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defon
3 years ago
11

Suppose you measure the absorbance of a yellow dye solution in a 1.00 cm cuvette. The absorbance of the solution at 427 nm is 0.

20. If the molar absorptivity of yellow dye at 427 nm is 27400 M–1cm–1, what is the concentration of the solution?
Chemistry
1 answer:
Dimas [21]3 years ago
4 0

Answer:

The concentration of the solution, C=7.2992\times 10^{-6} M

Explanation:

The absorbance of a solution can be calculated by Beer-Lambert's law as:

A=\varepsilon Cl

Where,  

A is the absorbance  of the solution

ɛ is the molar absorption coefficient (L.mol^{-1}.cm^{-1})

C is the concentration (mol^{-1}.L^{-1})

l is the path length of the cell in which sample is taken (cm)

Given,

A = 0.20

ɛ = 27400 M^{-1}.cm^{-1}

l = 1 cm

Applying in the above formula for the calculation of concentration as:

A=\varepsilon Cl

0.20= 27400\times C\times 1

C = \frac{0.20}{27400\times 1} M

So , concentration is:

C=7.2992\times 10^{-6} M

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The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. A
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Answer:

Following are the answer to this question:

Explanation:

The value of pH solution is =5.17 So, the p^{OH}:

p^{OH}=14-56.17

      =8.823

The volume of the NH_{3} = 40.00 ml  

convert into the liter= 0.040L

The value of the concentrated NH_{3} =0.10 M

The volume of the NH_{4}Cl= 50.00 ml

convert into the liter= 0.050L

 The value of concentrated NH_{4}Cl= 0.10 M

The volume of the H_{2}So_{4}= 30 ml

convert into the liter= 0.030L  

The value of concentrated H_2So_4=0.05 M

Calculating total volume=(0.40+0.050+0.030)

                                       =0.120 L

calculating the new concentrated value of NH_3 = \frac{0.10\times 0.040}{0.120}= 0.33 \ M

calculating the new concentrated value of NH_4Cl= \frac{0.050\times 0.10}{0.120}= 0.04166 \ Mcalculating the new concentrated value of H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M when 1 mol H_2So_4 produced 2 mols H^{+} so, 0.0125 in H_2So_4produced:

=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}

create the ICE table:    

NH_3    \ \ \ \ \ \ \ \     + H^{+}  \ \ \ \ \ \ \longrightarrow NH_4^{+}                    

I (m)       0.033(m)            0.025                       0.04166

C            -0.025                 -0.025                       + 0.025  

E            8.3\times 10^{-3}     0                    0.0667

now calculating pH:

when ph= 8.83:

P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769

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