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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?

Short Answer

Expert verified
The required volume of Zn虏鈦 for Co虏鈦 back titration is estimated to be 28.22 mL.

Step by step solution

01

Calculate the moles of EDTA used in excess

First, we need to determine the moles of EDTA that were added to both the Co虏鈦 and Ni虏鈦 in the initial solution. Given the volume and molarity, we can calculate it as follows:\[ n_{ ext{EDTA,total}} = 25.00 ext{ mL} \times 0.03872 ext{ M} - 0.02127 ext{ M} \times 23.54 ext{ mL} \]Convert milliliters to liters in the calculation.
02

Calculate total moles of metal ions originally present

In this step, we'll determine the total moles of metal ions (Co虏鈦 and Ni虏鈦) originally in the sample. Since EDTA binds with metal ions in a 1:1 ratio, the total moles of metal ions will be equal to the moles of EDTA initially added minus the moles of Zn虏鈦 used for back titration:\[ n_{ ext{metal,total}} = 0.03872 \times 0.02500 - 0.02127 \times 0.02354 \]
03

Determine moles of Ni虏鈦 captured initially

The 2.000 mL sample that passed through the ion-exchange column mainly contained Ni虏鈦 ions. We treated this with the same amount of EDTA, which required 25.63 mL of Zn虏鈦 for back titration. Calculate the amount of EDTA that reacted with Ni虏鈦:\[ n_{ ext{EDTA,Ni}} = 0.03872 \times 0.02500 - 0.02127 \times 0.02563 \]
04

Calculate moles of Ni虏鈦 in the initial solution

Now find the moles of Ni虏鈦 in the initial 1.000 mL solution using the moles from the 2.000 mL sample. Since the 2.000 mL sample reflected just the Ni虏鈦 ions, the initial 1.000 mL sample should have half the moles found at the exchange column:\[ n_{ ext{Ni,total}} = \frac{n_{ ext{EDTA,Ni}}}{2} \]
05

Calculate moles of Co虏鈦 in the initial solution

Subtract the total moles of Ni虏鈦 found from the total moles of metal ions to determine the moles of Co虏鈦 in the initial solution:\[ n_{ ext{Co,total}} = n_{ ext{metal,total}} - n_{ ext{Ni,total}} \]
06

Calculate Zn虏鈦 required for Co虏鈦 back titration

Lastly, determine how much Zn虏鈦 is required to back titrate the Co虏鈦. We already treated Co虏鈦 with the same 25.00 mL of EDTA. The moles of Zn虏鈦 needed will be equal to the excess EDTA moles:\[ V_{ ext{Zn,Co}} = \frac{0.03872 \times 0.02500 - n_{ ext{Co,total}}}{0.02127} \]

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Key Concepts

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

EDTA Titration
EDTA titration is a common analytical chemistry technique used to determine the concentration of metal ions in a solution. EDTA stands for ethylenediaminetetraacetic acid, a chelating agent that forms stable complexes with metal ions. This is done by the EDTA molecule bonding with the metal ions at various coordination sites, effectively "trapping" the ions.

Key points about EDTA titration include:
  • Chelation: EDTA binds with metal ions in a 1:1 stoichiometry, which means one molecule of EDTA binds to one metal ion.
  • pH dependence: The effectiveness of EDTA is highly pH-dependent, which typically requires the titration to be conducted in buffered solutions.
  • Indicators: Special indicators, like ion-sensitive dyes, are often used to detect the endpoint of the titration, where all metal ions are bound by EDTA.
  • Versatility: EDTA can be used for a wide variety of metal ions, making it a versatile tool for analytical chemists.
Understanding EDTA titration helps explain how it was utilized in the exercise to determine the amount of Co虏鈦 and Ni虏鈦 in a given solution. The reaction progresses until all metal ions are sequestered by the EDTA present, allowing for accurate determination of their concentrations.
Back Titration
Back titration is an analytical method used when the direct titration of a sample is not possible or practical. This indirect technique involves adding an excess of a standard reagent to the analyte, allowing the reaction to complete, and then titrating the remaining reagent with another titrant to determine the amount in excess.

Benefits of back titration include:
  • Dealing with complex samples: It's useful for determining substances present in compounds that don't have clear endpoints or are insoluble in the reaction medium.
  • Improving reaction conditions: It can sometimes offer more favorable reaction conditions by allowing the primary reaction to proceed more completely.
  • Indirect measurement: Especially helpful for substances that are difficult to titrate directly.
In the provided exercise, back titration was performed using zinc ions ( Zn虏鈦 ) to react with the excess EDTA after it had complexed with all available metal ions in the sample. By measuring the amount of Zn虏鈦 used, it is possible to calculate how much EDTA was not bound to the metal ions initially present. This information reveals how much Co虏鈦 and Ni虏鈦 were in the original sample.
Ion-Exchange Chromatography
Ion-exchange chromatography is a separation technique useful for distinguishing between ions and polar molecules based on their charge properties. In the context of the exercise, this approach helped separate the Co虏鈦 from Ni虏鈦 .

