Question

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

Short answer

The molarity of \(\text{NH}_4^+\) is 0.00799 M and the molarity of \(\text{PO}_4^{3-}\) is 0.001483 M.

Step-by-step solution

- Calculate moles of each compound

First, calculate the number of moles of \(\text{NH}_4\text{NO}_3\) and \(\text{(NH}_4\text{)}_3\text{PO}_4\). Use the formula: \(\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}\). \(\text{Molar mass of NH}_4\text{NO}_3 = 14 + 4 + 14 + 3 \times 16 = 80 \text{ g/mol}\) and \(\text{molar mass of (NH}_4\text{)}_3\text{PO}_4 = 3 \times \text{(14 + 4)} + 31 + 4 \times 16 = 149 \text{ g/mol}\).

- Calculate moles of \(\text{NH}_4\text{NO}_3\)

Using the molar mass of \(\text{NH}_4\text{NO}_3\) and its given mass: \(\frac{5.66 \text{ g}}{80 \text{ g/mol}} = 0.07075 \text{ moles of NH}_4\text{NO}_3\).

- Calculate moles of \(\text{(NH}_4\text{)}_3\text{PO}_4\)

Using the molar mass of \(\text{(NH}_4\text{)}_3\text{PO}_4\) and its given mass: \(\frac{4.42 \text{ g}}{149 \text{ g/mol}} = 0.02966 \text{ moles of (NH}_4\text{)}_3\text{PO}_4\).

- Determine moles of \(\text{NH}_4^+\) ions

Each molecule of \(\text{NH}_4\text{NO}_3\) provides one \(\text{NH}_4^+\) ion, and each molecule of \(\text{(NH}_4\text{)}_3\text{PO}_4\) provides three \(\text{NH}_4^+\) ions. Therefore, the total moles of \(\text{NH}_4^+\) is \((0.07075 \text{ mol} \times 1) + (0.02966 \text{ mol} \times 3) = 0.07075 + 0.08898 = 0.15973 \text{ moles}\).

- Determine moles of \(\text{PO}_4^{3-}\) ions

Each molecule of \(\text{(NH}_4\text{)}_3\text{PO}_4\) provides one \(\text{PO}_4^{3-}\) ion. Therefore, the total moles of \(\text{PO}_4^{3-}\) is \(\text{0.02966 moles}\).

- Calculate molarities

Molarity is defined as \(\frac{\text{moles of solute}}{\text{volume of solution in liters}}\). For \(\text{NH}_4^+\): \(\frac{0.15973 \text{ moles}}{20.0 \text{ L}} = 0.00799 \text{ M}\). For \(\text{PO}_4^{3-}\): \(\frac{0.02966 \text{ moles}}{20.0 \text{ L}} = 0.001483 \text{ M}\).

Key concepts

Molarity Calculation

Molarity (M) is a common way to express the concentration of a solution. It represents the number of moles of solute per liter of solution. The formula for molarity is easy to remember: \[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \] \ In the given exercise, we need to determine the molarities of \( \text{NH}_4^+ \) and \( \text{PO}_4^{3-} \) ions in a solution created by dissolving specific amounts of \( \text{NH}_4\text{NO}_3 \) and \( \text{(NH}_4\text{)}_3\text{PO}_4 \) in 20.0 liters of water. To do this, we first convert the mass of each compound to moles and then use the volume of the solution to find the molarity of the resulting ions.

Chemical Compounds Illustrated

Chemical compounds are substances made up of two or more different elements bonded together. In this exercise, we deal with ammonium nitrate (\( \text{NH}_4\text{NO}_3 \)) and ammonium phosphate (\( \text{(NH}_4\text{)}_3\text{PO}_4 \)). These compounds dissolve in water, releasing ions into the solution. \ To calculate the moles, you need the molar mass of each compound. For instance, the molar mass of \( \text{NH}_4\text{NO}_3 \) is calculated as: \[ \text{(14 (N) + 4 (H) + 14 (N) + 3 x 16 (O))} = 80 \text{ g/mol} \] and for \( \text{(NH}_4\text{)}_3\text{PO}_4 \) it is: \[ \text{3 x (14 (N) + 4 (H)) + 31 (P) + 4 x 16 (O)} = 149 \text{ g/mol} \] \ Ammonium nitrate provides one \( \text{NH}_4^+ \) ion per molecule, while ammonium phosphate provides three \( \text{NH}_4^+ \) ions and one \( \text{PO}_4^{3-} \) ion per molecule. Understanding these dissociations is key to solving the molarity.

Ionic Concentration

Ionic concentration tells us how much of each ion is present in a solution. Once ammonium nitrate and ammonium phosphate dissolve, they release \( \text{NH}_4^+ \) and \( \text{PO}_4^{3-} \) ions into the solution. \ To find the \( \text{NH}_4^+ \) concentration, we need to count total moles of \( \text{NH}_4^+ \) ions from both compounds: \[ \text{From NH}_4\text{NO}_3: 0.07075 \text{ moles} \] and \[ \text{From (NH}_4\text{)}_3\text{PO}_4: 0.02966 \text{ moles x 3} = 0.08898 \text{ moles} \] \ So, total moles of \( \text{NH}_4^+ \) is: \[ 0.07075 \text{ moles} + 0.08898 \text{ moles} = 0.15973 \text{ moles} \] \Similarly, for \( \text{PO}_4^{3-} \), from \( \text{(NH}_4\text{)}_3\text{PO}_4 \), it is \[ 0.02966 \text{ moles} \] \ These values help us find the molarity of each ion.

Fertilizer Solution

In agriculture, fertilizers are used to enhance soil nutrient content. Fertilizer solutions often contain mixtures of various chemical compounds that dissolve to provide essential ions like \( \text{NH}_4^+ \) (ammonium) and \( \text{PO}_4^{3-} \) (phosphate). \ The mentioned problem involves such a solution, set up by a florist. Dissolving fertilizers correctly ensures the right concentration of nutrients, crucial for plant health. \ In our exercise, the resulting solution combines ammonium nitrate and ammonium phosphate, providing both nitrogen (from \( \text{NH}_4^+ \)) and phosphorus (from \( \text{PO}_4^{3-} \)). These are vital macronutrients required for plant growth.

Molecular Mass Significance

Molecular mass, or molar mass, represents the mass of one mole of a substance and is expressed in g/mol. Calculating molecular mass allows us to convert between grams of a substance and moles, which is essential for molarity calculation. \ In the exercise, the molecular mass of \( \text{NH}_4\text{NO}_3 \) and \( \text{(NH}_4\text{)}_3\text{PO}_4 \) were calculated to be \( 80 \text{ g/mol} \) and \( 149 \text{ g/mol} \) respectively: \ When given the mass of a substance, we can find the moles using: \[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \] This crucial step enables us to calculate the concentration (molarity) of ions in the solution accurately.