Question

Liquid ammonia autoionizes like water: $$ 2 \mathrm{NH}_{3}(l) \longrightarrow \mathrm{NH}_{4}^{+}(a m)+\mathrm{NH}_{2}^{-}(a m) $$ where \((a m)\) represents solvation by \(\mathrm{NH}_{3}\). (a) Write the ion-product constant expression, \(K_{\text {am }}\) (b) What are the strongest acid and base that can exist in \(\mathrm{NH}_{3}(l) ?\) (c) \(\mathrm{HNO}_{3}\) and \(\mathrm{HCOOH}\) are leveled in \(\mathrm{NH}_{3}(l) .\) Explain with equations. (d) At the boiling point of ammonia \(\left(-33^{\circ} \mathrm{C}\right), K_{\text {unt }}=5.1 \times 10^{-27}\) Calculate \(\left[\mathrm{NH}_{4}^{+}\right]\) at this temperature. (c) Pure sulfuric acid also autoionizes. Write the ion-product constant expression, \(K_{\text {sulf }}\), and find the concentration of the conjugate base at \(20^{\circ} \mathrm{C}\left(K_{\mathrm{sulf}}=2.7 \times 10^{-4} \mathrm{at} 20^{\circ} \mathrm{C}\right)\)

Short answer

a) \ \[ K_{\text{am}} = [\mathrm{NH}_4^+][\mathrm{NH}_2^-] \] b) Strongest acid: \( \mathrm{NH}_4^+ \); strongest base: \( \mathrm{NH}_2^- \). c) \( \mathrm{HNO}_3 \) and \( \mathrm{HCOOH} \) level to \( \mathrm{NH}_4^+ \) in \( \mathrm{NH}_3 \). d) \( [\mathrm{NH}_4^+] = 7.14 \times 10^{-14} \text{M} \) e) \( K_{\text{sulf}} = [\mathrm{HSO}_4^-][\mathrm{H}_3\mathrm{SO}_4^+] \), \( [\mathrm{HSO}_4^-] = 1.64 \times 10^{-2} \text{M} \).

Step-by-step solution

Writing the Ion-Product Constant Expression

The ion-product constant expression, \( K_{\text {am }} \), represents the equilibrium concentration of the ions produced by the autoionization of liquid ammonia. The equilibrium reaction is: \ \[ 2 \mathrm{NH}_{3}(l) \longrightarrow \mathrm{NH}_{4}^{+}(am) + \mathrm{NH}_{2}^{-}(am) \] \ Hence, \( K_{\text{am}} \) is given by: \ \[ K_{\text{am}} = [\mathrm{NH}_{4}^{+}(am)][\mathrm{NH}_{2}^{-}(am)] \]

Identifying the Strongest Acid and Base in Liquid Ammonia

The strongest acid that can exist in liquid ammonia is the ammonium ion, \( \mathrm{NH}_4^+ \) as any stronger acid would completely transfer its proton to \( \mathrm{NH}_3 \). The strongest base is the amide ion, \( \mathrm{NH}_2^- \), as any stronger base would completely deprotonate \( \mathrm{NH}_3 \).

Explaining Acid-Base Leveling in Liquid Ammonia

Acid-base leveling occurs when strong acids or bases are converted to the strongest possible acid or base in a specific solvent. For example: \ \[ \mathrm{HNO}_3 + \mathrm{NH}_3 \rightarrow \mathrm{NH}_4^+ + \mathrm{NO}_3^- \] \ \[ \mathrm{HCOOH} + \mathrm{NH}_3 \rightarrow \mathrm{NH}_4^+ + \mathrm{HCOO}^- \] Both reactions show the strong acids (\( \mathrm{HNO}_3 \) and \( \mathrm{HCOOH} \)) are leveled to \( \mathrm{NH}_4^+ \) when dissolved in \( \mathrm{NH}_3 \).

Calculating \( [\mathrm{NH}_4^+] \) at Boiling Point of Ammonia

Given \( K_{\text{am}}=5.1 \times 10^{-27} \) at \( -33^{\circ}\text{C} \), and assuming complete dissociation of \( \mathrm{NH}_4^+ \) and \( \mathrm{NH}_2^- \), the ion concentrations are equal: \ \[ [\mathrm{NH}_4^+] = [\mathrm{NH}_2^- ] \] \ Hence, \ \[ [\mathrm{NH}_4^+]^2 = 5.1 \times 10^{-27} \] \ Solving for \( [\mathrm{NH}_4^+] \) we get: \ \[ [\mathrm{NH}_4^+] = (5.1 \times 10^{-27})^{\frac{1}{2}} \approx 7.14 \times 10^{-14} \text{M} \]

Writing the Ion-Product Constant Expression for Sulfuric Acid

For the autoionization of pure sulfuric acid, the reaction is: \ \[ 2 \mathrm{H}_2\mathrm{SO}_4 \rightleftharpoons \mathrm{HSO}_4^- + \mathrm{H}_3\mathrm{SO}_4^+ \] \ The ion-product constant expression \( K_{\text{sulf}} \) is: \ \[ K_{\text{sulf}} = [\mathrm{HSO}_4^- ][\mathrm{H}_3\mathrm{SO}_4^+] \]

