2. For the reaction, 2CrO\" + 2H+ Cr207-2 + H20, the equilibrium constant is 4.2

Question

2. For the reaction, 2CrO" + 2H+ Cr207-2 + H20, the equilibrium constant is 4.2 x 1014, Molar absorptivities for the chromium species are: E1 (Cr04-2) (M.1 cm-1) 1.84 x 10 4.81 x 103 1.88 x 103 Ez (Cr20,-) (M-1 cm -1) 10.7 x 10 7.28 x 102 1.89 x 102 a (nm) 345 370 400 A solution was prepared from solid potassium dichromate, K2Cr20, using pH buffer 5.60 as solvent. The concentration was 4.00 x 104 M in dichromate, before equilibrium. Assuming absorbances are additive, calculate the theoretical absorbance, in a 1.00 cm cell, for each solution at each wavelength.

Explanation / Answer

The equilibrium constant for the reaction is given as 4.2*1014. The pH of the solution is 5.60, i.e, pH = -log [H+] = 5.60; therefore, [H+] = antilog (-pH) = antilog (-5.60) = 2.51*10-6 M.

Write down the expression for the equilibrium constant for the reaction as below.

K = [Cr2O72-]/[CrO42-]2[H+]2

The equilibrium concentration of dichromate, Cr2O72- is 4.00*10-4 M; find out the equilibrium concentration of CrO42- at pH 5.60.

4.2*1014 = (4.00*10-4)/[CrO42-]2(2.51*10-6)2

====> [CrO42-]2 = (4.00*10-4)/(4.2*1014).(2.51*10-6)2 = 1.51*10-7

====> [CrO42-] = 3.88*10-4

The equilibrium concentration of CrO42- is 3.88*10-4 M.

Take the first entry as an example. The absorbances are additive; hence write

A1 = [(1.84*103 M-1cm-1)*(3.88*10-4 M) + (10.7*102 M-1cm-1)*(4.00*10-4 M)]*(1.00 cm) = 1.14192 1.14 (ans).

Next, take the second entry.

A2 = [(4.81*103 M-1cm-1)*(3.88*10-4 M) + (7.28*102 M-1cm-1)*(4.00*10-4 M)]*(1.00 cm) = 2.15748 2.16 (ans).

Finally take the third entry.

A3 = [(1.88*103 M-1cm-1)*(3.88*10-4 M) + (1.89*102 M-1cm-1)*(4.00*10-4 M)]*(1.00 cm) = 0.80504 0.80 (ans).

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