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Chapter 4: Problem 70

In each of the following pairs, indicate which has the higher concentration of I \(^{-}\) ion: (a) \(0.10 \mathrm{M} \mathrm{BaI}_{2}\) or \(0.25 \mathrm{M}\) KI solution, (b) \(100 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) KI solution or \(200 \mathrm{~mL}\) of \(0.040 \mathrm{M} \mathrm{ZnI}_{2}\) solution, (c) \(3.2 \mathrm{M}\) HI solution or a solution made by dissolving \(145 \mathrm{~g}\) of NaI in water to make \(150 \mathrm{~mL}\) of solution.

Short Answer

Expert verified
In summary, for each of the following pairs, the solution with a higher concentration of I⁻ ions is: (a) \(0.25\, \mathrm{M}\) KI solution (b) \(0.10\, \mathrm{M}\) KI solution (c) the solution made by dissolving \(145\, \mathrm{g}\) of NaI in water to make \(150\, \mathrm{mL}\) of solution.

Step by step solution

01

(a) Compare BaI₂ and KI concentrations

To find which solution has a higher I⁻ concentration, we must consider the balanced chemical equations for the dissolution of BaI₂ and KI in water. For BaI₂: \(\mathrm{BaI}_2 \rightarrow \mathrm{Ba^{2+}} + 2\mathrm{I^-}\) For KI: \(\mathrm{KI} \rightarrow \mathrm{K^+} + \mathrm{I^-}\) In a \(0.10 \mathrm{M}\) BaI₂ solution, one mole of BaI₂ produces 2 moles of I⁻ ions, so the I⁻ concentration will be: \(0.10 \, \mathrm{M} * 2 = 0.20 \, \mathrm{M}\) In a \(0.25\, \mathrm{M}\) KI solution, one mole of KI produces one mole of I⁻ ions, so the I⁻ concentration will be: \(0.25\, \mathrm{M} * 1 = 0.25\, \mathrm{M}\) Since \(0.25\, \mathrm{M} > 0.20\, \mathrm{M}\), the KI solution has a higher I⁻ concentration in this pair.
02

(b) Compare KI and ZnI₂ Solutions

We'll again consider the balanced chemical equations for the dissolution of KI and ZnI₂ in water. For KI: \(\mathrm{KI} \rightarrow \mathrm{K^+} + \mathrm{I^-}\) For ZnI₂: \(\mathrm{ZnI}_2 \rightarrow \mathrm{Zn^{2+}} + 2\mathrm{I^-}\) First, we need to convert the volumes of these two solutions to liters. \(100\, \mathrm{mL} = 0.1\, \mathrm{L}\), and \(200\, \mathrm{mL} = 0.2\, \mathrm{L}\) Next, we'll calculate the moles of I⁻ ions in both solutions: For \(0.1\, \mathrm{L}\) of \(0.10\, \mathrm{M}\) KI solution: \(\mathrm{moles \, of \, I^-} = 0.1\, \mathrm{L} × 0.10\, \mathrm{M} = 0.01\, \mathrm{mol}\) For \(0.2\, \mathrm{L}\) of \(0.040\, \mathrm{M}\) ZnI₂ solution: \(\mathrm{moles \, of \, I^-} = 0.2\, \mathrm{L} × 0.040\, \mathrm{M} × 2 = 0.016\, \mathrm{mol}\) Now we can calculate the concentrations of I⁻ ions: For the KI solution: \(\mathrm{I^- \, concentration} = \frac{0.01\, \mathrm{mol}}{0.1\, \mathrm{L}} = 0.10\, \mathrm{M}\) For the ZnI₂ solution: \(\mathrm{I^- \, concentration} = \frac{0.016\, \mathrm{mol}}{0.2\, \mathrm{L}} = 0.08\, \mathrm{M}\) Since \(0.10\, \mathrm{M} > 0.08\, \mathrm{M}\), the KI solution has a higher I⁻ concentration in this pair.
03

