Chapter 13, Problem 13.149P

Interpretation Introduction

Interpretation:

The molarities of ${\text{NH}}_4^{+}$ and ${\text{PO}}_4^{3-}$ are to be determined.

Concept introduction:

Molarity is defined as the number of moles of solute that are dissolved in one litre of solution. It is represented by $M$ and its unit is $\text{mol/L}$.

The formula to calculate the molarity of the solution is as follows:

$\text{Molarity}=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{volume }\left(\text{L}\right)\text{of}\text{}\text{solution}}$ (1)

Answer to Problem 13.149P

The molarities of ${\text{NH}}_4^{+}$ and ${\text{PO}}_4^{3-}$ are $7.98\times {10}^{-3}\text{}M$ and $1.48\times {10}^{-3}\text{}M$.

Explanation of Solution

The dissociation of ${\text{NH}}_4{\text{NO}}_3$ occurs as follows:

${\text{NH}}_4{\text{NO}}_3\rightleftharpoons {\text{NH}}_4^{+}+{\text{NO}}_3^{-}$

One mole of ${\text{NH}}_4{\text{NO}}_3$ dissociates to give one mole of ${\text{NH}}_4^{+}$ and one mole of ${\text{NO}}_3^{-}$.

The dissociation of ${\left({\text{NH}}_4\right)}_3{\text{PO}}_4$ occurs as follows:

${\left({\text{NH}}_4\right)}_3{\text{PO}}_4\rightleftharpoons 3{\text{NH}}_4^{+}+{\text{PO}}_4^{3-}$

One mole of ${\left({\text{NH}}_4\right)}_3{\text{PO}}_4$ dissociates to give three moles of ${\text{NH}}_4^{+}$ and one mole of ${\text{PO}}_4^{3-}$.

The formula to calculate the moles of ${\text{NH}}_4^{+}$ in ${\text{NH}}_4{\text{NO}}_3$ is as follows:

$\text{Moles}\text{of}{\text{NH}}_4^{+}=\left(\frac{\text{Given mass of}\text{}{\text{NH}}_4{\text{NO}}_3}{\text{Molar mass of}\text{}{\text{NH}}_4{\text{NO}}_3}\right)\left(\frac{1\text{mol}\text{of}{\text{NH}}_4^{+}}{1\text{mol}\text{of}\text{}{\text{NH}}_4{\text{NO}}_3}\right)$ (2)

Substitute $5.56\text{g}$ for the given mass of ${\text{NH}}_4{\text{NO}}_3$ and $80.05\text{g/mol}$ for the molar mass of ${\text{NH}}_4{\text{NO}}_3$ in equation (2).

$\begin{array}{c}\text{Moles}\text{of}{\text{NH}}_4^{+}=\left(5.56\text{g}\right)\left(\frac{1\text{}\text{mol}}{80.05\text{g}}\right)\left(\frac{1\text{mol}\text{of}{\text{NH}}_4^{+}}{1\text{mol}\text{of}\text{}{\text{NH}}_4{\text{NO}}_3}\right)\\ =7.07058\times {10}^{-2}\text{}\text{mol}\end{array}$

The formula to calculate the moles of ${\text{NH}}_4^{+}$ in ${\left({\text{NH}}_4\right)}_3{\text{PO}}_4$ is as follows:

$\begin{array}{c}\text{Moles}\text{of}{\text{NH}}_4^{+}=\left(\frac{\text{Given mass of}\text{}{\left({\text{NH}}_4\right)}_3{\text{PO}}_4}{\text{Molar mass of}\text{}{\left({\text{NH}}_4\right)}_3{\text{PO}}_4}\right)\\ \left(\frac{3\text{mol}\text{of}{\text{NH}}_4^{+}}{1\text{mol}\text{of}\text{}{\left({\text{NH}}_4\right)}_3{\text{PO}}_4}\right)\end{array}$ (3)

Substitute $4.42\text{g}$ for the given mass of ${\left({\text{NH}}_4\right)}_3{\text{PO}}_4$ and $149.10\text{g/mol}$ for the molar mass of ${\left({\text{NH}}_4\right)}_3{\text{PO}}_4$ in equation (3).

