Q147CP - Chemistry: Molecular Nature Of Matter And Change

Question

A florist prepares a solution of nitrogen-phosphorus fertilizer by dissolving $\mathbf{\text{5.66 g}}$ of ${\mathbf{\text{NH}}}_{\mathbf{\text{4}}}{\mathbf{\text{NO}}}_{\mathbf{\text{3}}}$ and $\mathbf{\text{4.42 g}}$ of ${\left({\mathbf{NH}}_4\right)}_3{\mathbf{PO}}_4$  in enough water to make $\mathbf{\text{20.0 L}}$ of solution. What are the molarities of ${{\mathbf{NH}}_4}^{+}$  and ${{\mathbf{PO}}_4}^{3-}$ of in the solution?

Step-by-Step Solution

Answer

In $20\text{ L}$ solution, the molarity of ${{\text{NH}}_{\text{4}}}^{\text{+}}$ and ${{\text{PO}}_{\text{4}}}^{\text{3-}}$ is $0.0078\raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.$ and $0.0014\raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.$ respectively.

A concept:

In ammonium nitrate only one ${{\mathbf{NH}}_4}^{+}$  is present but in ammonium phosphate three  ${{\mathbf{NH}}_4}^{+}$ are present.

Moles of NH 4 NO 3 and ( NH 4 ) 3 PO 4 :

A florist prepares a solution of nitrogen-phosphorus fertilizer by dissolved  $5.66\text{ g}$ of  ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ and $\text{4.42 g}$ of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{3}}{\text{PO}}_{\text{4}}$.

So, $5.66\text{ g}$  and  $\text{4.42 g}$ are the given masses of   and   respectively.

 

Molar mass of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$  is  $80\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$.

And molar mass of  ${\left({\text{NH}}_{\text{4}}\right)}_{\text{3}}{\text{PO}}_{\text{4}}$ is $\text{149 }\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$.

Therefore, 

$\begin{array}{rcl}{\text{moles\hspace{0.17em}of\hspace{0.17em}NH}}_{\text{4}}{\text{NO}}_{\text{3}}& =& \frac{{\text{mass\hspace{0.17em}of\hspace{0.17em}NH}}_{\text{4}}{\text{NO}}_{\text{3}}}{{\text{molecular\hspace{0.17em}mass\hspace{0.17em}of\hspace{0.17em}NH}}_{\text{4}}{\text{NO}}_{\text{3}}}\\ & =& \frac{\text{5.66 g}}{\text{80}{\displaystyle \raisebox{1ex}{$\text{ g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}\\ & =& 0.070\text{ moles}\end{array}$ 

$\begin{array}{rcl}\text{moles\hspace{0.17em}of\hspace{0.17em}}{\left({\text{NH}}_{\text{4}}\right)}_{\text{3}}{\text{PO}}_{\text{4}}& =& \frac{\text{mass\hspace{0.17em}of\hspace{0.17em}}{\left({\text{NH}}_{\text{4}}\right)}_{\text{3}}{\text{PO}}_{\text{4}}}{\text{molecular\hspace{0.17em}mass\hspace{0.17em}of\hspace{0.17em}}{\left({\text{NH}}_{\text{4}}\right)}_{\text{3}}{\text{PO}}_{\text{4}}}\\ & =& \frac{4.42\text{ g}}{149{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}\\ & =& 0.029\text{ moles}\end{array}$

 

Hence, the moles of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$  and  ${\left({\text{NH}}_{\text{4}}\right)}_{\text{3}}{\text{PO}}_{\text{4}}$ are $0.070\text{ moles}$ and  $0.029\text{ moles}$ respectively.

Molarity of NH 4 + and PO 4 3 - :

First, calculate the molarity for ${{\text{NH}}_{\text{4}}}^{\text{+}}$. In ammonium nitrate only one ${{\text{NH}}_{\text{4}}}^{\text{+}}$  is present but in ammonium phosphate three ${{\text{NH}}_{\text{4}}}^{\text{+}}$ are present. So, there is $0.070\text{ moles}$ of ${{\text{NH}}_{\text{4}}}^{\text{+}}$ in ammonium nitrate and  $0.029\text{ moles}$ of ${{\text{NH}}_{\text{4}}}^{\text{+}}$ in ammonium phosphate. So, the total moles of  ${{\text{NH}}_{\text{4}}}^{\text{+}}$ is equal to the addition of moles of ${{\text{NH}}_{\text{4}}}^{\text{+}}$in ammonium nitrate and moles of  ${{\text{NH}}_{\text{4}}}^{\text{+}}$ in ammonium phosphate, because in ammonium phosphate there is three ${{\text{NH}}_{\text{4}}}^{\text{+}}$ are present so moles are multiplied by three.

Hence, 

$\begin{array}{rcl}{{\text{Total\hspace{0.17em}moles\hspace{0.17em}of\hspace{0.17em}NH}}_{\text{4}}}^{\text{+}}& =& 0.070\text{ moles}+3\left(0.029\text{ moles}\right)\\ & =& (0.070+0.087)\text{ moles}\\ & =& 0.157\text{ moles}\end{array}$

Now, given the volume of solution is $20\text{ L}$. So, molarity can be calculated as follow.

$\begin{array}{rcl}\text{molarity}& =& \frac{{{\text{moles\hspace{0.17em}of\hspace{0.17em}NH}}_{\text{4}}}^{\text{+}}}{\text{volume\hspace{0.17em}of\hspace{0.17em}solution}}\\ & =& \frac{0.157\text{ mol}}{20\text{ L}}\\ & =& 0.0078\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.\end{array}$ 

 

Similarly, you can calculate the molarity of  ${{\text{PO}}_{\text{4}}}^{\text{3-}}$. $0.029\text{ moles}$ are present in ammonium phosphate.

$\begin{array}{rcl}\text{molarity}& =& \frac{{{\text{moles\hspace{0.17em}of\hspace{0.17em}PO}}_{\text{4}}}^{\text{3-}}}{\text{volume\hspace{0.17em}of\hspace{0.17em}solution}}\\ & =& \frac{0.029\text{ mol}}{20\text{ L}}\\ & =& 0.0014\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.\end{array}$ 

 

Hence, the molarity of  ${{\text{NH}}_{\text{4}}}^{\text{+}}$ and ${{\text{PO}}_{\text{4}}}^{\text{3-}}$  is $0.0078\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.$  and $0.0014\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.$ respectively.