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Chapter 4: Problem 38

The reaction of \(\mathrm{SnCl}_{2}(a q)\) with \(\mathrm{RhCl}_{3}(a q)\) in aqueous \(\mathrm{HCl}\) yields a red solution of a 1: 1 Rh-Sn compound. If a solution prepared by adding \(150 \mathrm{mL}\) of a \(0.272 \mathrm{m} M\) aqueous solution of \(\mathrm{SnCl}_{2}\) to \(50 \mathrm{mL}\) of an aqueous solution of \(8.5 \mathrm{mg} \mathrm{RhCl}_{3}\) has an absorbance of \(0.85,\) as measured in a \(1.00 \mathrm{cm}\) cell, what is the molar absorptivity of the red compound?

Short Answer

Expert verified
Answer: The molar absorptivity of the red Rh-Sn compound is 4187 M鈦宦� cm鈦宦�.

Step by step solution

01

Find the amount of moles of SnCl2 and RhCl3

First, we need to find the amount of moles of \(\mathrm{SnCl}_{2}\) and \(\mathrm{RhCl}_{3}\) in the given solutions. Amount of moles of \(\mathrm{SnCl}_{2} = \mathrm{Volume} \times \mathrm{Concentration} = 150 \mathrm{mL} \times 0.272 \mathrm{m} \mathrm{M} = 40.8 \mathrm{mmol}\) Amount of moles of \(\mathrm{RhCl}_{3} = \dfrac{\mathrm{Mass}}{\mathrm{Molar \ Mass}} = \dfrac{8.5 \mathrm{mg}}{209.24 \mathrm{g/mol}} = 0.0406 \mathrm{mmol}\)
02

Find the limiting reactant and moles of red compound formed

Now, we'll find the limiting reactant and calculate the moles of the red compound formed in the reaction. Since the molar ratio of Rh-Sn compound to both reactants is 1:1, we can deduce that \(\mathrm{RhCl}_{3}\) is the limiting reactant since we have fewer moles of it than SnCl2. Therefore, the moles of red Rh-Sn compound formed will be equal to the moles of \(\mathrm{RhCl}_{3}\): Moles of red Rh-Sn compound = 0.0406 mmol
03

Calculate the total volume of the solution

Before we can find the concentration of the Rh-Sn compound in the solution, we need to calculate the final volume of the solution after mixing. Total Volume = Volume of SnCl2 solution + Volume of RhCl3 solution = 150 mL + 50 mL = 200 mL
04

Determine the concentration of the red compound

We can now determine the concentration of the red Rh-Sn compound in the solution. Concentration of Rh-Sn compound = \(\dfrac{\mathrm{Moles}}{\mathrm{Volume}} = \dfrac{0.0406 \mathrm{mmol}}{200 \mathrm{mL}} = 0.000203 \mathrm{M}\)
05

Apply Beer's Law to find molar absorptivity

Finally, we will use Beer's Law, which states that \(\mathrm{Absorbance} = \epsilon \times \mathrm{Concentration} \times \mathrm{Pathlength}\), to determine the molar absorptivity, \(\epsilon\), of the red Rh-Sn compound. Rearrange Beer's Law equation to solve for \(\epsilon\): \(\epsilon = \dfrac{\mathrm{Absorbance}}{\mathrm{Concentration} \times \mathrm{Pathlength}}\) Molar absorptivity of red Rh-Sn compound = \(\dfrac{0.85}{0.000203 \mathrm{M} \times 1.00 \mathrm{cm}} = 4187 \mathrm{M^{-1} cm^{-1}}\) The molar absorptivity of the red Rh-Sn compound is 4187 \(\mathrm{M^{-1} cm^{-1}}\).

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Key Concepts

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

Molar Absorptivity
Molar absorptivity, often symbolized as \( \epsilon \), is an important parameter in spectrophotometry. It is a measure of how strongly a chemical species absorbs light at a given wavelength. In the context of Beer鈥檚 Law, it connects absorbance with concentration and path length.
  • Beer鈥檚 Law equation: \( \text{Absorbance} = \epsilon \times \text{Concentration} \times \text{Pathlength} \)
  • Molar absorptivity has units of \( \text{M}^{-1} \text{cm}^{-1} \).
  • A higher molar absorptivity indicates a compound absorbs more light, making it easier to detect at lower concentrations.

