Question

Calculate the number of moles of each of the ions in the following solutions. (a) \(18.5 \mathrm{~mL}\) of \(0.402 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\) (b) \(30.0 \mathrm{~mL}\) of \(0.359 \mathrm{MAl}_{2}\left(\mathrm{SO}_{4}\right)_{3}\)

Short answer

Solution (a) contains 0.00743 moles of \(\mathrm{NH}_4^+\) ions and 0.00371 moles of \(\mathrm{CO}_3^{2-}\) ions; solution (b) contains 0.0108 moles of \(\mathrm{Al}^{3+}\) ions and 0.0324 moles of \(\mathrm{SO}_4^{2-}\) ions.

Step-by-step solution

Convert Volume to Liters

For solution (a), convert volume from milliliters to liters by dividing by 1000. For solution (b), do the same thing to get the volume in liters.

Calculate Moles of Compound

For each solution, use the molarity (M) and the volume in liters (L) to calculate the moles of the compound by multiplying them together.

Determine Moles of Each Ion

For solution (a), use the stoichiometry of \(\left(\mathrm{NH}_4\right)_2\mathrm{CO}_3\) to find the moles of \(\mathrm{NH}_4^+\) and \(\mathrm{CO}_3^{2-}\) ions. For solution (b), use the stoichiometry of \(\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3\) to find the moles of \(\mathrm{Al}^{3+}\) and \(\mathrm{SO}_4^{2-}\) ions.

Perform the Calculation for (a)

Calculate the moles of \(\left(\mathrm{NH}_4\right)_2\mathrm{CO}_3\) and then multiply by the number of ions per formula unit to find the number of moles of \(\mathrm{NH}_4^+\) and \(\mathrm{CO}_3^{2-}\) ions.

Perform the Calculation for (b)

Calculate the moles of \(\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3\) and then multiply by the number of ions per formula unit to find the number of moles of \(\mathrm{Al}^{3+}\) and \(\mathrm{SO}_4^{2-}\) ions.

Key concepts

Molarity and Volume Relation

Understanding the relationship between molarity and volume is crucial when dealing with solutions in chemistry. Molarity, represented as M, is defined as the number of moles of solute dissolved per liter of solution. To put it simply, it indicates the concentration of a solution. Now, why does this matter? Imagine you have a recipe that calls for a specific concentration of salt in water. The molarity tells you exactly how much salt is present in each liter of water.

When you are asked to calculate the moles of ions in a particular volume of a solution, you need to grasp this concept. Let's say you're given an 18.5 mL solution with a molarity of 0.402 M. To find the number of moles of ions, you first convert the volume from milliliters to liters. Why? Because molarity is given in moles per liter, so consistency in units is key. Following this, you multiply the molarity by the volume in liters to find the moles of the compound. Now, you're probably thinking, 'Okay, but what about the ions?' This leads us to the next concept: stoichiometry of ions.

Stoichiometry of Ions

Diving deeper into our solution analysis, we arrive at stoichiometry, which is the field of chemistry that pertains to the quantitative relationships of substances as they participate in chemical reactions. For ions, stoichiometry deals with how the ions are arranged within compounds and how this arrangement influences the number of each type of ion in a solution.

Consider our example of \(\mathrm{NH}_4\)_2\mathrm{CO}_3\. The subscript '2' indicates that for every single carbonate \(\mathrm{CO}_3^{2-}\) ion, there are two ammonium \(\mathrm{NH}_4^+\) ions. To find the moles of ions, you'd first calculate the moles of the entire compound as described previously. Then, multiply those moles by the ion ratio from the compound's stoichiometry. For instance, if you have a mole of \(\mathrm{NH}_4\)_2\mathrm{CO}_3\, you'd actually have 2 moles of \(\mathrm{NH}_4^+\) ions and 1 mole of \(\mathrm{CO}_3^{2-}\) ions. This step is essential for accurately determining the composition of ions in your solution.

Chemical Solution Analysis

Chemical solution analysis incorporates both the concepts of molarity and stoichiometry to comprehensively understand a solution's composition. It's like investigating a mixture to determine what’s in it and how much of each component is present. When you extend this to ions in solutions, especially those resulting from dissolving ionic compounds, you're looking at how these compounds break apart and the concentration of each resulting ion.

For example, when analyzing \(\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3\), knowing its stoichiometry reveals that one formula unit releases two \(\mathrm{Al}^{3+}\) ions and three \(\mathrm{SO}_4^{2-}\) sulfate ions. With this in hand, and knowing the molarity of your solution, you can engage in a step-by-step calculation to determine the exact number of moles of each ion present in a given volume. This technique is vital in fields such as medicine, where the concentration of electrolytes in a solution can be critical for patient health. By mastering chemical solution analysis, students equip themselves with a fundamental tool for their future work in chemistry and related disciplines.