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?

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 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 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.

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, 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.