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Chapter 4: Problem 10

An aqueous solution contains \(1.2 \mathrm{~m} M\) of total ions. (a) If the solution is \(\mathrm{NaCl}(a q),\) what is the concentration of chloride ion? (b) If the solution is \(\mathrm{FeCl}_{3}(a q),\) what is the concentration of chloride ion? [Section 4.5\(]\)

Short Answer

Expert verified
(a) In the NaCl(aq) solution, the concentration of chloride ions (Cl鈦�) is 0.6 M. (b) In the FeCl鈧�(aq) solution, the concentration of chloride ions (Cl鈦�) is 0.36 M.

Step by step solution

01

Case (a): Sodium chloride (NaCl) solution

In NaCl(aq) solution, sodium chloride dissociates into its constituent ions: sodium ions (Na鈦�) and chloride ions (Cl鈦�). The dissociation of one mole of NaCl produces one mole of Na鈦� and one mole of Cl鈦� ions, so the ratio between NaCl and Cl鈦� is 1:1. Given the total ion concentration in the solution is 1.2 M, we can write this as the sum of the concentrations of Na鈦� and Cl鈦� ions: \[1.2 ~\text{M} = [Na^+] + [Cl^-]\] Since the ratio between NaCl and Cl鈦� is 1:1, their concentrations are equal: \[[Na^+] = [Cl^-]\] Substituting this back into the equation, we can determine the concentration of chloride ions: \[1.2 ~\text{M} = 2[Cl^-]\] Now, solve for [Cl鈦籡: \[[Cl^-] = \frac{1.2 ~\text{M}}{2} = 0.6 ~\text{M}\] So, the concentration of chloride ions in the NaCl(aq) solution is 0.6 M.
02

Case (b): Iron(III) chloride (FeCl鈧�) solution

In FeCl鈧�(aq) solution, iron(III) chloride dissociates into its constituent ions: iron(III) ions (Fe鲁鈦�) and chloride ions (Cl鈦�). The dissociation of one mole of FeCl鈧� produces one mole of Fe鲁鈦� and three moles of Cl鈦� ions, so the ratio between FeCl鈧� and Cl鈦� is 1:3. Given the total ion concentration in the solution is 1.2 M, we can write this as the sum of the concentrations of Fe鲁鈦� and Cl鈦� ions: \[1.2 ~\text{M} = [Fe^{3+}] + 3[Cl^-]\] Since the ratio between FeCl鈧� and Cl鈦� is 1:3, their concentrations are related as: \[[Fe^{3+}] = \frac{1}{3}[Cl^-]\] Substituting this back into the equation, we can determine the concentration of chloride ions: \[1.2 ~\text{M} = \frac{1}{3}[Cl^-] + 3[Cl^-]\] Combine the Cl鈦� terms: \[1.2 ~\text{M} = \frac{10}{3}[Cl^-]\] Now, solve for [Cl鈦籡: \[[Cl^-] = \frac{1.2 ~\text{M}}{\frac{10}{3}} = \frac{1.2 ~\text{M} \cdot 3}{10} = 0.36 ~\text{M}\] So, the concentration of chloride ions in the FeCl鈧�(aq) solution is 0.36 M.

