Answers
 

Question	Answer	Question	Answer	Question	Answer
102	4	106	2	110	3
103	2	107	1	111	2
104	5	108	2	112	4
105	2	109	4

Explanation

Pivotal to understanding pK is an understanding of both pH and equilibria.

If (1)         HA ⇌ A^- + H⁺
Where AH is the acid (can donate a H⁺ ion), and A^- is the base (can accept a H⁺ ion)

And (2)     pH = pK + log10 ([Base] / [Acid])

Then the pK of a group is the pH at which half of the population of the group is protonated, and half is deprotonated.
Thus, [Base] / [Acid] = 1. Log10 1 = 0. And so pH = pK.

(1)    is an equilibrium equation. Thus changes in the [H⁺] will affect the ratio of [base] : [acid].
        A [H⁺] higher than the pK will result in the equilibrium shifting to the left, so increasing the proportion of acid.
        A [H⁺] lower than the pK will result in the equilibria shifting to the right, increasing the proportion of base.

For example, the carboxylic acid group of an example molecule has a H⁺ dissociation represented by:
                                      -COOH ⇌ -COO-+ H⁺ and a pK of 4.2.
This means that at pH 4.2, [-COOH] = [-COO^-]. At a neutral pH (~7), the [H⁺] is much lower than at pH 4.2, thus the basic form -COO^- will predominate. At a highly acidic pH (pH <2), the acid form -COOH will predominate.

The equilibrium is also true of amine groups, -CHNH₃⁺⇌CHNH₂ + H⁺. The pK for amine groups are usually around 9.
At pH ~ 9, [acid] = [base].

I.e. [-CHNH₃⁺] = [-CHNH₂]. At a highly alkaline pH (~11), the [H⁺] is much lower than at pH 9, thus the basic form -CHNH₂ will predominate. At a neutral pH (~7) or acidic pH, the acid form -CHNH₃⁺ will predominate.

Q 109.    explains how the ionisation of a functional group changes over a pH range. Remember the pH scale. One pH unit describes a change in one order of magnitude of [H⁺] (ie a factor of 10). Thus pH 2 has 10 x [H⁺] of pH 3.

                CH₃CH₂.COOH  ⇌ CH₃CH₂.COO^- + H⁺         pK = 4.6

Answer 5. (At pH 4.6 [ethanoic acid] = [ethanoate]). This is true. Consider the definition of pK. Note that ethanoic acid is the H⁺ donor while ethanoate is the base, the H⁺ acceptor.

Answer 1. (At pH 3.6 [ethanoic acid] = 10 × [ethanoate]). At pH 3.6, the [H⁺] is 10 x that of the pK. This increase in [H⁺] will push the equilibrium over to the left by a factor of 10. This answer is correct.

Answer 3. (At pH 5.6 [ethanoic acid] = 0.1 × [ethanoate]). At pH 5.6, the [H⁺] is 0.1 x that of the pK. This decrease in [H⁺] will push the equilibrium over to the right by a factor or 10. This answer is correct.

Answer 2. (The buffer works well at pH 4.6) is correct, as at this pH there are equal amounts of acid and base to donate and accept H⁺.

Answer 4. (The buffer works well at pH 7.0) is incorrect. A pH 7 is over 2 pH units away from the pK. Ie the [H⁺] is less than 10^-2 times that of the pK. This means that [ethanoic acid] < 0.01 × [ethanoate]. There is very little acid left to donate H⁺ to a solution to buffer a further increase in pH.

Molecules functional well as buffers within one pH unit either side of the functional group's pK. This is because within this pH range there are significant quantities of both acid and base to buffer any change in [H⁺]. If a molecule has two functional groups that have pKs within a couple of pH units of each other, it will act as a good buffer over a larger pH range. If the pH increases so that one group begins to loose its buffering capacity, the ratio of [acid] to [base] of the other group will shift so that it will begin to act as a buffer. Examples of such buffers are oxalic acid, H₃PO₄ and succinic acid (Q 109,110 & 111).