Question

Acidity of Thiols (Section 10.9F) Thiols are weak acids, \(\mathrm{p} \mathcal{K}_{,}\)10-11, bue considerably stronger than alcohols, pK_ 15.5-18. $$ \mathrm{Ca}_{3} \mathrm{CH}_{2} \mathrm{SH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{~S}^{-}+\mathrm{H}_{5} \mathrm{O}^{+} \quad \mathrm{pK}=10.5 $$

Short answer

Based on the step by step solution provided above, analyze the acidity behavior of thiols in water by writing an acid-base equilibrium equation, and determining the equilibrium concentrations of the thiol, its conjugate base, and hydronium ion. Consider a thiol with a pK value of 10.5 and use this value to calculate the equilibrium constant, \(\mathcal{K}_\mathrm{a}\). Simplify the equilibrium expression and solve for x to find the concentrations of all species at equilibrium.

Step-by-step solution

Write the acid-base equilibrium equation and the equilibrium expression

The given acid-base reaction is: $$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{SH}+\mathrm{H}_{2} \mathrm{O}\rightleftharpoons \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{S}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} $$ The equilibrium expression for this reaction is given by: $$ \mathcal{K}_\mathrm{a} = \frac{([\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{S}^{-}][\mathrm{H}_{3}\mathrm{O}^{+}])}{([\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{SH}][\mathrm{H}_{2}\mathrm{O}])} $$

Determine the pK value and convert it to \(\mathcal{K}_\mathrm{a}\)

We are given the pK value as 10.5. To find the \(\mathcal{K}_\mathrm{a}\) value, we can use the relationship: $$ pK_\mathrm{a} = -\log{\mathcal{K}_\mathrm{a}} $$ Solving for \(\mathcal{K}_\mathrm{a}\), we get: $$ \mathcal{K}_\mathrm{a} = 10^{-pK_\mathrm{a}} = 10^{-10.5} = 3.16 \times 10^{-11} $$

Simplify the equilibrium expression for thiols

Since water is a solvent in this reaction, its concentration remains virtually constant throughout the reaction. Therefore, we can simplify the equilibrium expression by combining the constant concentrations of water with the equilibrium constant. $$ \mathcal{K'} = \mathcal{K}_\mathrm{a}[\mathrm{H}_{2}\mathrm{O}] = \frac{([\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{S}^{-}][\mathrm{H}_{3}\mathrm{O}^{+}])}{[\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{SH}]} $$ Now, we have a simplified equilibrium expression involving only the concentrations of the thiol and its conjugate base.

Determine the concentrations at equilibrium

Since we don't have any initial concentrations, let's assume an initial concentration of the thiol, \([\mathrm{CH_{3}CH_{2}SH}]_0\). At equilibrium, the concentrations will be: \([\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{S}^{-}] = x\) \([\mathrm{H}_{3}\mathrm{O}^{+}] = x\) \([\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{SH}] = [\mathrm{CH_{3}CH_{2}SH}]_0 - x\) Now, we can substitute these into the simplified equilibrium expression: $$ \mathcal{K'} = \frac{x^2}{([\mathrm{CH_{3}CH_{2}SH}]_0 - x)} $$

Solve for x and calculate concentrations at equilibrium

In this step, you can solve for x using the simplified equilibrium expression and the value of \(\mathcal{K'}\). Once x is determined, you can find the equilibrium concentrations of all species: \([\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{S}^{-}]_\mathrm{eq} = x\) \([\mathrm{H}_{3}\mathrm{O}^{+}]_\mathrm{eq} = x\) \([\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{SH}]_\mathrm{eq} = [\mathrm{CH_{3}CH_{2}SH}]_0 - x\) These equilibrium concentrations will give you an understanding of how the thiol behaves as a weak acid in the presence of water.

Key concepts

Acid-Base Equilibrium

Understanding the acid-base equilibrium is crucial for exploring the acidic or basic nature of a compound. It represents the balance between the proton donors (acids) and proton acceptors (bases) in a reaction. When a thiol, such as ethanethiol (\texttt{CH}_3\texttt{CH}_2\texttt{SH}), reacts with water, it can donate a proton, becoming its conjugate base (\texttt{CH}_3\texttt{CH}_2\texttt{S}^-).

This reaction is reversible and will reach a point of equilibrium where the rates of the forward and backward reactions are equal:
\
\(\begin{align*}\texttt{CH}_3\texttt{CH}_2\texttt{SH} + \texttt{H}_2\texttt{O} \rightleftharpoons \texttt{CH}_3\texttt{CH}_2\texttt{S}^- + \texttt{H}_3\texttt{O}^+\end{align*}\)
At this equilibrium point, the concentration of each species remains constant over time. For students, it's essential to remember that the equilibrium condition doesn't mean the reactions have stopped, but rather that they occur at the same rate in each direction.

Equilibrium Constant (Ka)

The acidity of thiols is characterized by their equilibrium constant, known as the acid dissociation constant (\texttt{Ka}). This constant quantifies the strength of an acid in a particular solvent. In our case, the solvent is water. The \texttt{Ka} value is defined by the ratio of the concentration of the products to the concentration of the reactants at equilibrium:

\[\mathcal{K}_\texttt{a} = \frac{([\texttt{CH}_3\texttt{CH}_2\texttt{S}^-][\texttt{H}_3\texttt{O}^+])}{([\texttt{CH}_3\texttt{CH}_2\texttt{SH}][\texttt{H}_2\texttt{O}])}\]

Since the concentration of water (\texttt{H}_2\texttt{O}) stays relatively constant, the equilibrium constant is usually simplified to:\[\mathcal{K'} = \mathcal{K}_\texttt{a}[\texttt{H}_2\texttt{O}]\]This makes it easier to solve equilibrium problems because only the concentrations of the thiol and its conjugate base are considered. A larger \texttt{Ka} value indicates a stronger acid, as it implies a greater concentration of ions at equilibrium.

pKa Value

The \texttt{pKa} value is another way to express the acidity of a substance. It is the negative logarithm of the acid dissociation constant (\texttt{Ka}):

\[\texttt{pKa} = -\log(\mathcal{K}_\texttt{a})\]

For thiols, this value helps us compare their acidity to other types of compounds, as it's easier to work with than the raw \texttt{Ka} values. Lower \texttt{pKa} values indicate higher acid strength. With thiols having \texttt{pKa} values between 10 to 11, they are more acidic than alcohols (\texttt{pKa} 15.5-18) but still considered weak acids. This relative scale of acidity is helpful when predicting reaction outcomes, especially in organic synthesis or biochemistry where the reaction environment's pH plays a crucial role.