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Key Concepts

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

Sodium Chloride
Sodium chloride, commonly known as table salt, is an iconic example when exploring ionic compounds and their behaviors in solutions. When sodium chloride (\(\mathrm{NaCl}\)) is dissolved in water, it dissociates completely into its constituent ions: sodium ions (\(\mathrm{Na^+}\)) and chloride ions (\(\mathrm{Cl^-}\)).
The process of dissociation is straightforward as each molecule of sodium chloride splits to produce one sodium ion and one chloride ion, creating an equal proportion of ions in the solution.
  • The dissociation equation: \(\mathrm{NaCl (s)} 鈫� \mathrm{Na^+ (aq)} + \mathrm{Cl^- (aq)}\)
  • The ratio of sodium ions to chloride ions is 1:1.
  • This makes calculating concentrations in solution quite simple: if you know the total ionic concentration is 1.2 M, you can split it equally.
In this example, because the total ion concentration equals the sum of both ions, we divide by two to find that the concentration of chloride ions is 0.6 M.
Iron(III) Chloride
Iron(III) chloride, or ferric chloride (\(\mathrm{FeCl_3}\)), provides a fascinating look into ionic compounds with multiple ions per molecule. When dissolved, each unit of \(\mathrm{FeCl_3}\) separates into one iron(III) ion (\(\mathrm{Fe^{3+}}\)) and three chloride ions (\(\mathrm{Cl^-}\)).
This 1:3 dissociation ratio means there are significantly more chloride ions than ferric ions in a solution.
  • The dissociation equation: \( \mathrm{FeCl_3 (s)} 鈫� \mathrm{Fe^{3+} (aq)} + 3 \mathrm{Cl^- (aq)} \)
  • This shows for every mole of \( \mathrm{FeCl_3} \) in the solution, there are three moles of chloride ions.
Knowing the total ionic concentration (1.2 M), we identify the contribution of each type of ion considering the dissociation ratio. The chloride ions are significantly abundant, hence needing careful calculation. The concentration of chloride ions in this example is calculated to be 0.36 M, illustrating their prolific quantity in the solution relative to the iron(III) ions.
Dissociation in Aqueous Solutions
Dissociation in aqueous solutions is a fundamental chemical concept that describes how ionic compounds dissolve in water. This process occurs when a solid ionic compound separates into its individual ions.
The solvent molecules, usually water, surround and isolate these ions, allowing them to drift apart and interact freely in the solution.
  • Ionic compounds dissociate according to fixed stoichiometric ratios determined by their chemical formulas.
  • For example, sodium chloride splits into equal parts, whereas iron(III) chloride releases three chloride ions for every one ferric ion.
  • This dissociation influences the resulting ion concentrations in a solution, affecting various properties, such as conductivity and reactivity.
Understanding dissociation is key to mastering how aqueous solutions behave, helping to predict how ions will affect chemical reactions in solutions. It's crucial when calculating ionic concentrations, often determining the stoichiometry needed to achieve desired concentrations in practical and experimental chemistry.

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Most popular questions from this chapter

Copper exists in the form of \(\mathrm{CuFeS}_{2}\) in copper ore. Copper is isolated in a two-step process. First, CuFeS \(_{2}\) is heated with \(\mathrm{SiO}_{2}\) in the presence of oxygen to form copper(I) sulfide, \(\mathrm{CuS:} 2 \mathrm{CuFeS}_{2}+2 \mathrm{SiO}_{2}(s)+4 \mathrm{O}_{2}(g) \longrightarrow \mathrm{Cu}_{2} \mathrm{~S}(s)+\) \(2 \mathrm{FeSiO}_{3}(s)+3 \mathrm{SO}_{2}(g) . \mathrm{Cu}_{2} \mathrm{~S}\) is then heated with oxygen to form copper and \(\mathrm{SO}_{2}(g)\). (a) Write the balanced chemical equation for the second reaction. (b) Which atoms from which compounds are being oxidized, and which atoms from which compounds are being reduced? (c) How many grams of copper would be isolated from \(85.36 \mathrm{~g}\) of \(\mathrm{CuFeS}_{2}\) in copper ore?

The commercial production of nitric acid involves the following chemical reactions: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ (a) Which of these reactions are redox reactions? (b) In each redox reaction identify the element undergoing oxidation and the element undergoing reduction. (c) How many grams of ammonia must you start with to make \(1000.0 \mathrm{~L}\) of a \(0.150 \mathrm{M}\) aqueous solution of nitric acid? Assume all the reactions give \(100 \%\) yield.

As \(\mathrm{K}_{2} \mathrm{O}\) dissolves in water, the oxide ion reacts with water molecules to form hydroxide ions. (a) Write the molecular and net ionic equations for this reaction. (b) Based on the definitions of acid and base, what ion is the base in this reaction? (c) What is the acid in the reaction? (d) What is the spectator ion in the reaction?

A person suffering from hyponatremia has a sodium ion concentration in the blood of \(0.118 \mathrm{M}\) and a total blood volume of \(4.6 \mathrm{~L}\). What mass of sodium chloride would need to be added to the blood to bring the sodium ion concentration up to \(0.138 \mathrm{M}\), assuming no change in blood volume?

The average adult human male has a total blood volume of 5.0 L. If the concentration of sodium ion in this average individual is \(0.135 \mathrm{M}\), what is the mass of sodium ion circulating in the blood?
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