indicate whether the following would increase decrease or have no effect on the solubility of copper (ii) carbonate (ksp - 2.5 x 10^-10) when compared to the solubility of water
1. dissolve it in an acidic solution
2. dissolve it in ammonia, forming a complex
3. dissolve it in Cu(CH3COO)2
4 answers
To clarify: the Ksp is not negative, it is 2.5 x 10^-10. the negative sign is unintentional
This is Le Chatelier's Principle. What do you not understand about. D it? