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Chapter 18: Q3TY (page 440)

0.10mMKMnO4has an absorbance maximum of 0.26 at 525 nm in a

1.000-cm cell. Find the molarabsorptiveand the concentration of a solution whose

absorbance is 0.52 at 525 nm in the same cell.

Short Answer

Expert verified

The value of molar absorptive is2600M-1cm-1

The value of solution absorbance is0.2mM

Step by step solution

01

Calculate molar absorptive:

Equation to calculate ε is,

ε=Abc →(1)

Given,

c=0.10mMb=1.000cmA=0.26

Substitute in equation (1),

ε=0.261.000cm0.10mM10-3M1mMε=2600M-1cm-1

The value of molar absorptive is 2600M-1cm-1

02

Calculate solution concentration:

Equation for concentration is c=Abc

Given,

b=1.000cmA=0.26ε=2600M-1cm-1

Substitute in above equation,

c=0.521.000cm2600M-1cm-1=0.0002M103mM1Mc=0.2mM

The value of solution absorbance is0.2mM

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Most popular questions from this chapter

Semi-xylenol orange is a yellow compound at pH 5.9but turns red when it reacts with Pb2+. A 2.025 - mLsample of semixylenol orange at pH 5.9was titrated with7.515×10-4MPb(NO3)2with the following results:

Total role="math" localid="1663647483742" μLPb2+added

Absorbance at 490nmwavelength

Total μLPb2+added

Absorbance at490nmwavelength

0.0

6.0

12.0

18.0

24.0

30.0

36.0

0.227

0.256

0.286

0.316

0.345

0.370

0.399

42.0

48.0

54.0

60.0

70.0

80.0


0.425

0.445

0.448

0.449

0.450

0.447


Make a graph of absorbance versus microliters of Pb2+added. Be sure to correct the absorbances for dilution. Corrected absorbance is what would be observed if the volume were not changed from its initial value of 2.025Ml. Assuming that the reaction of semi orange with Pb2+has a 1 : 1stoichiometry, find the molarity of semi-xylenol orange in the original solution.

Nitrite ionNO-2, is a preservative for bacon and other foods, but it is potentially carcinogenic. A spectrophotometric determination ofNO-2makes use of the following reactions

Here is an abbreviated procedure for the

determination:

1. To 50.0ml of unknown solution containing nitrite is added 1.00mL of sulfanilic acid solution.

2. After 10min , 2.00mL of 1 –a minonaphthalene solution and 1.00 mL of buffer are added.

3. After 15 min, the absorbance is read at 520 nm in a 5.00-cm cell.

The following solutions were analyzed:

A. 50.0 mL of food extract known to contain no nitrite (that is, a negligible amount); final absorbance =0.153.

B. 50.0mL of food extract suspected of containing nitrite; final absorbance \(=0.622\).

C. Same as B, but with 10.0μLof7.50×10−3MNaNO2added to the 50.0-mL sample; final absorbance =0.967.

(a) Calculate the molar absorptivity, of the colored product. Remember that a \(5.00\)-cm cell was used.

(b) How many micrograms ofNO-2were present in 50.0mL of food extract?

Gold nanoparticles (Figure 17-31) can be titrated with the oxidizing agent TCNQ in the presence of excess ofBr2-to oxidize Au(0)"toAuBr2-in deaerated toluene. Gold atoms in the interior of the particle are Au(0) . Gold atoms bound to C12H25S- (dodecanethiol) ligands on the surface of the particle are Au(I) and are not titrated.

The table gives the absorbance at $856 0.700 mL of 1.00×10-4MTCNQ+0.05M(C8H17)4N+Br-in toluene is titrated with gold nanoparticles (1.43 g/Lin toluene) from a microsyringe with a Teflon-coated needle. Absorbance in the table has already been corrected for dilution.

(a) Make a graph of absorbance versus volume of titrant and estimate the equivalence point. Calculate the Au(0) in 1.00 g of nanoparticles.

(b) From other analyses of similarly prepared nanoparticles, it is estimated that 25 % of the mass of the particle is dodecanethiol ligand. Calculate mmol of localid="1667559229564" C12H25Sin 1.00 g of nanoparticles.

(c) The Au(I) content of 1.00 g of nanoparticles should be 1.00 mass of Au(0) - mass of C12H25S. Calculate the micromoles of Au(I) in 1.00 g of nanoparticles and the mole ratio Au(I) :. In principle, this ratio should be 1: 1. The difference is most likely because C12H25Swas not measured for this specific nanoparticle preparation.

18-8: What is an absorption spectrum?

The absorbance of a 2.31×10-5M solution of a compound is 0.822 at a wavelength of 266nm in a 1.00-cm cell. Calculate the molar absorptivity at 266nm.
See all solutions

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