOH = 14 - pH = 14 - 10.10 = 3.90

[OH-] = 10-pOH = 10-3.90 = 1.26 x 10-4

[0.00013] [0.00013]

=1.7 x 10-6

[0.00987]

nh 3 h 2 o nh 4 oh c 0 02m ph

R

I

C

E

[x] [x]

Kb =

=

[0.020]

NH3 + H2O  NH4+ + OH– c=0.02M; pH - ?

NH3

NH4+

OH–

1

1

1

0.020

0

0

-x

+x

+x

0.020-x

x

x

pOH = -log[OH-] = 3.22

pH = 14 - pOH = 10.78

x2

= 1.8 x 10-5

x= 6.0 x 10-4

[0.020]

ka summary
Ka summary
  • Ka follows the pattern of other “K” equations
  • I.e. for HA(aq) + H2O(l)  H3O+(aq) + A–(aq)
  • Ka = [H3O+][A–] / [HA]
  • Notice that H2O is ignored because it is liquid
  • HA cannot be ignored because it is aqueous
  • This is different than with Ksp. In Ksp, solids could only be in solution as ions
  • Acidscanbeinsolutionwhetherionizedornot
  • The solubility of acids makes sense if you think back to the partial charges in HCl for ex.
ka summary29
Ka summary
  • Generally Ka tells you about acid strength
  • Strong acids have high Ka values
  • A “strong” acid is an acid that completely ionizes. E.g. HCl + H2O  H3O+ + Cl–
  • A “weak” acid is an acid that doesn’t ionize completely. E.g. HF + H2O  H3O+ + F–
  • Note: don’t get confused between strength and concentration. 1 M HCN has a smaller [H+], thus a higher pH, than 0.001 M HCl
  • In general: Ka < 10–3 Weak acid

10–3 < Ka < 1 Moderate acid

Ka > 1 Strong acid