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An aqueous solution contains 1.2 \(\mathrm{mM}\) of total ions. (a) If the solution is NaCl(aq), 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
The concentration of chloride ions in the NaCl(aq) solution is 1.2 mM, while in the FeCl3(aq) solution it is 0.9 mM.

Step by step solution

01

Making sense of the given data

The exercise tells us that the aqueous solution contains 1.2 mM of total ions. That means the sum of the concentrations of each type of ion in the solution is equal to 1.2 mM.
02

Determine chloride ion concentration for NaCl(aq)

In this case, the dissociation of NaCl(aq) in water is given by the following equation: \[NaCl \rightarrow Na^+ + Cl^-\] Each formula unit of NaCl dissociates into one Na+ ion and one Cl- ion. Since the total ion concentration of the solution is 1.2 mM, and NaCl gives one chloride ion per formula unit, the concentration of chloride ions will be equal to the total ion concentration. Therefore, the concentration of chloride ion in the NaCl(aq) solution is 1.2 mM.
03

Determine chloride ion concentration for FeCl3(aq)

In this case, the dissociation of FeCl3(aq) in water is given by the following equation: \[FeCl_{3} \rightarrow Fe^{3+}+3 Cl^-\] Each formula unit of FeCl3 dissociates into one Fe3+ ion and three Cl- ions. Let the concentration of Fe3+ ion be x mM. Since 3 chloride ions are produced per formula unit of FeCl3, the concentration of chloride ions will be 3x mM. The sum of the concentrations of the two ions should be equal to the total ion concentration, which is 1.2 mM. So we have the following equation: \[x+3x=1.2\]
04

Solve for x

Solve the equation from Step 3 for x: \[4x=1.2\] \[x=\frac{1.2}{4}\] \[x=0.3\, mM\] Finally, we can calculate the concentration of Cl- ions: Concentration of Cl- ions = 3x = 3 × 0.3 mM = 0.9 mM The concentration of chloride ions in the FeCl3(aq) solution is 0.9 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 a liquid mixture where water acts as the solvent. The term "aqueous" means that a substance is dissolved in water. This type of solution is very common in chemistry because water is a universal solvent. It can dissolve a wide variety of substances due to its polar nature.

In an aqueous solution:
  • Solutes can be ions, molecules, or gases that dissolve in water.
  • Water molecules surround solute particles, allowing them to spread evenly throughout the solution.
These solutions are integral in chemical reactions as they allow for easy interaction between dissolved substances.

Understanding how substances dissolve and dissociate in aqueous solutions is crucial for solving problems related to ion concentrations, like finding out how much of a particular ion exists in the solution.
Chloride Ions
Chloride ions (\(\text{Cl}^-\)) are negatively charged particles found in many chemical compounds. They are formed when chlorine atoms gain an extra electron. In aquatic environments, such as aqueous solutions, chloride ions commonly result from the dissociation of various salts, including sodium chloride (NaCl) and iron(III) chloride (\(\text{FeCl}_3\)).

Key points about chloride ions:
  • They are important in processes such as electrical conductivity and cellular functions.
  • Chloride ions maintain osmotic balance in biological systems.
  • The concentration of chloride ions in a solution can affect the solution's properties, such as its electrical charge and reactivity.
In studies involving NaCl and \(\text{FeCl}_3\) solutions, calculating the chloride ion concentration allows chemists to understand and predict the behavior of these solutions in various contexts.
Dissociation Equations
Dissociation equations describe how compounds split into ions when they dissolve in water, forming an aqueous solution. These equations are essential to understanding how ions behave and interact in solutions.

Consider these dissociation examples:
  • Sodium chloride (NaCl) dissociates in water as \(\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-\).
  • Iron(III) chloride (\(\text{FeCl}_3\)) dissociates as \(\text{FeCl}_3 \rightarrow \text{Fe}^{3+} + 3\text{Cl}^-\).
For both reactions, each formula unit of the compound separates into individual ions.

These equations help determine how concentrations of ions are calculated in solutions. For example, knowing that \(\text{FeCl}_3\) produces three chloride ions per formula unit tells us that for every unit of \(\text{FeCl}_3\), three times that concentration is present in chloride ions. Understanding dissociation equations is fundamental for predicting and calculating exact ion concentrations in any given solution.

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Most popular questions from this chapter

Bronze is a solid solution of \(\mathrm{Cu}(\mathrm{s})\) and \(\mathrm{Sn}(\mathrm{s})\) ; solutions of metals like this that are solids are called alloys. There is a range of compositions over which the solution is considered a bronze. Bronzes are stronger and harder than either copper or tin alone. (a) \(\mathrm{A} 100.0\) -g sample of a certain bronze is 90.0\(\%\) copper by mass and 10.0\(\%\) tin. Which metal can be called the solvent, and which the solute? (b) Based on part (a), calculate the concentration of the solute metal in the alloy in units of molarity, assuming a density of 7.9 \(\mathrm{g} / \mathrm{cm}^{3}\) . (c) Suggest a reaction that you could do to remove all the tin from this bronze to leave a pure copper sample. Justify your reasoning.

Uranium hexafluoride, UF\(_{6}\), is processed to produce fuel for nuclear reactors and nuclear weapons. UF\(_{6}\) is made from the reaction of elemental uranium with \(\mathrm{ClF}_{3},\) which also produces \(\mathrm{Cl}_{2}\) as a by-product. (a) Write the balanced molecular equation for the conversion of U and \(\mathrm{ClF}_{3}\) into UF \(_{6}\) and \(\mathrm{Cl}_{2}\) . (b) Is this a metathesis reaction? (c) Is this a redox reaction?

(a) How many milliliters of a stock solution of 6.0 \(\mathrm{MHNO}_{3}\) would you have to use to prepare 110 \(\mathrm{mL}\) of 0.500 \(\mathrm{M} \mathrm{HNO}_{3} ?\) (b) If you dilute 10.0 \(\mathrm{mL}\) of the stock solution to a final volume of \(0.250 \mathrm{L},\) what will be the concentration of the diluted solution?

A 1.248 -g sample of limestone rock is pulverized and then treated with 30.00 mL of 1.035\(M\) HCl solution. The excess acid then requires 11.56 \(\mathrm{mL}\) of 1.010 \(\mathrm{M}\) NaOH for neutralization. Calculate the percentage by mass of calcium carbonate in the rock, assuming that it is the only substance reacting with the HCl solution.

(a) A strontium hydroxide solution is prepared by dissolving 12.50 g of \(\operatorname{Sr}(\mathrm{OH})_{2}\) in water to make 50.00 \(\mathrm{mL}\) of solution. What is the molarity of this solution? (b) Next the strontium hydroxide solution prepared in part (a) is used to titrate a nitric acid solution of unknown concentration. Write a balanced chemical equation to represent the reaction between strontium hydroxide and nitric acid solutions. (c) If 23.9 mL of the strontium hydroxide solution was needed to neutralize a 37.5 mL aliquot of the nitric acid solution, what is the concentration (molarity) of the acid?

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