Q18.184CP - Chemistry: Molecular Nature Of Matter And Change

Question

Nitrogen is discharged from wastewater treatment facilities into rivers and streams, usually as ${\text{NH}}_{\text{3}}$ and ${\mathbf{\text{NH}}}_{\mathbf{4}}$ :
${\mathbf{\text{NH}}}_{\mathbf{\text{3}}}{\mathbf{\text{(aq) + H}}}_{\mathbf{\text{2}}}\mathbf{\text{O(l)}}\mathbf{\rightleftharpoons }{\mathbf{\text{NH}}}_{\mathbf{\text{4}}}^{\mathbf{\text{ + }}}{\mathbf{\text{(aq) + OH}}}^{\mathbf{\text{ - }}}\mathbf{\text{(aq)}}\mathbf{ }\mathbf{ }\mathbf{ }{\mathbf{\text{K}}}_{\mathbf{\text{b}}}{\mathbf{\text{ = 1.76×10}}}^{\mathbf{\text{ - 5}}}$

One strategy for removing it is to raise the pH and “strip” the ${\mathbf{\text{NH}}}_{\mathbf{\text{3}}}$ from solution by bubbling air through the water. 

(a) At pH=7.00, what fraction of the total nitrogen in solution is  ${\mathbf{\text{NH}}}_{\mathbf{\text{3}}}$, defined as  $\left[{\text{NH}}_{\text{3}}\right]\mathbf{\text{/}}\left(\left[{\text{NH}}_{\text{3}}\right]\text{ + }\left[\text{N}{{\text{H}}_{\text{4}}}^{\text{ + }}\right]\right)$ 

 (b) What is the fraction at pH 10.00 ? 

(c) Explain the basis of ammonia stripping.

Step-by-Step Solution

Answer

a) The fraction of the total nitrogen in solution is, $5.65\times {10}^{-3}$

b) The fraction at pH =10.00 is 0.850

c) The basis for ammonia stripping is in the explanation

Step 1: To find the fraction of the total nitrogen in solution

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

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

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

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

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

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

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

Step 2: To calculate the [ H 3 O  +  ]

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

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

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

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

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

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

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 $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]{\text{and K}}_{\text{a}}$

$\frac{\left[{\text{NH}}_{\text{3}}\right]}{\left[{\text{NH}}_{\text{4}}^{\text{ + }}\right]\text{ + }\left[{\text{NH}}_{\text{3}}\right]}\text{ = } \frac{{\text{K}}_{\text{a}}}{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]{\text{ + K}}_{\text{a}}} \text{ = } \frac{{\text{5.68×10}}^{\text{ - 10}}}{{\text{1×10}}^{\text{ - 7}}{\text{ + 5.68×10}}^{\text{ - 10}}}\frac{\left[{\text{NH}}_{\text{3}}\right]}{\left[{\text{NH}}_{\text{4}}^{\text{ + }}\right]\text{ + }\left[{\text{NH}}_{\text{3}}\right]}\text{ = } {\text{5.65×10}}^{\text{ - 3}}$

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

Step 3: To find the fraction at pH 10.00

(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.

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

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

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

Hence the fraction at pH =10.00 is 0.850

Step 4: To explain the basis of ammonia stripping

(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.