CHM 2046C Module 11 Chapter 14 Exam

Part C: Derivation of Kw Kh Kb and Ka Chapter 14 10 points

Write the ionization reaction of water, then develop the special K for water:

Kw from the Kc:

 

H2O + H2O H3O+1 + OH1-

 

[H3O1+]1 [OH1-]1

Kc = ----------------------

[ H2O ]2

 

 

 

 

Since the ionization of water occurs only once in 10 billion collisions, the concentration of water at 55.5 M is constant and so is the square of the water concentration: [ H2O ]2

 

Kc [ H2O ]2 = [H3O1+] [OH1-]

 

A constant times a constant is a constant, which is the special constant

Kw= Kc [ H2O ]2

 

Therefore: Kw = [H3O1+] [OH1-]

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Write the hydrolysis reaction for Sodium acetate, NaC2H3O2 , when it is dissolved in water. Write the Kh (Kb ) expression for this reaction:

 

NaC2H3O2 (s) + H2O (l) Na1+(aq) + C2H3O21-(aq)

 

The Sodium ion does not react with water:

Na1+(aq) + HOH No reaction (salt of a strong base)

 

The acetate ion does react with water because Acetic Acid is a weak acid:

C2H3O21- (aq) + H2O (l) HC2H3O2 (aq) + OH1- (aq)

 

The hydrolysis equilibrium expression for this reaction is:

 

[HC2H3O2] [OH 1-]

Kh = ----------------------

[C2H3O21-]

 

 

Write the ionization equilibrium expression for the Ka of acetic acid: HC2H3O2 and the Kw expression for water.

 

H2O + H2O == H3O 1+ + OH 1-

 

Kw = [H3O 1+] [OH 1-]

 

HC2H3O2(aq) + H2O (l) H3O+1 (aq) + C2H3O2-1 (aq)

 

[ C2H3O21-] [H3O1+]

Ka = -----------------------------

[ HC2H3O2]

 

 

.

 

 

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Derive the Kb expression from the Ka and the Kw expressions

Since: Kw = [H3O 1+] [OH 1-]

 

Therefore: [H3O 1+] = Kw / [OH 1-]

 

Now:

 

[ C2H3O2-1] [H3O+1]

Ka = -----------------------------

[ HC2H3O2]

 

 

 

 

 

Substituting [H3O 1+] = Kw / [OH 1-]

Into the Ka expressions for acetic acid:

 

[ C2H3O2-1] Kw / [OH 1-] [ C2H3O2-1] Kw

Ka = ----------------------------- or --------------------------

[ HC2H3O2] [ HC2H3O2] [OH 1-]

 

 

 

 

 

 

Cross multiplying:

Ka[ HC2H3O2] [OH 1-] = [ C2H3O2-1] Kw

 

Isolating the constants:

Kw [HC2H3O2] [OH 1-]

----- = ----------------------

Ka [C2H3O21-]

 

 

Therefore a constant divided by a constant is a constant, Kh or Kb

Kw [HC2H3O2] [OH 1-]

Kb = ----- = ----------------------

Ka [C2H3O21-]

QED

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There is the following connection:

Ka x Kb = Kw

Using the ionization reaction of HCN in water, demonstrate the development of the above formula from the Ka of HCN and its conjugate base Kb of CN-

 

HCN(aq) + H2O (l) H3O+1 (aq) + CN-1 (aq)

 

[CN 1-] [H3O1+]

Ka = -----------------------

[HCN]

 

 

 

 

 

or

 

Ka [CN 1-]

---------- = --------------

[H3O1+] [HCN]

 

 

 

 

 

CN 1- (aq) + H2O (l) HCN (aq) + OH1- (aq)

 

The hydrolysis equilibrium expression for this reaction is:

[HCN] [OH 1-]

Kh = ----------------------

[CN 1-]

Cross multiplying:

Kh[CN 1-] = [HCN] [OH 1-]

 

Cross multiplying again (Kh is Kb):

 

[CN 1-] [OH 1-]

--------- = -------------

[HCN] Kb

 

 

 

 

 

 

From the Ka expression:

 

Ka [CN 1-]

---------- = --------------

[H3O1+] [HCN]

 

 

 

 

 

 

From the hydrolysis expression:

 

[CN 1-] [OH 1-]

--------- = -------------

[HCN] Kb

 

Therefore:

 

Ka [OH 1-]

---------- = ---------

[H3O1+] Kb

 

 

 

 

 

Cross multiplying:

KaKb = [H3O1+] [OH 1-]

 

Since

 

Kw = [H3O1+] [OH 1-]

 

Therefore:

 

Kw = Ka Kb

 

qed

 

Read Chapter 14 Section 14.4; Section 14.13 in the McMurry text.