Core aspects of ion-exchange chromatography include:
  • Stationary phase: Typically, an ion-exchange resin that can either capture or repel ions based on their charges relative to the resin.
  • Mobile phase: A liquid that passes through the column, to which the analytes are added. This helps move the ions through the system.
  • Selectivity: Different ions will have varying affinities for the ion-exchange resin, allowing for selective separation. In this case, Co虏鈦 is retained longer than Ni虏鈦 .
  • Dynamic Process: Conditions such as resin type and column length can be optimized for target breaches and separations.
By using ion-exchange chromatography in the exercise, it was easy to first separate the Ni虏鈦 from Co虏鈦 before performing the EDTA titrations on these individually, providing clarity on which metal ions were being analyzed at any given step.
Xylenol Orange Endpoint
Xylenol orange is a dye used as an indicator in complexometric titrations involving metal ions. This dye changes color once it reacts with metal ions, providing a clear visual endpoint for the titration process.

Features of xylenol orange as an endpoint indicator:
  • Color change: The color change usually indicates the endpoint of the titration; for xylenol orange, this transition is sharp and significant, making it a reliable choice.
  • pH sensitivity: The color change depends on the pH level, meaning precise pH control is essential for an accurate endpoint.
  • Compatibility: Suitable for various metal ions, especially effective with Zn虏鈦 , making it ideal for zinc-based back titrations.
In the exercise described, xylenol orange was used to determine the endpoint of the back titration with zinc ions, effectively identifying when all the excess EDTA had reacted. This ensures that the calculation regarding the amount of metal ions complexed initially is accurate, as the endpoint precisely marks the necessary titration completion point.

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Most popular questions from this chapter

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.

Purification by reprecipitation and predominant species of polyprotic acids. To measure oxygen isotopes in \(\mathrm{SO}_{4}^{2-}\) for geologic studies, \(\mathrm{SO}_{4}^{2-}\) was precipitated with excess \(\mathrm{Ba}^{2+}{ }^{24}\) In the presence of \(\mathrm{HNO}_{3}, \mathrm{BaSO}_{4}\) precipitate is contaminated by \(\mathrm{NO}_{3}^{-}\). The solid can be purified by washing, redissolving in the absence of \(\mathrm{HNO}_{3}\), and reprecipitating. For purification, \(30 \mathrm{mg}\) of \(\mathrm{BaSO}_{4}\) were dissolved in \(15 \mathrm{~mL}\) of \(0.05 \mathrm{M} \mathrm{DTPA}\) (Figure 11-4) in \(1 \mathrm{M} \mathrm{NaOH}\) with vigorous shaking at \(70^{\circ} \mathrm{C} . \mathrm{BaSO}_{4}\) was reprecipitated by adding \(10 \mathrm{M} \mathrm{HCl}\) dropwise to obtain pH \(3-4\) and allowing the mixture to stand for \(1 \mathrm{~h}\). The solid was washed twice by centrifuging, removing the liquid, and resuspending in deionized water. The molar ratio \(\mathrm{NO}_{3}^{-} / \mathrm{SO}_{4}^{2-}\) was reduced from \(0.25\) in the original precipitate to \(0.001\) after two cycles of dissolution and reprecipitation. What are the predominant species of sulfate and DTPA at \(\mathrm{pH} 14\) and \(\mathrm{pH} 3\) ? Explain why \(\mathrm{BaSO}_{4}\) dissolves in DTPA in \(1 \mathrm{M} \mathrm{NaOH}\) and then reprecipitates when the \(\mathrm{pH}\) is lowered to \(3-4\).

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) \(60.0 \mathrm{~mL}\)

Indirect EDTA determination of cesium. Cesium ion does not form a strong EDTA complex, but it can be analyzed by adding a known excess of \(\mathrm{NaBil}_{4}\) in cold concentrated acetic acid containing excess \(\mathrm{NaI}\). Solid \(\mathrm{Cs}_{3} \mathrm{Bi}_{2} \mathrm{I}_{9}\) is precipitated, filtered, and removed. The excess yellow \(\mathrm{BiI}_{4}^{-}\)is then titrated with EDTA. The end point occurs when the yellow color disappears. (Sodium thiosulfate is used in the reaction to prevent the liberated \(\mathrm{I}^{-}\)from being oxidized to yellow aqueous \(\mathrm{I}_{2}\) by \(\mathrm{O}_{2}\) from the air.) The precipitation is fairly selective for \(\mathrm{Cs}^{+}\). The ions \(\mathrm{Li}^{+}, \mathrm{Na}^{+}, \mathrm{K}^{+}\), and low concentrations of \(\mathrm{Rb}^{+}\)do not interfere, although \(\mathrm{Tl}^{+}\)does. Suppose that \(25.00 \mathrm{~mL}\) of unknown containing \(\mathrm{Cs}^{+}\)were treated with \(25.00 \mathrm{~mL}\) of \(0.08640\) \(\mathrm{M} \mathrm{NaBiI}_{4}\) and the unreacted \(\mathrm{BiI}_{4}^{-}\)required \(14.24 \mathrm{~mL}\) of \(0.0437 \mathrm{M}\) EDTA for complete titration. Find the concentration of \(\mathrm{Cs}^{+}\)in the unknown.

Pyrocatechol violet (Table 11-3) is to be used as a metal ion indicator in an EDTA titration. The procedure is as follows: 1\. Add a known excess of EDTA to the unknown metal ion. 2\. Adjust the \(\mathrm{pH}\) with a suitable buffer. 3\. Back-titrate the excess chelate with standard \(\mathrm{Al}^{3+}\). From the following available buffers, select the best buffer, and then state what color change will be observed at the end point. Explain your answer. (a) \(\mathrm{pH}\) 6-7 (b) \(\mathrm{pH} 7-8\) (c) \(\mathrm{pH}\) 8-9 (d) \(\mathrm{pH}\) 9-10

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