A solution of Compound X has a transmittance of 15.8% at 475 nm in a 2.00 cm cel

Question

A solution of Compound X has a transmittance of 15.8% at 475 nm in a 2.00 cm cell. The molar absorptivity of Compound X at this wavelength is 5.47 times 10^4 L/mol middot cm. What is the concentration of compound X in the solution? 1.95 times 10^-6 M 3.92 times 10^-6 M 7.32 times 10^-6 M 7.84 times 10^-6 M At 450 and 575 nm, the indicator species HIn and In- have the tabulated molar absorptivities. An unknown solution of this indicator had measured absorbance of 0.120 at 450 nm and 0.040 at 575 nm in a 1.00-cm cell. Calculate the total concentration of the indicator. 1.03 times 10^-5 M 1.03 times 10^-6 M 3.30 times 10^-5 M 2.06 times 10^-5 M

Explanation / Answer

4.

% Transmittance = 15.8 %

Transmittance, T = 15.8 / 100 = 0.158

Absorbance, A = log10 1 / T

A = log10 (1 / 0.158) = 0.8

Absorbance, A = e L c

Molar absorptivity, e = 5.47 x 104 L /mol-cm

L = 2 cm

Concentration, c = A / (e L)

c = 0.8 / (5.47 x 104 * 2)

c = 7.32 x 10-6 M

5.

HIn --> In- + H+

Absorbance, A = e1 L1 c1 + e2 L2 c2

At 450 nm

A = 0.12 = 650 * 1 * c1 + 21500 * 1 * c2

650 c1 + 21500 c2 = 0.12 ... (1)

At 575 nm

A = 0.04 = 7200 * 1 * c1 + 975 * 1 * c2

7200 c1 + 975 c2 = 0.04 ... (2)

[Eqn (1) * 7200 - Eqn (2) * 650] gives,

(154800000 - 633750) c2 = 864 - 26

c2 = 5.4357 x 10-6 M

c1 = 4.82 x 10-6 M

Total concentration of indicator, c1 + c2 = 1.03 x 10-5 M

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