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Chapter 4: Problem 29

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes. a. \(0.100\) mole of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) in \(100.0 \mathrm{~mL}\) of solution b. \(2.5\) moles of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in \(1.25 \mathrm{~L}\) of solution c. \(5.00 \mathrm{~g}\) of \(\mathrm{NH}_{4} \mathrm{Cl}\) in \(500.0 \mathrm{~mL}\) of solution d. \(1.00 \mathrm{~g} \mathrm{~K}_{2} \mathrm{PO}_{4}\) in \(250.0 \mathrm{~mL}\) of solution

Short Answer

Expert verified
In summary, the concentrations of all ions present in the given solutions are: a. \(\mathrm{Ca}^{2+}\): \(1.00\,\text{M}\) and \(\mathrm{NO}_{3}^{-}\): \(2.00\,\text{M}\) b. \(\mathrm{Na}^{+}\): \(4.0\,\text{M}\) and \(\mathrm{SO}_{4}^{2-}\): \(2.0\,\text{M}\) c. \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{Cl}^{-}\): both \(0.1868\,\text{M}\) d. \(\mathrm{K}^{+}\): \(0.0459\,\text{M}\) and \(\mathrm{PO}_{4}^{3-}\): \(0.0229\,\text{M}\)

Step by step solution

01

Write the ionization equations and stoichiometry

For each electrolyte, write the equation for complete ionization in water and determine the stoichiometry (i.e., the number of moles of each ion produced per mole of electrolyte). a. \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} \rightarrow \mathrm{Ca}^{2+} + 2\mathrm{NO}_{3}^{-}\) b. \(\mathrm{Na}_{2}\mathrm{SO}_{4} \rightarrow 2\mathrm{Na}^{+} + \mathrm{SO}_{4}^{2-}\) c. \(\mathrm{NH}_{4}\mathrm{Cl} \rightarrow \mathrm{NH}_{4}^{+} + \mathrm{Cl}^{-}\) d. \(\mathrm{K}_{2}\mathrm{PO}_{4} \rightarrow 2\mathrm{K}^{+} + \mathrm{PO}_{4}^{3-}\)
02

Calculate the concentrations of ions

Use the given volume of each solution and the amount of electrolyte dissolved to calculate the concentration of each ion. a. The concentration of \(\mathrm{Ca}^{2+}\) in the solution is: \[\frac{0.100\,\text{mole}}{0.100\,\text{L}} = 1.00\,\text{M}\] The concentration of \(\mathrm{NO}_{3}^{-}\) in the solution is: \[\frac{2 \times 0.100\,\text{mole}}{0.100\,\text{L}} = 2.00\,\text{M}\] b. The concentration of \(\mathrm{Na}^{+}\) in the solution is: \[\frac{2 \times 2.5\,\text{moles}}{1.25\,\text{L}} = 4.0\,\text{M}\] The concentration of \(\mathrm{SO}_{4}^{2-}\) in the solution is: \[\frac{2.5\,\text{moles}}{1.25\,\text{L}} = 2.0\,\text{M}\] c. First, we need to find the moles of \(\mathrm{NH}_{4} \mathrm{Cl}\) using its molar mass which is \(53.49\,\text{g/mol}\). \[\frac{5.00\,\text{g}}{53.49\,\text{g/mol}} = 0.0934\,\text{moles}\] Then, the concentration of both \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{Cl}^{-}\) in the solution: \[\frac{0.0934\,\text{moles}}{0.500\,\text{L}} = 0.1868\,\text{M}\] d. First, we need to find the moles of \(\mathrm{K}_{2}\mathrm{PO}_{4}\) using its molar mass which is \(174.18\,\text{g/mol}\). \[\frac{1.00\,\text{g}}{174.18\,\text{g/mol}} = 0.00574\,\text{moles}\] Then, the concentration of \(\mathrm{K}^{+}\) in the solution is: \[\frac{2 \times 0.00574\,\text{moles}}{0.250\,\text{L}} = 0.0459\,\text{M}\] The concentration of \(\mathrm{PO}_{4}^{3-}\) in the solution is: \[\frac{0.00574\,\text{moles}}{0.250\,\text{L}} = 0.0229\,\text{M}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Strong Electrolytes
Understanding strong electrolytes is crucial when calculating ion concentrations in a solution. Strong electrolytes are substances that completely dissociate into ions when dissolved in water. Common examples include table salt (NaCl), hydrochloric acid (HCl), and many salts containing alkali metal ions or the nitrate ion. These substances yield high concentrations of ions in solution, which contribute to the solution's ability to conduct electricity.

When a strong electrolyte dissolves in water, each formula unit of the substance breaks apart into its constituent ions. For instance, \(Ca(NO_3)_2\) dissociates completely to produce one calcium ion (\(Ca^{2+}\)) and two nitrate ions (\(NO_3^-\)) for each formula unit that dissolves. This complete ionization is key to understanding the resulting molarity of the ions in the solution, as the number of ions produced will directly affect the electrical conductivity and chemical reactivity of the solution.

