Question

Calculate the number of moles of the indicated ion present in each of the following solutions. a. \(\mathrm{Na}^{+}\) ion in \(1.00 \mathrm{L}\) of \(0.251 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution. b. \(\mathrm{Cl}^{-}\) ion in 5.50 L of \(0.10 \mathrm{M} \mathrm{FeCl}_{3}\) solution c. \(\mathrm{NO}_{3}^{-}\) ion in \(100 . \mathrm{mL}\) of \(0.55 \mathrm{M} \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) solution d. \(\mathrm{NH}_{4}^{+}\) ion in \(250 .\) mL of \(0.350 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) solution

Short answer

a. Moles of \(\mathrm{Na}^{+}\) ions = 0.502 moles b. Moles of \(\mathrm{Cl}^{-}\) ions = 1.65 moles c. Moles of \(\mathrm{NO}_{3}^{-}\) ions = 0.110 moles d. Moles of \(\mathrm{NH}_{4}^{+}\) ions = 0.175 moles

Step-by-step solution

Calculate the number of moles of Na2SO4

Using the given concentration and volume, we can calculate the number of moles of Na2SO4: Moles of Na2SO4 = concentration × volume Moles of Na2SO4 = 0.251 M × 1.00 L = 0.251 moles

Calculate the number of moles of Na+

Since there are 2 Na+ ions in every Na2SO4 molecule, we can now calculate the moles of Na+ ions: Moles of Na+ ions = 2 × moles of Na2SO4 Moles of Na+ ions = 2 × 0.251 moles = 0.502 moles b. Calculate the number of moles of \(\mathrm{Cl}^{-}\) ion in 5.50 L of \(0.10 \mathrm{M} \mathrm{FeCl}_{3}\) solution

Calculate the number of moles of FeCl3

Using the given concentration and volume, we can calculate the number of moles of FeCl3: Moles of FeCl3 = concentration × volume Moles of FeCl3 = 0.10 M × 5.50 L = 0.550 moles

Calculate the number of moles of Cl-

Since there are 3 Cl- ions in every FeCl3 molecule, we can now calculate the moles of Cl- ions: Moles of Cl- ions = 3 × moles of FeCl3 Moles of Cl- ions = 3 × 0.550 moles = 1.65 moles c. Calculate the number of moles of \(\mathrm{NO}_{3}^{-}\) ion in \(100 . \mathrm{mL}\) of \(0.55 \mathrm{M} \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) solution

Calculate the number of moles of Ba(NO3)2

First, we need to convert the volume from mL to L: 100 mL = 0.100 L. Using the given concentration and volume, we can calculate the number of moles of Ba(NO3)2: Moles of Ba(NO3)2 = concentration × volume Moles of Ba(NO3)2 = 0.55 M × 0.100 L = 0.055 moles

Calculate the number of moles of NO3-

Since there are 2 NO3- ions in every Ba(NO3)2 molecule, we can now calculate the moles of NO3- ions: Moles of NO3- ions = 2 × moles of Ba(NO3)2 Moles of NO3- ions = 2 × 0.055 moles = 0.110 moles d. Calculate the number of moles of \(\mathrm{NH}_{4}^{+}\) ion in \(250 .\) mL of \(0.350 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) solution

Calculate the number of moles of (NH4)2SO4

First, we need to convert the volume from mL to L: 250 mL = 0.250 L. Using the given concentration and volume, we can calculate the number of moles of (NH4)2SO4: Moles of (NH4)2SO4 = concentration × volume Moles of (NH4)2SO4 = 0.350 M × 0.250 L = 0.0875 moles

Calculate the number of moles of NH4+

Since there are 2 NH4+ ions in every (NH4)2SO4 molecule, we can now calculate the moles of NH4+ ions: Moles of NH4+ ions = 2 × moles of (NH4)2SO4 Moles of NH4+ ions = 2 × 0.0875 moles = 0.175 moles

Key concepts

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships of the substances as they participate in chemical reactions. Essentially, it tells us how much of each substance is needed to react with another, and how much product is formed. It is a fundamental concept for understanding how to balance chemical equations and to calculate quantities needed or produced in a reaction.

For example, when we say that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water, we are talking about stoichiometric relationships. In the solutions provided in the textbook exercise, stoichiometry helps us in knowing how many ions will be present when certain compounds dissolve in water. We translate the molarity of a solution and its volume into moles of the solute, and using the mole ratios from the formula of the compound, we can find the moles of the desired ion.

Molarity Calculations

Molarity is a measure of chemical concentration expressing the number of moles of a solute per liter of solution. It is denoted by the symbol 'M'. To calculate molarity, we use the formula:
\[ \text{Molarity} = \frac{\text{number of moles of solute}}{\text{volume of solution in liters}} \]

In the given exercises, molarity calculations are the starting point to find the number of moles of ions in solution. In each part of the exercise, the molarity of the compound is used along with the volume of the solution to find the total moles of solute, from which we can then find the specific ions' moles.

Chemical Concentration

Chemical concentration refers to the amount of a substance present in a given quantity of a mixture or solution. It can be expressed in various ways, such as molarity, molality, mass percent, or volume percent. Understanding concentration is crucial when preparing solutions for reactions, as well as when performing titrations and other analytical techniques.

Molarity, which represents moles of solute per liter of solution, is commonly used in chemistry for its convenience in stoichiometric calculations in aqueous solutions. It helps us predict the amount of reactant and product in a chemical reaction. In the textbook exercises, the concentrations are given in molarity, which directly correlates to the number of moles of ions when we dissolve ionic compounds in water.

Ions in Solution

When ionic compounds dissolve in water, they disassociate into their constituent ions. The number of these ions in solution is of interest in various chemical applications such as precipitations, acid-base reactions, and electrochemistry. Knowing how to calculate the number of moles of these ions is crucial.

In the provided exercises, for instance, we are asked to calculate the number of moles of specific ions in solution. The number of each ion type is contingent on the mole ratio that comes from the chemical formula of the compound. For example, one mole of \(Na_2SO_4\) will provide two moles of \(Na^+\) ions, and one mole of \(FeCl_3\) will provide three moles of \(Cl^-\) ions. This concept of ion disassociation in solution is fundamental for understanding solutions' behavior in electrochemical cells and in reactions that occur in an aqueous environment.