Advance Study Assignment: Relative Stabilities of Complex Ions and Precipitates

Question

Advance Study Assignment: Relative Stabilities of Complex Ions and Precipitates Prepared from

Solutions of Copper(II)

1. In testing the relative stabilities of Cu(II) species using a well plate, a student adds 6 drops 1 M NH3 to 6 drops 0.1 M Cu(NO3)2. He observes that a blue precipitate initially forms, but that in excess NH3 the precipitate dissolves and the solution turns blue. Addition of 6 drops 1 M NaOH to the dark-blue solution results in the formation of a blue precipitate.

What is the formula of Cu(II) species in the dark-blue solution?

What is the formula of the blue precipitate present after addition of 1 M NaOH?

Which species is more stable in equal concentrations of NH3 and OH, the one in Part a or the one in Part b?

= 5 × 1012 = 7 × 109

2. Given the following two reactions and their equilibrium constants:
Cu(H O) 2+ (aq) + 4 NH (aq) Cu(NH ) 2+ (aq) + 4 H O K

2433421 Cu(H O) 2+ (aq) + CO 2 CuCO (s) + 4 H O K

243322 a. Evaluate the equilibrium constant for the reaction (see discussion):

CuCO (s) + 4 NH (aq) Cu(NH ) 2+(aq) + CO 2 (aq) 33343

b. If 1 M NH3 were added to some solid CuCO3 in a test tube containing 1 M Na2CO3, what, if any- thing, would happen? Explain your reasoning.

Explanation / Answer

ANSWER:

Dear candidate you have two questions, but as per guidelines one question one time. So here we answer Q1.

a) the formulla for dark blue ppt is Cu(NH3)42+. It is formed when xcess of ammonia is added to Cu(NO3)2.

b) The formula of the blue ppt after addition of NaOH is Cu(OH)2.

c) In equal concentrations of OH- and NH3   Cu(OH)2 is more stable than Cu(NH3)42+.

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