CHM 2046C Module 11 Chapter 14 Exam

Part C: Derivation of K_w K_h K_b and K_a Chapter 14 10 points

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

K_w from the K_c:

H₂O + H₂O ó H₃O⁺¹ + OH^1-

[H₃O¹⁺]¹ [OH^1-]¹

K_c = ----------------------

[ 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

K_c [ H₂O ]² = [H₃O¹⁺] [OH^1-]

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

K_w= K_c [ H₂O ]²

Therefore: K_w = [H₃O¹⁺] [OH^1-]

---------------------------------------

Write the hydrolysis reaction for Sodium acetate, NaC₂H₃O_{2 ,} when it is dissolved in water. Write the K_h (K_b ) expression for this reaction:

NaC₂H₃O₂ (s) + H₂O (l) ó Na¹⁺⁽aq) + C₂H₃O₂^1-(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:

C₂H₃O₂^1- (aq) + H₂O (l) ó HC₂H₃O₂ (aq) + OH^1- (aq)

The hydrolysis equilibrium expression for this reaction is:

[HC₂H₃O₂] [OH ^1-]

K_h = ----------------------

[C₂H₃O₂^1-]

Write the ionization equilibrium expression for the K_a of acetic acid: HC₂H₃O₂ and the K_w expression for water.

H₂O + H₂O ç==è H₃O ¹⁺ + OH ^1-

K_w = [H₃O ¹⁺] [OH ^1-]

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

[ C₂H₃O₂^1-] [H₃O¹⁺]

K_a = -----------------------------

[ HC₂H₃O₂]

.

----------------------------

Derive the K_b expression from the K_a and the K_w expressions

Since: K_w = [H₃O ¹⁺] [OH ^1-]

Therefore: [H₃O ¹⁺] = K_w / [OH ^1-]

Now:

[ C₂H₃O₂^-1] [H₃O⁺¹]

K_a = -----------------------------

[ HC₂H₃O₂]

Substituting [H₃O ¹⁺] = K_w / [OH ^1-]

Into the K_a expressions for acetic acid:

[ C₂H₃O₂^-1] K_w / [OH ^1-] [ C₂H₃O₂^-1] K_w

K_a = ----------------------------- or --------------------------

[ HC₂H₃O₂] [ HC₂H₃O₂] [OH ^1-]

Cross multiplying:

K_a[ HC₂H₃O₂] [OH ^1-] = [ C₂H₃O₂^-1] K_w

Isolating the constants:

K_w [HC₂H₃O₂] [OH ^1-]

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

K_a [C₂H₃O₂^1-]

Therefore a constant divided by a constant is a constant, K_h or K_b

K_w [HC₂H₃O₂] [OH ^1-]

K_b = ----- = ----------------------

K_a [C₂H₃O₂^1-]

_QED

-------------------------------

There is the following connection:

K_a x K_b = K_w

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

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

[CN ^1-] [H₃O¹⁺]

K_a = -----------------------

[HCN]

or

K_a [CN ^1-]

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

[H₃O¹⁺] [HCN]

CN ^1- (aq) + H₂O (l) ó HCN (aq) + OH^1- (aq)

The hydrolysis equilibrium expression for this reaction is:

[HCN] [OH ^1-]

K_h = ----------------------

[CN ^1-]

Cross multiplying:

K_h[CN ^1-] = [HCN] [OH ^1-]

Cross multiplying again (K_h is K_b):

[CN ^1-] [OH ^1-]

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

[HCN] K_b

From the K_a expression:

K_a [CN ^1-]

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

[H₃O¹⁺] [HCN]

From the hydrolysis expression:

[CN ^1-] [OH ^1-]

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

[HCN] K_b

Therefore:

K_a [OH ^1-]

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

[H₃O¹⁺] K_b

Cross multiplying:

K_a∙ K_{b =} [H₃O¹⁺] [OH ^1-]

Since

K_w = [H₃O¹⁺] [OH ^1-]

Therefore:

K_w = K_a∙ K_b

_qed

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