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Chapter 0: QDE (page 1)

The metal ion indicator xylenol orange (Table 12-3) is yellow at $pH 6 (位_{\max} = 439 nm)$ .The spectral changes that occur as ${VO}^{2 +}$ is added to the indicator atpH6 are shown here. The mole ratio ${VO}^{2 +}$ /xylenol orange at each point is

Suggest a sequence of chemical reactions to explain the spectral changes, especially the isosbestic points at457 and 528nm.
Absorption spectra for the reaction of xylenol orange with ${VO}^{2 +}$ atpH 6.0

Short Answer

Expert verified

A sequence of chemical reactions to explain the spectral changes, especially the isosbestic points at 457 and528nm is

$X ({VO}^{2 +}) + Y (xylenol orange) 鈫� XY XY + X 鈫� X_{2} Y$

Step by step solution

01

Define isosbestic point:

During the physical change or chemical change, the total absorbance sample will not change and it will have a specific wavelength or frequency is called isosbestic points.

02

Determine the sequence of chemical reactions from the given isosbestic points:

Hence, when ${VO}^{2 +}$ is added, corresponding to the traces1-9,the peak at 439nm decreases while one arises at 485nm with isosbestic points at457nm.

Observe the traces 10-16, the peak at 485 nm decreases while one arises at566 nm with isosbestic point at5258nm.

Thus, we write the reaction as,

role="math" localid="1663649615719" $X ({VO}^{2 +}) + Y (xylenol orange) 鈫� XY XY + X 鈫� X_{2} Y$

Therefore, the sequences of chemical reaction about the spectral changes are,

$X ({VO}^{2 +}) + Y (xylenol orange) 鈫� XY XY + X 鈫� X_{2} Y$

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Most popular questions from this chapter

For a solution of ${Ni}^{2 +}$ and ethylenediamine, the following

equilibrium constants apply atrole="math" localid="1663385095626" $20^{掳}$ c:

role="math" localid="1663386154645" ${Ni}^{2} + H_{2} {NCH}_{2} {CH}_{2} {NH}_{2} Ni {(cn)}^{2 +} \log K_{1} = 7.25 Eilhylenediamine (abbreviated en) Ni {(en)}^{2} + cn Ni {(en)}_{2}^{2 +} \log K_{2} = 6.32 Ni {(en)}^{2} + cn Ni {(en)}_{3}^{2 +} \log K_{3} = 4.49$

Calculate the concentration of free role="math" localid="1663386355087" ${Ni}^{2 +}$ in a solution prepared by mixing

0.100 mol of en plus 1.00 mL of role="math" localid="1663386332852" $0.0100 {MNi}^{2 +}$ and diluting to 1.00 L with dilute base (which keeps all the enin its unprotonated form). Assume that nearly allnickel is in the form $Ni {(cn)}_{3}^{2 +} so [Ni {(cn)}_{3}^{2 +}] = 1.00 x 10^{- 5} M$ .Calculate the concentrations of

$Ni {(en)}^{2 +} and Ni {(en)}_{3}^{2 +}$ to verify that they are negligible incomparison with $Ni {(en)}_{3}^{2 +}$ .

(a) What is the difference between absorption and adsorption?

(b) How is an inclusion different from an occlusion?

What does the designation 200/400mesh mean on a bottle of chromatography stationary phase? What is the size range of such particles? Which particles are smaller,100/200or 200-400mesh?

25. Write a mass balance for a solution of $F e_{2} ({SO}_{4})_{3}$ if the species

are ${Fe}^{3 +}, Fe {(OH)}^{2 +}, Fe {(OH)}_{2}^{+}, {FeSO}_{4}^{+}, {SO}_{4}^{2 -}$ androle="math" localid="1663402048017" ${SO}_{4}^{2 -}$ .

To pre-concentrate cocaine and benzoylecgonine from river water described at the opening of this chapter, solid-phase extraction was carried out at $p H 2^{}$ using the mixed-mode cation-exchange resin in Figure 28-19. After passing $500 m L$ of river water through $60 m g$ of resin, the retained analytes were eluted first with $2 m L$ of $C H_{3} O H_{}$ and then with ammonia solution in. Explain the purpose of using for retention and dilute ammonia for elution.

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