Ionization of Weak Acids & Weak Bases (College Board AP Chemistry)

• Weak acids will only partially ionize in aqueous solutions 
  • Examples include ethanoic acid and carbonic acid 
• An equilibrium is established between the un-ionized acid and its conjugate base (HA represents the acid molecule):

HA (aq) + H₂O (l) ⇌  A^- (aq) + H₃O⁺ (aq)

OR

HA (aq) ⇌  A^- (aq) + H⁺ (aq)

• The position of the equilibrium lies to the left whereby most of the acid molecules remain unionized 

 Diagram to show the ionization of a weak acid

The diagram shows the partial dissociation of a weak acid in aqueous solution

• The equations above can be used to write the following equilibrium expression:

${\mathit{K}}_{\mathbf{a}}\mathbf{ }\mathbf{=}\mathbf{ }\frac{\mathbf{\left[}{\mathbf{A}}^{\mathbf{-}}\mathbf{\right]}\left[H^{+}\right]}{\mathbf{\left[}\mathbf{HA}\mathbf{\right]}}$

• K_a is called the acid dissocation constant 
• In the context of ethanoic acid, CH₃COOH, the equation for the ionization of the acid would be:

CH₃COOH (aq) + H₂O (l) ⇌  CH₃COO^- (aq) + H₃O⁺ (aq)

OR

CH₃COOH (aq) ⇌  CH₃COO^- (aq) + H⁺ (aq)

• These could be used to deduce an equilibrium expression of:

$K_{\mathrm{a}} = \frac{\left[{\mathrm{CH}}_3{\mathrm{COO}}^{-}\right]\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]}$

• When writing the equilibrium expression for weak acids, we assume that the concentration of H⁺ (aq)  due to the ionisation of water is negligible
• K_a values are very small so we often use pK_a values when we are calculating pH values
• To convert between K_a and pK_a we do the following: 

pK_a = –logK_a   

K_a= 10^–pK_a

• The larger the K_a value the stronger the acid
• The larger the pK_a value the weaker the acid
• The pH of a weak acid solution can be determined from the initial acid concentration and the pK_a

• Weak bases will also partially ionize in aqueous solutions 
  • Examples include ammonia and methylamine 
• There will be a lower concentration of OH^– ions so the pH of a weak base is lower than that of a strong base 
• Again, an equilibrium is established between the un-ionized base and its conjugate acid with its position lying to the left so that a large proportion of base molecules remain un-ionizied 

 Diagram to show the ionization of a weak base

The diagram shows the partial dissociation of a weak base in aqueous solution

• The equilibrium established is:

B (aq) + H₂O (l) ⇌  BH⁺ (aq) + OH^- (aq)

• • The base dissociation constant, K_b is deduced from this as: 

${\mathit{K}}_{\mathrm{b}}\mathbf{ }\mathbf{=}\mathbf{ }\frac{\left[{\mathrm{BH}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\mathbf{\left[}\mathbf{B}\mathbf{\right]}}$

• In the context of 1-phenylmethanamine, C₆H₅CH₂NH₂ (aq) this dissociates as follows: 

C₆H₅CH₂NH₂ (aq) + H₂O (l) ⇌ C₆H₅CH₂NH₃⁺ (aq) + OH^– (aq)

• • This leads to the K_b  expression: 

${\mathit{K}}_{\mathrm{b}}\mathbf{ }\mathbf{=}\mathbf{ }\frac{\mathbf{[}{\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{5}}{\mathbf{CH}}_{\mathbf{2}}{{\mathbf{NH}}_{\mathbf{3}}}^{\mathbf{+}}\mathbf{\left]}\mathbf{\right[}{\mathbf{OH}}^{\mathbf{-}}\mathbf{]}}{\mathbf{\left[}{\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{5}}{\mathbf{CH}}_{\mathbf{2}}{\mathbf{NH}}_{\mathbf{2}}\mathbf{\right]}}$

• K_b values are very small so we often use pK_b values when we are calculating pH values
• To convert between K_b and pK_b we do the following: 

pK_b = –logKb   

K_b= 10^–pKb

• The pH of a weak base solution can be determined from the initial base concentration and the pK_b

• The percentage ionization of a weak acid / base represents the extent to which the acid / base dissociates into ions in a solution 
• To calculate the percentage ionization of a weak acid use:

$\% \mathrm{ionization} = \frac{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}{\mathrm{HA}} \mathrm{x} 100$

• To calculate the percentage ionization of a weak base (A-) use:

$\% \mathrm{ionization} = \frac{\left[{\mathrm{OH}}^{-}\right]}{{\mathrm{A}}^{-}} \mathrm{x} 100$