Calculating the Concentration of the Conjugate Base in Sulfuric Acid

Given \( K_{\text{sulf}} = 2.7 \times 10^{-4} \) at \( 20^{\circ}\text{C} \), the concentrations of \( \mathrm{HSO}_4^- \) and \( \mathrm{H}_3\mathrm{SO}_4^+ \) are equal: \ \[ [\mathrm{HSO}_4^-] = [\mathrm{H}_3\mathrm{SO}_4^+] \] \ Hence, \ \[ [\mathrm{HSO}_4^-]^2 = 2.7 \times 10^{-4} \] \ Solving for \( [\mathrm{HSO}_4^-] \) we get: \ \[ [\mathrm{HSO}_4^-] = (2.7 \times 10^{-4})^{\frac{1}{2}} \approx 1.64 \times 10^{-2} \text{M} \]

Key concepts

Ion-Product Constant

The ion-product constant, often denoted as \(K_{am}\) for liquid ammonia, is an equilibrium constant specific to the autoionization of a solvent. In liquid ammonia, the reaction can be described as:
\(2 \mathrm{NH}_{3}(l) \longrightarrow \mathrm{NH}_{4}^{+}(am) + \mathrm{NH}_{2}^{-}(am)\). This essentially measures the extent of this equilibrium.

The expression for the ion-product constant can be written as:
\[ K_{am} = [\mathrm{NH}_{4}^{+}(am)][\mathrm{NH}_{2}^{-}(am)] \]
The placement of these ions in brackets signifies their molar concentrations at equilibrium. This is analogous to the ion-product constant for water, \(K_w\), given as:
\[K_w = [\mathrm{H}^{+}][\mathrm{OH}^{-}] \]. This value of \(K_w\) for water changes with temperature, similar to \(K_{am}\) for ammonia.
Understanding the ion-product constant helps to estimate how many ions are present in the solvent, giving insights into the solvent's properties like conductivity and reactivity.

Acid-Base Leveling

Acid-base leveling is a concept which dictates that in a given solvent, all acids stronger than the solvent's characteristic acid will give a common strongest acid. Similarly, all bases stronger than the strongest base will become the same in the solvent. For example, in liquid ammonia (\(NH_3\)), the strongest base is \(NH_2^-\). This means that stronger bases will just convert to \(NH_2^-\). The strongest acid in \(NH_3\) is \(NH_4^+\). This explains why acids like \(HNO_3\) and \(HCOOH\) will result in the formation of \(NH_4^+\) in liquid ammonia.

Consider these equations:
\[\mathrm{HNO}_3 + \mathrm{NH}_3 \rightarrow \mathrm{NH}_4^+ + \mathrm{NO}_3^- \]
\[\mathrm{HCOOH} + \mathrm{NH}_3 \rightarrow \mathrm{NH}_4^+ + \mathrm{HCOO}^- \]
Both strong acids are leveled to the same product, \(NH_4^+\), in \(NH_3\). This leveling effect occurs because the solvent (in this case \(NH_3\)) dictates the strongest possible acid or base by autoionizing itself to its own ions.

Equilibrium Concentrations

In any chemical reaction at equilibrium, the concentrations of the reactants and the products stay constant over time. For the case of liquid ammonia, given \(K_{am}=5.1 \times 10^{-27}\) at the boiling point of ammonia \(-33^{\circ} \text{C}\), the equilibrium concentration of \(NH_4^+\) and \(NH_2^-\) ions can be calculated.
From the relationship:
\[ K_{am} = [\mathrm{NH}_4^+][\mathrm{NH}_2^-] \]
And knowing that in a 1:1 dissociation process, \([\mathrm{NH}_4^+] = [\mathrm{NH}_2^-]\), the equation simplifies to:
\[ [\mathrm{NH}_4^+]^2 = 5.1 \times 10^{-27} \]
Solving for the concentration:
\[ [\mathrm{NH}_4^+] = (5.1 \times 10^{-27})^{\frac{1}{2}} \approx 7.14 \times 10^{-14} \text{M} \]
This tiny concentration indicates that very little autoionization occurs at \(-33^{\circ} \text{C}\). These numbers help chemists predict and design reactions in these solvents by knowing the maximum ion concentration.

Strong Acids and Bases

The strongest acids and bases which can exist in a solvent are determined by the nature of the solvent's autoionization. In liquid ammonia, the strongest acid is \(NH_4^+\) and the strongest base is \(NH_2^-\).

This is due to the autoionization of \(NH_3\):
\[ 2 \mathrm{NH}_3(l) \longrightarrow \mathrm{NH}_4^{+}(am) + \mathrm{NH}_2^{-}(am) \]
No acid stronger than \(NH_4^+\) will remain as such, as they will donate protons to form \(NH_4^+\). Similarly, stronger bases than \(NH_2^-\) will take protons to form \(NH_2^-\).

This determines their reactivity in liquid ammonia. For instance, in water (\(H_2O\)), sulfuric acid (\(H_2SO_4\)) is very strong, but in \(NH_3\), it would act differently, becoming \(NH_4^+\). Understanding these relations can help predict outcomes in various chemical reactions, proving crucial for applied and theoretical chemistry.