$\begin{array}{c}\text{Moles}\text{of}{\text{NH}}_4^{+}=\left(4.42\text{g}\right)\left(\frac{1\text{}\text{mol}}{149.10\text{g}}\right)\left(\frac{3\text{mol}\text{of}{\text{NH}}_4^{+}}{1\text{mol}\text{of}\text{}{\left({\text{NH}}_4\right)}_3{\text{PO}}_4}\right)\\ =8.89336\times {10}^{-2}\text{}\text{mol}\end{array}$

The total moles of ${\text{NH}}_4^{+}$ is calculated as follows:

$\text{Moles}\text{of}{\text{NH}}_4^{+}=\text{Moles}\text{of}{\text{NH}}_4^{+}\text{}\text{in}{\text{NH}}_4{\text{NO}}_3+\text{Moles}\text{of}{\text{NH}}_4^{+}\text{}\text{in}{\left({\text{NH}}_4\right)}_3{\text{PO}}_4$ (4)

Substitute $7.07058\times {10}^{-2}\text{mol}$ for the moles of ${\text{NH}}_4{\text{NO}}_3$ and $8.89336\times {10}^{-2}\text{}\text{mol}$ for the moles of ${\left({\text{NH}}_4\right)}_3{\text{PO}}_4$ in equation (4).

$\begin{array}{c}\text{Moles}\text{of}{\text{NH}}_4^{+}=7.07058\times {10}^{-2}\text{mol}+8.89336\times {10}^{-2}\text{}\text{mol}\\ =15.96394\times {10}^{-2}\text{mol}\end{array}$

Substitute $15.96394\times {10}^{-2}\text{mol}$ for the amount of solute and $\text{20}\text{}\text{L}$ for the volume of the solution in equation (1) to calculate the molarity of ${\text{NH}}_4^{+}$.

$\begin{array}{c}\text{Molarity of}{\text{NH}}_4^{+}=\frac{15.96394\times {10}^{-2}\text{mol}}{\text{20}\text{}\text{L}}\\ =7.98197\times {10}^{-3}\text{}M\\ =7.98\times {10}^{-3}\text{}M\end{array}$

The formula to calculate the moles of ${\text{PO}}_4^{3-}$ in ${\left({\text{NH}}_4\right)}_3{\text{PO}}_4$ is as follows:

$\begin{array}{c}\text{Moles}\text{of}{\text{PO}}_4^{3-}=\left(\frac{\text{Given mass of}\text{}{\left({\text{NH}}_4\right)}_3{\text{PO}}_4}{\text{Molar mass of}\text{}{\left({\text{NH}}_4\right)}_3{\text{PO}}_4}\right)\\ \left(\frac{1\text{mol}\text{of}{\text{PO}}_4^{3-}}{1\text{mol}\text{of}\text{}{\left({\text{NH}}_4\right)}_3{\text{PO}}_4}\right)\end{array}$ (5)

Substitute $4.42\text{g}$ for the given mass of ${\left({\text{NH}}_4\right)}_3{\text{PO}}_4$ and $149.10\text{g/mol}$ for the molar mass of ${\left({\text{NH}}_4\right)}_3{\text{PO}}_4$ in equation (5).

$\begin{array}{c}\text{Moles}\text{of}{\text{PO}}_4^{3-}=\left(4.42\text{g}\right)\left(\frac{1\text{}\text{mol}}{149.10\text{g}}\right)\left(\frac{1\text{mol}\text{of}{\text{PO}}_4^{3-}}{1\text{mol}\text{of}\text{}{\left({\text{NH}}_4\right)}_3{\text{PO}}_4}\right)\\ =2.96445\times {10}^{-2}\text{}\text{mol}\end{array}$

Substitute $2.96445\times {10}^{-2}\text{mol}$ for the amount of solute and $\text{20}\text{}\text{L}$ for the volume of the solution in equation (1) to calculate the molarity of ${\text{PO}}_4^{3-}$.

$\begin{array}{c}\text{Molarity of}{\text{PO}}_4^{3-}=\frac{2.96445\times {10}^{-2}\text{mol}}{\text{20}\text{}\text{L}}\\ =1.482225\times {10}^{-3}\text{}M\\ =1.48\times {10}^{-3}\text{}M\end{array}$

Conclusion

The molarities of ${\text{NH}}_4^{+}$ and ${\text{PO}}_4^{3-}$ are $7.98\times {10}^{-3}\text{}M$ and $1.48\times {10}^{-3}\text{}M$.