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  ó    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)  ó  Na1+(aq) + C2H3O21-(aq)

The Sodium ion does not react with water:

Na1+(aq)  + HOH à 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)  ó  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            ç==è           H3O 1+      +      OH 1-

Kw  =  [H3O 1+] [OH 1-]

HC2H3O2(aq)   +   H2O (l)  ó    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)  ó    H3O+1 (aq)   +   CN-1 (aq)

 [CN 1-] [H3O1+] Ka = ----------------------- [HCN]

or

 Ka                 [CN 1-] ---------- = -------------- [H3O1+]      [HCN]

CN 1- (aq) +   H2O (l)  ó  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.