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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? \((\mathbf{b})\) If the solution is \(\mathrm{FeCl}_{3}(a q),\) what is the concentration of chloride ion?

Short Answer

Expert verified
(a) 0.6 mM Cl鈦� (b) 0.9 mM Cl鈦�

Step by step solution

01

Understanding the Problem

We are given a solution that contains a total of 1.2 mM of ions. We need to determine the concentration of chloride ions in the solution when it is a solution of \( \text{NaCl} \) and \( \text{FeCl}_3 \).
02

Analyze \( \text{NaCl} \) Dissociation

Sodium chloride (\( \text{NaCl} \)) dissociates in water to form sodium ions \((\text{Na}^+)\) and chloride ions \((\text{Cl}^-)\). Each formula unit of \( \text{NaCl} \) produces one \( \text{Cl}^- \). SInce \( \text{NaCl} \) contributes equally to both \( \text{Na}^+ \) and \( \text{Cl}^- \), if we have 1.2 mM total ions, each ion type (\( \text{Na}^+ \) and \( \text{Cl}^- \)) will contribute equally.
03

Calculate \( \text{Cl}^- \) Concentration from \( \text{NaCl} \)

In \( \text{NaCl} \), since 1 mole of \( \text{NaCl} \) results in 1 mole of \( \text{Cl}^- \), the concentration of \( \text{Cl}^- \) is half of the total ion concentration. So, \[ [\text{Cl}^-] = \frac{1.2 \text{ mM}}{2} = 0.6 \text{ mM} \].
04

Analyze \( \text{FeCl}_3 \) Dissociation

Iron(III) chloride (\( \text{FeCl}_3 \)) dissociates in water to form one \( \text{Fe}^{3+} \) ion and three \( \text{Cl}^- \) ions. This means each formula unit produces four ions total.
05

Calculate \( \text{Cl}^- \) Concentration from \( \text{FeCl}_3 \)

Since each \( \text{FeCl}_3 \) unit contributes three \( \text{Cl}^- \) ions and four total ions, 1 unit of \( \text{FeCl}_3 \) produces three units of chloride ions. For \( \text{FeCl}_3 \), \[ \frac{1.2 \text{ mM}}{4} = 0.3 \text{ mM} \text{ for } \text{FeCl}_3 \]. \( \text{Cl}^- \) concentration is \( 3 \times 0.3 \text{ mM} = 0.9 \text{ mM} \).

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Key Concepts

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

Aqueous Solution
An aqueous solution is simply a mixture where water acts as the solvent. It means that the solid, liquid, or gas component is fully dissolved in water. When substances like sodium chloride or iron(III) chloride dissolve, their ions integrate throughout the water, forming a homogenous mixture. Aqueous solutions have unique properties because water is a universal solvent, making it ideal for facilitating chemical reactions. Creating an aqueous solution involves:
  • Dissolving a solute in water
  • Forming a solution where compounds dissociate into ions
In these solutions, water doesn鈥檛 alter the compound structure but allows ions to disperse freely. This feature is essential for studying ionic concentrations.
Dissociation
In chemistry, dissociation refers to the process where a compound breaks down into its individual ions when dissolved in a solvent. For sodium chloride (NaCl), dissociation results in sodium (Na鈦�) and chloride (Cl鈦�) ions. Similarly, when iron(III) chloride (FeCl鈧�) dissolves, it splits into one Fe鲁鈦� ion and three Cl鈦� ions. This process is crucial in determining how many ions are present in a solution. Understanding dissociation is important because:
  • It helps determine how many of each ion type are in a solution
  • Enables calculations of ionic concentration based on formula units
The pattern of dissociation varies with each compound, impacting the total number of resulting ions.
Sodium Chloride
Sodium chloride, commonly known as table salt, is a compound made of sodium (Na) and chloride (Cl). When dissolved in water, each molecule dissociates to form one Na鈦� and one Cl鈦� ion. Thus, the concentration of each ion in a solution is directly proportional to the amount of sodium chloride dissolved. For example, if you have a total ion concentration of 1.2 mM, the concentration of Cl鈦� ions will be 0.6 mM because they contribute one Cl鈦� per Na鈦� ion. NaCl's dissociation attributes include:
  • Equal Na鈦� and Cl鈦� ions in solution
  • Direct correlation between dissolved NaCl and ion concentrations
This one-to-one ratio makes NaCl a straightforward compound for studying basic ionic solutions.
Iron(III) Chloride
Iron(III) chloride, or FeCl鈧�, is a chemical compound that dissociates into four ions. Upon dissolving in water, it produces one Fe鲁鈦� ion and three Cl鈦� ions. This results in multiple ions forming from a single compound unit, differing from simpler compounds like NaCl. Understanding FeCl鈧�'s dissociation is key to determining how many chloride ions are present in a solution. Unique aspects of FeCl鈧� include:
  • Dissociates into more ions than simpler salts
  • Creates a solution with a distinct ion ratio
In a scenario with a total ion concentration of 1.2 mM, the concentration of Cl鈦� ions becomes 0.9 mM due to this dissociation pattern. This variety in ion production allows for more complex calculations and applications in chemistry.

