The following equilibrium was studied spectrophotometrically: A+B <--> C where C

Question

The following equilibrium was studied spectrophotometrically: A+B <--> C

where C absorbs light at 455nm. one reaction mixture was formed by mixing 25.0 mL of .10 M of A with 25mL of .10 M B and allowed to equilibriate. the absorbance of the solution was found to be .5528. Based on the calibration curve of Y=26.1x + .081 (where absorbance is on y axis and concentration of product C is on the x axis) determine the equilibrium concentration of C, then using an ICE table format determine the valu of K.

Explanation / Answer

Given, absorbance = 0.5528

Now, using calibration curve of Y = 26.1x + 0.081

Substituting the value of Y = 0.5528,

0.5528 = 26.1[C] + 0.081

We will get the concentration of [C] = 0.0180 M

Now, while combining the reactant solutions, the concentration will be reduced to 0.050 M. The 0.0180 M of C needs the reaction of 0.0180 M of reactant A and 0.0180 M reactant B. Therefore, the equilibrium concentration of reactant A and reactant B will be twice the concentration of C

Equilibrium concentration of A = 0.032 M
Equilibrium concentration of B = 0.032 M
Equilibrium concentration of C = 0.018 M
Now, K = [C]/[A][B] = 0.018 / (0.032 x 0.032) = 17.6

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