Chapter 18, Problem 18.159P

(a)

Interpretation Introduction

Interpretation:

Liquid ammonia autoionizes like water:

  ${\text{2NH}}_{\text{3(l)}}\to {\text{NH}}_{\text{4}}^{\text{+}}{}_{\text{(am)}}{\text{+NH}}_{\text{2}}^{\text{-}}{}_{\text{(am)}}$

The ion-product constant expression, ${\text{K}}_{\text{am}}$ has to be written ($\text{am}$ represents solvation by ${\text{NH}}_{\text{3}}$).

Concept Introduction:

Equilibrium constant:

The relationship between the concentration of products and concentration of reactants in a chemical reaction at equilibrium is said to be equilibrium constant.  It is denoted by $\text{K}$.

For a reaction,

  $\text{xX + yY }\rightleftharpoons \text{ zZ}$

The expression of $\text{K}$ can be given as

  $\begin{array}{l}{\text{K}}_{\text{c}}\text{ = }\frac{{\text{[Z]}}^{\text{z}}}{{\text{[X]}}^{\text{x}}{\text{[Y]}}^{\text{y}}}\\ \\ \text{where,}\text{[X] = equilibrium concentration of X}\\ \\ \text{[Y] = equilibrium concentration of Y}\\ \\ \text{[Z] = equilibrium concentration of Z}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The strongest acid and base that can exist in ${\text{NH}}_{\text{3(l)}}$ has to be given.

(c)

Interpretation Introduction

Interpretation:

${\text{HNO}}_{\text{3}}\text{and}\text{HCOOH}$ are levelled in ${\text{NH}}_{\text{3(l)}}$.  It has to be explained with equations.

(d)

Interpretation Introduction

Interpretation:

At the boiling point of ammonia (${\text{-33}}^{\text{°}}\text{C}$), ${\text{K}}_{\text{am}}\text{=5}{\text{.1×10}}^{\text{-27}}$, the ${\text{[NH}}_{\text{4}}^{\text{+}}\text{]}$ has to be calculated at this temperature.

Concept Introduction:

Equilibrium constant:

The relationship between the concentration of products and concentration of reactants in a chemical reaction at equilibrium is said to be equilibrium constant.  It is denoted by $\text{K}$.

For a reaction,

  $\text{xX + yY }\rightleftharpoons \text{ zZ}$

The expression of $\text{K}$ can be given as

  $\begin{array}{l}{\text{K}}_{\text{c}}\text{ = }\frac{{\text{[Z]}}^{\text{z}}}{{\text{[X]}}^{\text{x}}{\text{[Y]}}^{\text{y}}}\\ \\ \text{where,}\text{[X] = equilibrium concentration of X}\\ \\ \text{[Y] = equilibrium concentration of Y}\\ \\ \text{[Z] = equilibrium concentration of Z}\end{array}$

(e)

Interpretation Introduction

Interpretation:

Pure sulfuric acid also autoionizes.  The ion-product constant expression, ${\text{K}}_{\text{sulf}}$ has to be written and the concentration of the conjugate base at ${\text{20}}^{\text{°}}\text{C}$ has to be calculated.  (${\text{K}}_{\text{sulf}}\text{ = 2}{\text{.7×10}}^{\text{-4}}\text{at}{\text{20}}^{\text{°}}\text{C}$)