Q18.148CP - Chemistry: Molecular Nature Of Matter And Change

Question

The disinfectant phenol, ${\mathbf{\text{C}}}_{\mathbf{6}}{\mathbf{\text{H}}}_{\mathbf{5}}\mathbf{\text{OH,}}$, has a${\mathbf{\text{p K}}}_{\mathbf{\text{a}}}$ of$\mathbf{\text{10.0}}$ in water, but$\mathbf{\text{14.4}}$ in methanol. 

(a) Why are the values different? 

(b) Is methanol a stronger or weaker base than water? 

(c) Write the dissociation reaction of phenol in methanol. 

(d) Write an expression for the autoionization constant of methanol.

Step-by-Step Solution

Answer

a) The values are different because depends on the acid-base character of the other reactant

b) The${\text{pK}}_{\text{a}}$ of methanol is greater than the ${\text{pK}}_{\text{a}}$of water, it is a weaker base than water.

c) The dissociation reaction is ${\text{C}}_6{\text{H}}_5\text{OH}+{\text{CH}}_3\text{OH}\rightleftharpoons {\text{C}}_6{\text{H}}_5{\text{O}}^{-}+{\text{CH}}_3{\text{OH}}_2^{+}$

d) An expression for the auto ionization constant of methanol is $\text{K=}\left[{\text{CH}}_{\text{3}}{\text{OH}}_{\text{2}}^{\text{+}}\right]\left[{\text{CH}}_{\text{3}}{\text{O}}^{\text{-}}\right]$

Step 1: To find why the values are different

(a)

${\text{pK}}_{\text{a}}$ shows how much an acid will dissociate. The lower the,${\text{pK}}_{\text{a}}$ the higher the value of${\text{K}}_{\text{a}}$ Higher${\text{K}}_{\text{a}}$ dissociates more. phenol have different ${\text{pK}}_{\text{a}}$values because the dissociation reaction depends on the strength of the other reactant. Hence It depends on the acid-base character of the other reactant.${\text{pK}}_{\text{a}}$

Step 2: To find the methanol is stronger or weaker

(b)

Phenol has${\text{pK}}_{\text{a}}$ =$10.0$in water and a${\text{pK}}_{\text{a}}$ =$14.4$in methanol. Since lower  ${\text{pK}}_{\text{a}}$indicates higher ${\text{K}}_{\text{a}}$ value and higher ${\text{K}}_{\text{a}}$dissociates more, the lower${\text{pK}}_{\text{a}}$ is the stronger base. This is because a stronger base helps in the dissociation of the acid. A weaker base will be as "stagnant" as the weak acid so less dissociation will happen.

Since the ${\text{pK}}_{\text{a}}$ of methanol is greater than the${\text{pK}}_{\text{a}}$ of water, it is a weaker base than water.

Step 3: Write the dissociation reaction of phenol in methanol

(c)

The dissociation reaction of phenol in methanol is written below.

${\text{C}}_6{\text{H}}_5\text{OH}+{\text{CH}}_3\text{OH}\rightleftharpoons {\text{C}}_6{\text{H}}_5{\text{O}}^{-}+{\text{CH}}_3{\text{OH}}_2^{+}$

The hydrogen from the OH group in phenol will transfer to the OH group of methanol.Hencethedissociation reaction is

${\text{C}}_6{\text{H}}_5\text{OH}+{\text{CH}}_3\text{OH}\rightleftharpoons {\text{C}}_6{\text{H}}_5{\text{O}}^{-}+{\text{CH}}_3{\text{OH}}_2^{+}$ 

Step 4: Write an expression for the auto ionization constant of methanol

(d)

First, write auto ionization reaction of methanol.

${\text{CH}}_3\text{OH}+{\text{CH}}_3\text{OH}\rightleftharpoons {\text{CH}}_3{\text{OH}}_2^{+}+{\text{CH}}_3{\text{O}}^{-}$

The hydrogen from the OH group in methanol will transfer to the OH group of the other methanol.

Now, write the K of the reaction.

$\begin{array}{l}\text{K =}\frac{\text{[ product ]}}{\text{[ reactant ]}}\\ \text{K=}\left[{\text{CH}}_{\text{3}}{\text{OH}}_{\text{2}}^{\text{+}}\right]\left[{\text{CH}}_{\text{3}}{\text{O}}^{\text{-}}\right]\end{array}$

The methanol is not included since it is liquid.

Hence an expression for the auto ionization constant of methanol is $\text{K=}\left[{\text{CH}}_{\text{3}}{\text{OH}}_{\text{2}}^{\text{+}}\right]\left[{\text{CH}}_{\text{3}}{\text{O}}^{\text{-}}\right]$