For students, it's essential to remember that strong electrolytes fully dissociate, therefore the concentration of ions will be proportional to the original concentration of the electrolyte. This forms the basis for the calculations required in these exercises.
Stoichiometry
Stoichiometry is a segment of chemistry that involves the calculation of the reactants and products in a chemical reaction. It allows us to predict the amounts of substances consumed and created in a given chemical process. In the context of ion concentration calculation for strong electrolytes, stoichiometry refers to the relationship between the amount of the initial electrolyte and the resulting ions once dissolved.

For instance, when \(Ca(NO_3)_2\) ionizes, the stoichiometry from the ionization equation tells us we get one \(Ca^{2+}\) ion and two \(NO_3^-\) ions. If we start with 0.100 moles of \(Ca(NO_3)_2\), stoichiometry dictates we will have 0.100 moles of \(Ca^{2+}\) ions and 0.200 moles of \(NO_3^-\) ions in the solution. By understanding the ratios in which the ions are produced, students can accurately determine the concentration of each ion in solut
Molarity
The concept of molarity is often used in the context of solution concentration. Molarity is defined as the number of moles of a solute divided by the volume of the solution in liters. It is commonly denoted as 'M' and is expressed as moles per liter (mol/L).

For example, if we dissolve 0.100 moles of \(Ca(NO_3)_2\) in 100.0 mL of water, since 100.0 mL is equivalent to 0.100 L, the concentration of \(Ca^{2+}\) is calculated using the formula \[\frac{0.100\,\text{moles}}{0.100\,\text{L}} = 1.00\,\text{M}\].

Molarity is a critical factor in understanding the properties of solutions, such as osmotic pressure, boiling point elevation, or freezing point depression. For students, mastering the calculation of molarity is a stepping stone to solving various chemistry problems, especially those involving reactions in aqueous solutions.
Ionization Equations
When it comes to ionization equations, these are the representations of the chemical process where a compound, typically an electrolyte, breaks down into its component ions when it is dissolved in water. Understanding ionization equations is important for students as they explicitly show the ratio in which ions are produced, guiding the subsequent calculations of their concentrations.

For example, the ionization equation for \(Ca(NO_3)_2\) is \(Ca(NO_3)_2 \rightarrow Ca^{2+} + 2NO_3^-\), which indicates that for every one mole of \(Ca(NO_3)_2\) that ionizes, one mole of \(Ca^{2+}\) and two moles of \(NO_3^-\) ions are formed. Similarly, for \(K_2PO_4\), the equation \(K_2PO_4 \rightarrow 2K^+ + PO_4^{3-}\) shows the formation of two potassium ions and one phosphate ion for each unit of \(K_2PO_4\) that dissolves.

These equations are essential for solving problems that involve predicting the outcome of electrolyte dissolution in water. By writing correct ionization equations and combining them with stoichiometry and molarity concepts, students can calculate the concentration of ions in any given solution effectively.

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Most popular questions from this chapter

Polychlorinated biphenyls (PCBs) have been used extensively as dielectric materials in electrical transformers. Because PCBs have been shown to be potentially harmful, analysis for their presence in the environment has become very important. PCBs are manufactured according to the following generic reaction: $$ \mathrm{C}_{12} \mathrm{H}_{10}+{ }_{n \mathrm{Cl}_{2}} \rightarrow \mathrm{C}_{12} \mathrm{H}_{10-n} \mathrm{Cl}_{n}+n \mathrm{HCl} $$ This reaction results in a mixture of \(\mathrm{PCB}\) products. The mixture is analyzed by decomposing the \(\mathrm{PCBs}\) and then precipitating the resulting \(\mathrm{Cl}^{-}\) as \(\mathrm{AgCl}\). a. Develop a general equation that relates the average value of \(n\) to the mass of a given mixture of \(\mathrm{PCBs}\) and the mass of \(\mathrm{AgCl}\) produced. b. A \(0.1947-g\) sample of a commercial PCB yielded \(0.4791 \mathrm{~g}\) of \(\mathrm{AgCl}\). What is the average value of \(n\) for this sample?

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The blood alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}\right)\) level can be determined by titrating a sample of blood plasma with an acidic potassium dichromate solution, resulting in the production of \(\mathrm{Cr}^{3+}(a q)\) and carbon dioxide. The reaction can be monitored because the dichromate ion \(\left(\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}\right)\) is orange in solution, and the \(\mathrm{Cr}^{3+}\) ion is green. The balanced equation is \(16 \mathrm{H}^{+}(a q)+2 \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}(a q) \longrightarrow\) \(4 \mathrm{Cr}^{3+}(a q)+2 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(t)\) This reaction is an oxidation-reduction reaction. What species is reduced, and what species is oxidized? How many electrons are transferred in the balanced equation above?

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