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Chapter 11: Problem 28

Describe what is done in a displacement titration and give an example.

Short Answer

Expert verified
In displacement titration, a substance displaces a titratable substance, which is then titrated. An example is using iodide to displace copper ions, followed by titration of iodine with thiosulfate.

Step by step solution

01

Understanding the Concept

In a displacement titration, a known amount of a compound (substance A) is reacted with a titrant (substance B). However, A is not directly titratable, so A reacts with another substance (substance C) to displace a different titratable component (substance D). Then a titration is performed to determine the concentration of substance D.
02

Initial Reaction Process

Begin by mixing compound A with an excess of compound C. Compound A will react with C and displace a known substance D from C. This reaction should proceed to completion.
03

Titration Reaction

The displaced substance D, which is titratable, is then titrated with a standard titrant solution (substance B) to determine its concentration. The volume of titrant used gives information on the amount of displaced substance D.
04

Example of Displacement Titration

A common example is the titration of a mixture containing copper ions. Copper ions can react with iodide ions to form copper iodide and free iodine, as shown in the reaction: \[ Cu^{2+} + 2I^- \rightarrow CuI + I_2 \]. The free iodine is then titrated with a thiosulfate solution to determine the amount of \( Cu^{2+} \).

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Key Concepts

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

Reaction Process
A displacement titration involves a fascinating reaction process that unfolds in multiple stages. Initially, we have a compound, known as compound A, which is not readily titratable on its own. To bring this compound into the spotlight, it is allowed to interact with another substance, let's call it compound C. During this interaction, an interesting chemical reaction occurs leading to the displacement of another substance, compound D. Compound D becomes essential as it is the one we can easily titrate later on.

The reaction process between A and C should continue until it is complete, meaning compound D is entirely displaced and is free from reactions with C. It's essential to ensure this reaction goes to completion for accurate results, as incomplete reactions could lead to erroneous data in the titration step. Therefore, controlling conditions like temperature, concentration, and reaction time is crucial.
Titration Example
Let's delve into an example to clarify the concept of displacement titration.

Imagine you have a sample containing a complex mixture, but your interest lies in the copper ions present. Copper ions (Cu虏鈦�) aren鈥檛 directly titrated; instead, they can displace another component in a reaction, like iodide ions (I鈦�). When mixed, copper ions work their chemical magic, forming copper iodide (CuI) and freeing up iodine (I鈧�).
  • The reaction proceeds as:
    • \[ Cu^{2+} + 2I^- \rightarrow CuI + I_2 \]
The free iodine then steps into the spotlight for the titration process. This intricate dance of ions and atoms exemplifies a displacement titration perfectly.
Copper Ions Titration
Titrating copper ions involves additional steps following the initial displacement reaction. Once iodine is liberated in the reaction between copper and iodide ions, it enters the titration phase, where a titrant, such as a thiosulfate solution, is employed to quantify the iodine (and, by proxy, the initial copper ions).

When iodine (\[ I_2 \]) is titrated with thiosulfate (\[ S_2O_3^{2-} \]), the following reaction takes place:
  • \[ 2S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^- \]
The completion of this titration allows chemists to determine exactly how much iodine was generated, which in turn provides the concentration of copper ions initially present.
  • Steps to remember:
    • React copper ions with iodide to produce iodine.
    • Titrate iodine with thiosulfate to measure copper content.
This method offers high accuracy and is a classic example of using displacement titration intelligently to analyze complex mixtures.

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Most popular questions from this chapter

A \(1.000-\mathrm{mL}\) sample of unknown containing \(\mathrm{Co}^{2+}\) and \(\mathrm{Ni}^{2+}\) was treated with \(25.00 \mathrm{~mL}\) of \(0.03872 \mathrm{M}\) EDTA. Back titration with \(0.02127 \mathrm{M} \mathrm{Zn}^{2+}\) at \(\mathrm{pH} 5\) required \(23.54 \mathrm{~mL}\) to reach the xylenol orange end point. A \(2.000-\mathrm{mL}\) sample of unknown was passed through an ion- exchange column that retards \(\mathrm{Co}^{2+}\) more than \(\mathrm{Ni}^{2+}\). The \(\mathrm{Ni}^{2+}\) that passed through the column was treated with \(25.00 \mathrm{~mL}\) of \(0.03872 \mathrm{M}\) EDTA and required \(25.63 \mathrm{~mL}\) of \(0.02127 \mathrm{M} \mathrm{Zn}^{2+}\) for back titration. The \(\mathrm{Co}^{2+}\) emerged from the column later. It, too, was treated with \(25.00 \mathrm{~mL}\) of \(0.03872\) M EDTA. How many milliliters of \(0.02127 \mathrm{M} \mathrm{Zn}^{2+}\) will be required for back titration?