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Most popular questions from this chapter

Federal regulations set an upper limit of 50 parts per million (ppm) of \(\mathrm{NH}_{3}\) in the air in a work environment [that is, 50 molecules of \(\mathrm{NH}_{3}(g)\) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing \(1.00 \times 10^{2} \mathrm{~mL}\) of \(0.0105 \mathrm{M} \mathrm{HCl}\). The \(\mathrm{NH}_{3}\) reacts with HCl according to: $$ \mathrm{NH}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q) $$ After drawing air through the acid solution for 10.0 min at a rate of \(10.0 \mathrm{~L} / \mathrm{min},\) the acid was titrated. The remaining acid needed \(13.1 \mathrm{~mL}\) of \(0.0588 \mathrm{M} \mathrm{NaOH}\) to reach the equivalence point. (a) How many grams of \(\mathrm{NH}_{3}\) were drawn into the acid solution? (b) How many ppm of \(\mathrm{NH}_{3}\) were in the air? (Air has a density of \(1.20 \mathrm{~g} / \mathrm{L}\) and an average molar mass of \(29.0 \mathrm{~g} / \mathrm{mol}\) under the conditions of the experiment.) \((\mathbf{c})\) Is this manufacturer in compliance with regulations?

In 2014 , a major chemical leak at a facility in West Virginia released \(28,390 \mathrm{~L}\) of MCHM (4-methylcyclohexylmethanol, \(\mathrm{C}_{8} \mathrm{H}_{16} \mathrm{O}\) ) into the Elk River. The density of MCHM is 0.9074 \(\mathrm{g} / \mathrm{mL}\). (a) Calculate the initial molarity of MCHM in the river, assuming that the first part of the river is \(2.00 \mathrm{~m}\) deep, \(90.0 \mathrm{~m}\) wide, and \(90.0 \mathrm{~m}\) long. (b) How much farther down the river would the spill have to spread in order to achieve a "safe" MCHM concentration of \(1.00 \times 10^{-4} \mathrm{M}\) ? Assume the depth and width of the river are constant and the concentration of MCHM is uniform along the length of the spill.

Separate samples of a solution of an unknown ionic compound are treated with dilute \(\mathrm{AgNO}_{3}, \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}\), and \(\mathrm{BaCl}_{2}\). Precipitates form in all three cases. Which of the following could be the anion of the unknown salt: \(\mathrm{Br}^{-}, \mathrm{CO}_{3}^{2-}, \mathrm{NO}_{3}^{-}\) ?

The average adult human male has a total blood volume of 5.0 \(\mathrm{L}\). 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?

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.
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