Q160CP - Chemistry: Molecular Nature Of Matter And Change

Question

Liquid ammonia auto ionizes like water:

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

where (am) represents solvation by ${\mathbf{\text{NH}}}_{\mathbf{\text{3}}}$. 

(a) Write the ion-product constant expression, Kam.

b) What are the strongest acid and base that can exist in ${\mathbf{\text{NH}}}_{\mathbf{\text{3}}}$ (l)? 

(c) ${\mathbf{\text{HNO}}}_{\mathbf{\text{3}}}$  and HCOOH are levelled in ${\mathbf{\text{NH}}}_{\mathbf{\text{3}}}$  (l). Explain with equations. 

(d) At the boiling point($\mathbf{-}\mathbf{33}\mathbf{°}\mathbf{C}$), ${\mathbf{\text{K}}}_{\mathbf{\text{am }}}{\mathbf{\text{ = 5.1×10}}}^{\mathbf{-}\mathbf{27}}$. Calculate [$\mathbf{\text{N}}{{\mathbf{\text{H}}}_{\mathbf{\text{4}}}}^{\mathbf{\text{ + }}}$] at this temperature. 

(e) Pure sulfuric acid also auto ionizes. Write the ion-product constant expression, ${\mathbf{\text{K}}}_{\mathbf{\text{sulf }}}$ , and find the concentration of the conjugate base at $\mathbf{20}\mathbf{°}\mathbf{C}$$\left({\text{K}}_{\text{sutf }}{\text{ = 2.7×10}}^{-4}\text{at 20°C}\right)$

Step-by-Step Solution

Answer

a) The ion product expression is ${\text{K}}_{\text{am}}\text{ = }\left[{\text{NH}}_{\text{4}}^{\text{ + }}\right]\left[{\text{NH}}_{\text{2}}^{\text{ - }}\right].$

b) The strongest acid and base than can exist in ${\text{NH}}_{\text{3}}{\text{ are NH}}_{\text{4}}^{\text{ + }}{\text{ and NH}}_{\text{2}}^{\text{ - }}\text{.}$

c) ${\text{HNO}}_3$ is a strong acid and HCOOH is weak acid in water

d) $\text{x = }\left[{\text{NH}}_{\text{4}}^{\text{ + }}\right]\text{ = }\left[{\text{NH}}_{\text{2}}^{\text{ - }}\right],$ then $\left[{\text{N}}_{\text{4}}^{\text{ + }}\right]{\text{ = 7.14×10}}^{-14}\text{ M}$

e) $x=\left[{\text{H}}_{\text{3}}{\text{SO}}_{\text{4}}^{\text{ + }}\right]\text{ = }\left[{\text{HSO}}_{\text{4}}^{\text{ - }}\right]\text{,then }\left[{\text{HSO}}_{\text{4}}^{\text{ - }}\right]\text{ = 0.016 M}$

Step 1(a): Write the ion-product constant expression

To write the Kam, we will not include the liquid state chemical species because it will remain constant.

$\begin{array}{rcl}& & {\text{K}}_{\text{am}}\text{ = }\frac{\text{[ product ]}}{\text{[ reactant ]}}\\ & & \text{ = }\left[{\text{NH}}_{\text{4}}^{\text{ + }}\right]\left[{\text{NH}}_{\text{2}}^{\text{ - }}\right].\end{array}$

Hence, the ion product expression is ${\text{K}}_{\text{am}}\text{ = }\left[{\text{NH}}_{\text{4}}^{\text{ + }}\right]\left[{\text{NH}}_{\text{2}}^{\text{ - }}\right].$ 

Step 2(b): Find the strongest acid and base

According to the reaction above, the strongest acid and base than can exist in ${\text{NH}}_{\text{3}}{\text{ are NH}}_{\text{4}}^{\text{ + }}{\text{ and NH}}_{\text{2}}^{\text{ - }}\text{.}$Just like how ${\text{H}}_3{\text{O}}^{+}$ and ${\text{OH}}^{-}$are the strongest acid and base that can exist in water.