Calculate \(\mathrm{pCo}^{2+1}\) at each of the following points in the titration of \(25.00 \mathrm{~mL}\) of \(0.02026 \mathrm{M} \mathrm{Co}^{2+}\) by \(0.03855 \mathrm{M}\) EDTA at \(\mathrm{pH}\) 6.00: (a) \(12.00 \mathrm{~mL}\); (b) \(V_{c}\); (c) \(14.00 \mathrm{~mL}\). 11-8. Consider the titration of \(25.0 \mathrm{~mL}\) of \(0.0200 \mathrm{M} \mathrm{MnSO}_{4}\) with \(0.0100 \mathrm{M}\) EDTA in a solution buffered to \(\mathrm{pH} 8.00\). Calculate \(\mathrm{pMn}^{2+}\) at the following volumes of added EDTA and sketch the titration curve: (a) \(0 \mathrm{~mL}\) (d) \(49.0 \mathrm{~mL}\) (g) \(50.1 \mathrm{~mL}\) (b) \(20.0 \mathrm{~mL}\) (e) \(49.9 \mathrm{~mL}\) (h) \(55.0 \mathrm{~mL}\) (c) \(40.0 \mathrm{~mL}\) (f) \(50.0 \mathrm{~mL}\) (i) \(60.0 \mathrm{~mL}\)

Microequilibrium constants for binding of metal to a protein. The iron- transport protein, transferrin, has two distinguishable metal-binding sites, designated a and b. The microequilibrium formation constants for each site are defined as follows: For example, the formation constant \(k_{\mathrm{la}}\) refers to the reaction \(\mathrm{Fe}^{3+}+\) transferrin \(\rightleftharpoons \mathrm{Fe}_{\mathrm{a}}\) transferrin, in which the metal ion binds to site a: $$ k_{1 \mathrm{a}}=\frac{\left[\mathrm{Fe}_{a} \text { transferrin }\right]}{\left[\mathrm{Fe}^{3+}\right][\text { transferrin }]} $$ (a) Write the chemical reactions corresponding to the conventional macroscopic formation constants, \(K_{1}\) and \(K_{2}\). (b) Show that \(K_{1}=k_{1 \mathrm{a}}+k_{1 \mathrm{~b}}\) and \(K_{2}^{-1}=k_{2 \mathrm{a}}^{-1}+k_{2 \mathrm{~b}}^{-1}\). (c) Show that \(k_{1 \mathrm{a}} k_{2 \mathrm{~b}}=k_{1 \mathrm{~b}} k_{2 \mathrm{a}}\). This expression means that, if you know any three of the microequilibrium constants, you automatically know the fourth one. (d) A challenge to your sanity. From the equilibrium constants below, find the equilibrium fraction of each of the four species in the diagram above in circulating blood that is \(40 \%\) saturated with iron (that is, Fe/transferrin \(=0.80\), because each protein can bind 2 Fe). Effective formation constants for blood plasma at \(\mathrm{pH} 7.4\) \begin{tabular}{ll} \hline\(k_{1 \mathrm{a}}=6.0 \times 10^{22}\) & \(k_{2 \mathrm{a}}=2.4 \times 10^{22}\) \\ \(k_{1 \mathrm{~b}}=1.0 \times 10^{22}\) & \(k_{2 \mathrm{~b}}=4.2 \times 10^{21}\) \\ \(K_{1}=7.0 \times 10^{22}\) & \(K_{2}=3.6 \times 10^{21}\) \\ \hline \end{tabular} The binding constants are so large that you may assume that there is negligible free \(\mathrm{Fe}^{3+}\). To get started, let's use the abbreviations [T] = \([\) transferrin \(],[\mathrm{FeT}]=\left[\mathrm{Fe}_{\mathrm{a}} \mathrm{T}\right]+\left[\mathrm{Fe}_{\mathrm{b}} \mathrm{T}\right]\), and \(\left[\mathrm{Fe}_{2} \mathrm{~T}\right]=\left[\mathrm{Fe}_{2} \operatorname{transferrin}\right] .\) Now we can write Mass balance for protein: \(\quad[\mathrm{T}]+[\mathrm{FeT}]+\left[\mathrm{Fe}_{2} \mathrm{~T}\right]=1\) Mass balance for iron: $$ \frac{[\mathrm{FeT}]+2\left[\mathrm{Fe}_{2} \mathrm{~T}\right]}{[\mathrm{T}]+[\mathrm{Fe} \mathrm{T}]+\left[\mathrm{Fe}_{2} \mathrm{~T}\right]}=[\mathrm{FeT}]+2\left[\mathrm{Fe}_{2} \mathrm{~T}\right]=0.8 $$ Combined equilibria: \(\frac{K_{1}}{K_{2}}=19_{.44}=\frac{[\mathrm{FeT}]^{2}}{[\mathrm{~T}]\left[\mathrm{Fe}_{2} \mathrm{~T}\right]}\) Now you have three equations with three unknowns and should be able to tackle this problem.

List four methods for detecting the end point of an EDTA titration.

A \(50.0-\mathrm{mL}\) solution containing \(\mathrm{Ni}^{2+}\) and \(\mathrm{Zn}^{2+}\) was treated with \(25.0 \mathrm{~mL}\) of \(0.0452 \mathrm{M}\) EDTA to bind all the metal. The excess unreacted EDTA required \(12.4 \mathrm{~mL}\) of \(0.0123 \mathrm{M} \mathrm{Mg}^{2+}\) for complete reaction. An excess of the reagent 2,3-dimercapto-1-propanol was then added to displace the EDTA from zinc. Another \(29.2 \mathrm{~mL}\) of \(\mathrm{Mg}^{2+}\) were required for reaction with the liberated EDTA. Calculate the molarity of \(\mathrm{Ni}^{2+}\) and \(\mathrm{Zn}^{2+}\) in the original solution.
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