To find the molar absorptivity of a compound, like the red Rh-Sn compound in our exercise, we rearrange the Beer鈥檚 Law equation to solve for \( \epsilon \). Substituting the known absorbance, concentration, and path length gives us the measure of how strongly our compound absorbs light.
Limiting Reactant
In any chemical reaction, the limiting reactant is the substance that is completely consumed first, stopping the reaction from proceeding further. This concept is crucial because it determines the maximum amount of product that can be formed.
  • To identify the limiting reactant, compare the mole ratio of reactants to the coefficients in the balanced chemical equation.
  • The reactant with the smallest number of moles relative to its stoichiometric requirement is the limiting reactant.

In the example provided, we compare the moles of \( \text{SnCl}_2 \) and \( \text{RhCl}_3 \). With a 1:1 ratio based on the reaction, \( \text{RhCl}_3 \) is the limiting reactant because we have fewer moles of it. This determines that an equal amount of the red Rh-Sn compound, in moles, can be formed.
Concentration Calculation
Calculating concentration in solutions is often necessary to understand the behavior of a mixture in a reaction. Concentration is generally expressed in molarity, which is defined as moles of solute per liter of solution.
  • The formula used is \( \text{Concentration} = \frac{\text{Moles of solute}}{\text{Total volume of solution (in liters)}} \).
  • It鈥檚 important to convert all volume measurements to liters when using this formula.

In the step-by-step solution, the final concentration of the Rh-Sn compound is calculated using the total volume of the combined solutions. This concentration value is then used in conjunction with Beer's Law to find the molar absorptivity. Understanding how to calculate concentration is key to progressing through these types of exercises and accurately applying Beer鈥檚 Law.

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Most popular questions from this chapter

Balance the following half-reactions by adding the appropriate number of electrons. Which are oxidation half-reactions and which are reduction half- reactions? a. \(\mathrm{Fe}^{2+}(a q) \rightarrow \mathrm{Fe}^{3+}(a q)\) b. \(\operatorname{Ag}\left[(s) \rightarrow \operatorname{Ag}(s)+\bar{I}^{-}(a q)\right.\) c. \(\mathrm{VO}_{2}^{+}(a q)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{VO}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell)\) d. \(\mathrm{I}_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 2 \mathrm{IO}_{3}^{-}(a q)+12 \mathrm{H}^{+}(a q)\)

Is a saturated solution always a concentrated solution? Explain.

Rank the following solutions on the basis of their ability to conduct electricity, starting with the most conductive: (a) \(1.0 \mathrm{M} \mathrm{NaCl} ;\) (b) \(1.2 \mathrm{M} \mathrm{KCl} ;\) (c) \(1.0 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) (d) \(0.75 M\) LiCl.

A food chemist determines the concentration of acetic acid in a sample of apple cider vinegar (see Problem 4.133 ) by acid-base titration. The density of the sample is \(1.01 \mathrm{g} / \mathrm{m} \mathrm{L}\) The titrant is \(1.002 M\) aOH. The average volume of titrant required to titrate \(25.00 \mathrm{mL}\) aliquots of the vinegar is \(20.78 \mathrm{mL} .\) What is the concentration of acetic acid in the vinegar? Express your answer the way a food chemist probably would: as percent by mass.

Toxic chromate can be precipitated from an aqueous solution by bubbling \(\mathrm{SO}_{2}\) through the solution. How many grams of \(\mathrm{SO}_{2}\) are required to treat \(3.0 \times 10^{8} \mathrm{L}\) of \(0.050 \mathrm{mM} \mathrm{CrO}_{4}^{-} ?\) $$\begin{aligned} &2 \mathrm{CrO}_{4}^{2-}(a q)+3 \mathrm{SO}_{2}(g)+4 \mathrm{H}^{+}(a q) \rightarrow\\\ &\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}(s)+2 \mathrm{H}_{2} \mathrm{O}(\ell) \end{aligned}$$
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