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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鈦� 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 BaI鈧� and KI concentrations

To find which solution has a higher I鈦� concentration, we must consider the balanced chemical equations for the dissolution of BaI鈧� and KI in water. For BaI鈧�: \(\mathrm{BaI}_2 \rightarrow \mathrm{Ba^{2+}} + 2\mathrm{I^-}\) For KI: \(\mathrm{KI} \rightarrow \mathrm{K^+} + \mathrm{I^-}\) In a \(0.10 \mathrm{M}\) BaI鈧� solution, one mole of BaI鈧� produces 2 moles of I鈦� ions, so the I鈦� 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鈦� ions, so the I鈦� 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鈦� concentration in this pair.
02

(b) Compare KI and ZnI鈧� Solutions

We'll again consider the balanced chemical equations for the dissolution of KI and ZnI鈧� in water. For KI: \(\mathrm{KI} \rightarrow \mathrm{K^+} + \mathrm{I^-}\) For ZnI鈧�: \(\mathrm{ZnI}_2 \rightarrow \mathrm{Zn^{2+}} + 2\mathrm{I^-}\) 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鈦� ions in both solutions: For \(0.1\, \mathrm{L}\) of \(0.10\, \mathrm{M}\) KI solution: \(\mathrm{moles \, of \, I^-} = 0.1\, \mathrm{L} 脳 0.10\, \mathrm{M} = 0.01\, \mathrm{mol}\) For \(0.2\, \mathrm{L}\) of \(0.040\, \mathrm{M}\) ZnI鈧� solution: \(\mathrm{moles \, of \, I^-} = 0.2\, \mathrm{L} 脳 0.040\, \mathrm{M} 脳 2 = 0.016\, \mathrm{mol}\) Now we can calculate the concentrations of I鈦� ions: For the KI solution: \(\mathrm{I^- \, concentration} = \frac{0.01\, \mathrm{mol}}{0.1\, \mathrm{L}} = 0.10\, \mathrm{M}\) For the ZnI鈧� 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鈦� 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鈦� ions in both solutions: For the HI solution, the I鈦� concentration is \(3.2\, \mathrm{M}\) because HI completely dissociates in water. For the NaI solution, we need to find the concentration of I鈦� 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鈦� 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 metal cadmium tends to form Cd \(^{2+}\) ions. The following observations are made: (i) When a strip of zinc metal is placed in \(\mathrm{CdCl}_{2}(a q),\) cadmium metal is deposited on the strip. (ii) When a strip of cadmium metal is placed in \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q),\) nickel metal is deposited on the strip. (a) Write net ionic equations to explain each of the preceding observations. (b) What can you conclude about the position of cadmium in the activity series? (c) What experiments would you need to perform to locate more precisely the position of cadmium in the activity series?

You are presented with a white solid and told that due to careless labeling it is not clear if the substance is barium chloride, lead chloride, or zinc chloride. When you transfer the solid to a beaker and add water, the solid dissolves to give a clear solution. Next a \(\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) solution is added and a white precipitate forms. What is the identity of the unknown white solid? \([\) Section 4.2\(]\)

Using the activity series (Table 4.5 ), write balanced chemical equations for the following reactions. If no reaction occurs, simply write NR. (a) Iron metal is added to a solution of copper(II) nitrate; (b) zinc metal is added to a solution of magnesium sulfate; (c) hydrobromic acid is added to tin metal; (d) hydrogen gas is bubbled through an aqueous solution of nickel(II) chloride; (e) aluminum metal is added to a solution of cobalt(II) sulfate.

(a) How many milliliters of a stock solution of \(6.0 \mathrm{M} \mathrm{HNO}_{3}\) 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?

What does it mean to say that ions are solvated when an ionic substance dissolves in water?
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