(c) Compare HI and NaI Solutions

We'll consider the balanced chemical equations for the dissolution of HI and NaI in water. For HI: \(\mathrm{HI} \rightarrow \mathrm{H^+} + \mathrm{I^-}\) For NaI: \(\mathrm{NaI} \rightarrow \mathrm{Na^+} + \mathrm{I^-}\) First, we need to find the number of moles of NaI given its mass: Given that the molar mass of NaI is approximately \(149\, \mathrm{g/mol}\): \(\mathrm{moles \, of \, NaI} = \frac{145\, \mathrm{g}}{149\, \mathrm{g/mol}} \approx 0.97\, \mathrm{mol}\) Next, calculate the concentration of I⁻ ions in both solutions: For the HI solution, the I⁻ concentration is \(3.2\, \mathrm{M}\) because HI completely dissociates in water. For the NaI solution, we need to find the concentration of I⁻ ions based on the provided mass and volume: \(\mathrm{I^- \, concentration} = \frac{0.97\, \mathrm{mol}}{0.150\, \mathrm{L}} \approx 6.47\, \mathrm{M}\) Since \(6.47\, \mathrm{M} > 3.2\, \mathrm{M}\), the NaI solution has a higher I⁻ concentration in this pair.

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Key Concepts

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

Dissolution of Compounds
The process of dissolving a compound in a solvent, such as water, to produce ions is referred to as dissolution. When a compound dissolves, it separates into its constituent ions. For instance, when solid barium iodide (BaI2) dissolves in water, it dissociates into one barium ion (Ba2+) and two iodide ions (I-). Similarly, potassium iodide (KI) gives one potassium ion (K+) and one iodide ion (I-) upon dissolution.

Dissolution is crucial in determining the concentration of a specific ion in solution. It's important for students to remember that not all compounds dissociate in the same ratio. For example, one mole of BaI2 produces two moles of iodide ions, while one mole of KI produces only one mole of iodide. Understanding this concept allows for a better grasp of ionic concentrations in solution.
Chemical Equation Balancing
A balanced chemical equation ensures that the law of conservation of mass is upheld, indicating that the number of atoms of each element is the same on both sides of the equation. This is vital for calculating the changes in concentration when compounds dissolve in solution.

For instance, the dissolution of barium iodide (BaI2) is represented by \( \mathrm{BaI}_2 \rightarrow \mathrm{Ba^{2+}} + 2\mathrm{I^-} \), which means that for each formula unit of BaI2 that dissolves, there is one Ba2+ ion and two I- ions produced. Balancing chemical equations gives a clear picture of these stoichiometric relationships and is essential for accurately calculating ion concentrations.
Molar Concentration Calculations
Molar concentration, often represented as molarity and expressed in moles per liter (M), is a measure of the concentration of a solute in a solution. To determine molar concentration, you need the number of moles of the solute and the volume of the solution in liters.

Calculating molar concentrations involves using the following formula: \( \text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} \). For example, to find the iodide ion concentration from a solid NaI dissolved in water, you first find the moles of NaI using the mass given and its molar mass. Then, you relate this to the volume of solution, which gives the molarity of the iodide ions. It is crucial to take into account that some compounds can produce more than one mole of an ion per mole of compound dissolved, affecting the final concentration.

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Most popular questions from this chapter

The newest US standard for arsenate in drinking water, mandated by the Safe Drinking Water Act, required that by January \(2006,\) public water supplies must contain no greater than 10 parts per billion (ppb) arsenic. If this arsenic is present as arsenate, \(\mathrm{AsO}_{4}{ }^{3-},\) what mass of sodium arsenate would be present in a \(1.00-\mathrm{L}\) sample of drinking water that just meets the standard? Parts per billion is defined on a mass basis as $$\mathrm{ppb}=\frac{\mathrm{g} \text { solute }}{\mathrm{g} \text { solution }} \times 10^{9}$$

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.

A sample of \(1.50 \mathrm{~g}\) of lead(II) nitrate is mixed with \(125 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) sodium sulfate solution. (a) Write the chemical equation for the reaction that occurs. (b) Which is the limiting reactant in the reaction? (c) What are the concentrations of all ions that remain in solution after the reaction is complete?

Using solubility guidelines, predict whether each of the following compounds is soluble or insoluble in water: (a) \(\mathrm{MgBr}_{2}\), (b) \(\mathrm{PbI}_{2}\), (c) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\) (e) \(\mathrm{ZnSO}_{4}\). (d) \(\mathrm{Sr}(\mathrm{OH})_{2}\),

Write balanced molecular and net ionic equations for the reactions of (a) manganese with dilute sulfuric acid, (b) chromium with hydrobromic acid, (c) tin with hydrochloric acid, (d) aluminum with formic acid, HCOOH.
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