Answer to Question #128058 in General Chemistry for Rich

Question

A 10 sample of 0.150 M Cu(NO3)2 is mixed with 10.0 mL of 0.100 M NH3 and 80 mL of water and the solution is allowed to reach equilibrium. The solution at equilibrium has an absorbance of 0.455 at 600 nm. calculate Kc for the reaction between Cu2+ and NH3. Show your work very clearly to receive credit.

Answer 1

Consider the reaction between Cu²⁺ and NH₃:

Cu²⁺ + 4NH₃ => Cu(NH₃)₄²⁺

K_c = [Cu(NH₃)₄²⁺]/[Cu²⁺][NH₃]⁴, where the [X] -- the equilibrium molar concentration of the ion or molecule X

[Cu(NH₃)₄²⁺] is supposed to be found according to the Beer-Lambert law:

[Cu(NH₃)₄²⁺] = $A / (\epsilon * l)$ , where A -- absorbance, $\epsilon$ -- molar attenuation coefficient, $l$ -- optical path length.

From the data given in the task we know the absorbance (0.455), but don't know the molar attenuation coefficient and the optical path length.

[Cu²⁺] = 0.015 M, because the initial solution was diluted 10 times from 10 ml to 100 ml.

[NH₃] = 0.01 M for the same reason.

Answer: the final formula is:

$K_c = 0.455 / (\epsilon * l * 0.015 * 0.01^4)$

The K_c can be easily calculated using this formula, knowing the molar attenuation coefficient (can be calculated experimentally, or found in the corresponding books or tables); the optical path length (the width of the solution layer when the spectre was recorded).