Cu\(^{2+}\)(aq) + 4NH\(_3\)(g) ↔ [Cu(NH\(_3\))\(_4\)]\(^{2+}\)(aq)
In the reaction above, what is the effect of precipitating Cu\(^{2+}\)(aq) as CuS(s)?
NH3(g) concentration will decrease
more NH3(g)
The equilibrium will shift to the right
There will be no effect
Explanation
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Here is an explanation:
precipitating Cu^[2+] as CuS will reduce its conc thereby shiftin the equil.position to the left where more of NH3 will be generated.
option B is the correct answer
REF: check: lamlad, Page 172-15
When Copper is removed as a solid, the system will have to shift to the left to oppose the change caused by copper leaving the system [Le chatelier's principle].....Therfore, NH3 will be produced since copper is no longer in the system
precipitating Cu^[2+] as CuS will reduce its conc thereby shiftin the equil.position to the left where more of NH3 will be generated
Effect of removing Cu²⁺:
According to Le Châtelier’s Principle, if you remove a reactant (Cu²⁺), the equilibrium will shift to the left to produce more Cu²⁺, in order to counter the change.
So let’s go through the options:
A. NH₃(g) concentration will decrease → False, because the shift is to the left, NH₃ will actually be released, not consumed.
B. More NH₃(g) → True, because NH₃ is a reactant, and the shift to the left releases it.
C. The equilibrium will shift to the right → False, it shifts left.
D. There will be no effect → False, removing Cu²⁺ does affect equilibrium.
✅ Correct Answer: B. more NH₃(g)
Let me know if you want a visual or further explanation!
I think its D, bcos the copper ion present in CuS is still Cu2+, so i don't think there would be any effect.