Hence, the strongest acid and base than can exist in  _{${\text{NH}}_{\text{3}}{\text{ are NH}}_{\text{4}}^{\text{ + }}{\text{ and NH}}_{\text{2}}^{\text{ - }}\text{.}$}

Step 3: Explain with equation

First, write the acid-base reaction of ${\text{HNO}}_{\text{3}}{\text{ and HCOOH in NH}}_{\text{3}}.$

$$\begin{gathered} {\text{HNO}}_{\text{3}}{\text{ + NH}}_{\text{3}}\to {\text{NO}}_{\text{3}}^{\text{ - }}{\text{ + NH}}_{\text{4}}^{\text{ + }} \\ {\text{HCOOH + NH}}_{\text{3}}\to {\text{HCOO}}^{\text{ - }}{\text{ + NH}}_{\text{4}}^{\text{ + }}\text{n} \end{gathered}$$

The following reactants are being levelled in ${\text{NH}}_{\text{3}}$ because they both reacted and produced  ${\text{NH}}_4^{+},$which also indicate that they are equally strong in ${\text{NH}}_{\text{3}}$ 

Hence,  ${\text{HNO}}_3$is a strong acid and HCOCH is weak acid in water.

Step 4: To Calculate [ N H 4  +  ] at this temperature

First, note that $\left[{\text{NH}}_{\text{4}}^{\text{ + }}\right]\text{ = }\left[{\text{NH}}_{\text{2}}^{\text{ - }}\right].$

Since they are produced equally by the auto ionization of ${\text{NH}}_3,$  we can designate the two as x.

$\begin{array}{rcl}& & {\text{K}}_{\text{am}}\text{ = }\left[{\text{NH}}_{\text{4}}^{\text{ + }}\right]\left[{\text{NH}}_{\text{2}}^{\text{ - }}\right]\\ & & {\text{ = x}}^{\text{2}}\\ & & \text{ x = }\sqrt{{\text{K}}_{\text{am}}}\\ & =& \sqrt{5.1\times {10}^{-27}}\\ & =& 7.14\times {10}^{-14}\end{array}$

Since , $\text{x} \text{ = }\left[{\text{NH}}_{\text{4}}^{\text{ + }}\right]\text{ = }\left[{\text{NH}}_{\text{2}}^{\text{ - }}\right]$then $\left[{\text{N}}_{\text{4}}^{\text{ + }}\right] {\text{ = 7.14×10}}^{-14}\text{M}$ 

Step 5(e): Fnd the concentration of the conjugate base

First, write the auto ionization reaction.

${\text{2H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(l)}\to {\text{H}}_{\text{3}}{\text{SO}}_{\text{4}}^{\text{ + }}{\text{(sulf ) + HSO}}_{\text{4}}^{\text{ - }}\text{(sulf )}$

To write the ${\text{K}}_{\text{sulf }},$ we will not include the liquid state chemical species because it will remain constant.

$\begin{array}{rcl}& & {\text{K}}_{\text{sulf }}\text{ = }\frac{\text{[ product ]}}{\text{[ reactant ]}}\\ & & \text{ = }\left[{\text{H}}_{\text{3}}{\text{SO}}_{\text{4}}^{\text{ + }}\right]\left[{\text{HSO}}_{\text{4}}^{\text{ - }}\right].\end{array}$

Note that $\left[{\text{H}}_{\text{3}}{\text{SO}}_{\text{4}}^{\text{ + }}\right]\text{ = }\left[{\text{HSO}}_{\text{4}}^{\text{ - }}\right]$ since they are produced equally by the auto ionization of ${\text{H}}_2{\text{SO}}_4$. So we can designate the two as x.

$\begin{array}{rcl}& & {\text{K}}_{\text{sulf }}\text{ = }\left[{\text{NH}}_{\text{4}}^{\text{ + }}\right]\left[{\text{NH}}_{\text{2}}^{\text{ - }}\right]\\ & & {\text{ = x}}^{\text{2}}\\ & & \text{ x = }\sqrt{{\text{K}}_{\text{sulf }}}\\ & =& \sqrt{2.7\times {10}^{-4}}\\ & =& 0.016\end{array}$

Since $\text{x} \text{ = } \left[{\text{H}}_{\text{3}}{\text{SO}}_{\text{4}}^{\text{ + }}\right]\text{ = }\left[{\text{HSO}}_{\text{4}}^{\text{ - }}\right]\text{, then }\left[{\text{HSO}}_{\text{4}}^{\text{ - }}\right]\text{ = 0.016 M.}$