91影视

(a)$\text{pH}\text{=}\text{7.00}$

${\text{NH}}_3+{\text{H}}_2\text{O}鈬�{\text{NH}}_4^{+}+{\text{OH}}^{-}$

Write the dissociation equation for the${\text{NH}}_4^{+}$

${\text{NH}}_4^{+}+{\text{H}}_2\text{O}鈬�{\text{NH}}_3+{\text{H}}_3{\text{O}}^{+}$

Write and calculate the${\text{K}}_{\text{a}}$ expression from the${\text{K}}_{\text{b}}$ of${\text{NH}}_3$

$$\begin{gathered} {\text{K}}_{\text{a}}\text{=}\frac{\text{[ product ]}}{\text{[ reactant ]}} \\ {\text{K}}_{\text{a}}\text{=}\frac{\left[{\text{NH}}_{\text{3}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[{\text{NH}}_{\text{4}}^{\text{+}}\right]} \end{gathered}$$

$$\begin{gathered} {\text{K}}_{\text{w}}\text{=}{\text{K}}_{\text{a}}{\text{脳碍}}_{\text{b}} \\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{b}}} \\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{1脳10}}^{\text{- 14}}}{{\text{1.76脳10}}^{\text{- 5}}} \\ {\text{K}}_{\text{a}}{\text{= 5.68脳10}}^{\text{- 10}} \end{gathered}$$

Now calculate the $\left[{\text{H}}_3{\text{O}}^{+}\right]$from the given pH.

$$\begin{gathered} \text{pH =}\text{- log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right] \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}{\text{10}}^{\text{- pH}} \\ \text{=}{\text{10}}^{\text{- 7.00}} \end{gathered}$$

Since we know the value of $\left[{\text{H}}_3{\text{O}}^{+}\right]$and ${\text{K}}_{\text{a}}$,let us try to derive the

$\frac{\left[{\text{NH}}_{\text{3}}\right]}{\left[{\text{NH}}_{\text{3}}\right]\text{+}\left[{\text{NH}}_{\text{4}}\right]}{\text{from theK}}_{\text{a}}$

${\text{K}}_{\text{a}}\text{=}\frac{\left[{\text{NH}}_{\text{3}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[{\text{NH}}_{\text{4}}^{\text{+}}\right]}$

$\frac{\left[{\text{NH}}_4^{+}\right]}{\left[{\text{NH}}_3\right]}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]}{{\text{K}}_a}$

To have an addition element, let us try to add a factor equivalent to 1 on both sides: denominator.

$\frac{\left[{\text{NH}}_{\text{4}}^{\text{+}}\right]}{\left[{\text{NH}}_{\text{3}}\right]}\text{+}\frac{\left[{\text{NH}}_{\text{3}}\right]}{\left[{\text{NH}}_{\text{3}}\right]}\text{=}鈥�\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{{\text{K}}_{\text{a}}}\text{+}\frac{{\text{K}}_{\text{a}}}{{\text{K}}_{\text{a}}}\frac{\left[{\text{NH}}_{\text{4}}^{\text{+}}\right]\text{+}\left[{\text{NH}}_{\text{3}}\right]}{\left[{\text{NH}}_{\text{3}}\right]}\text{=}\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{+K}}_{\text{a}}}{{\text{K}}_{\text{a}}}$

Reverse both reactions to get the equation

$\frac{\left[{\text{NH}}_3\right]}{\left[{\text{NH}}_4^{+}\right]+\left[{\text{NH}}_3\right]}=\frac{{\text{K}}_{\text{a}}}{\left[{\text{H}}_3{\text{O}}^{+}\right]+{\text{K}}_a}$

Since we obtained the formula for$\frac{\left[{\text{NH}}_3\right]}{\left[{\text{NH}}_3\right]+\left[{\text{NH}}_4\right]}$one can use the values for$\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{andK}}_{\text{a}}$

role="math" localid="1663320945978" $\frac{\left[{\text{NH}}_{\text{3}}\right]}{\left[{\text{NH}}_{\text{4}}^{\text{+}}\right]\text{+}\left[{\text{NH}}_{\text{3}}\right]}\text{=}\frac{{\text{K}}_{\text{a}}}{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{+K}}_{\text{a}}}\text{=}\frac{{\text{5.68脳10}}^{\text{- 10}}}{{\text{1脳10}}^{\text{- 7}}{\text{+ 5.68脳10}}^{\text{- 10}}}\frac{\left[{\text{NH}}_{\text{3}}\right]}{\left[{\text{NH}}_{\text{4}}^{\text{+}}\right]\text{+}\left[{\text{NH}}_{\text{3}}\right]}\text{=}{\text{5.65脳10}}^{\text{- 3}}$

Hence the fraction of the total nitrogen in solution is,$5.65脳{10}^{-3}$

(b)$\text{pH = 10}$

Since we already obtained the formula for$\frac{\left[{\text{NH}}_3\right]}{\left[{\text{NH}}_3\right]+\left[{\text{NH}}_4\right]}$ and the value for${\text{K}}_{\text{a}}$ , solve for the$\left[{\text{H}}_3{\text{O}}^{+}\right]$ of the solution first.

$$\begin{gathered} \text{pH =}\text{- log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right] \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}{\text{10}}^{\text{- pH}} \\ \text{pH =}{\text{10}}^{\text{- 10}} \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{= 1脳10}}^{\text{- 10}} \end{gathered}$$

Lastly, solve for the$\frac{\left[{\text{NH}}_{\text{3}}\right]}{\left[{\text{NH}}_{\text{3}}\right]\text{+}\left[{\text{NH}}_{\text{4}}\right]}$.

$\frac{\left[{\text{NH}}_{\text{3}}\right]}{\left[{\text{NH}}_{\text{4}}^{\text{+}}\right]\text{+}\left[{\text{NH}}_{\text{3}}\right]}\text{=}\frac{{\text{K}}_{\text{a}}}{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{+K}}_{\text{a}}}\text{=}\frac{{\text{5.68脳10}}^{\text{- 10}}}{{\text{1脳10}}^{\text{- 10}}{\text{+ 5.68脳10}}^{\text{- 10}}}\frac{\left[{\text{NH}}_{\text{3}}\right]}{\left[{\text{NH}}_{\text{4}}^{\text{+}}\right]\text{+}\left[{\text{NH}}_{\text{3}}\right]}\text{=}\text{0.850}$

Hence the fraction at pH =10.00 is 0.850

(c)

As observed in the result above, when the pH is increased, the $\frac{\left[{\text{NH}}_{\text{3}}\right]}{\left[{\text{NH}}_{\text{3}}\right]\text{+}\left[{\text{NH}}_{\text{4}}\right]}$also increases. At high pH, the${\text{OH}}^{-}$ is abundant which will make the reaction shift to the reactant side producing more ammonia gas, stripping ammonia from the solution and into the air. The fraction of ${\text{NH}}_3$in the solution will approach because the ammonium ions in the solution will be consumed to produce the gas. However, this does not mean that the${\text{NH}}_3$ in the solution increased. The basis is that ammonium ions were consumed to be converted to ammonia gas and not the other way around.

Hence the basis for ammonia stripping is in the explanation.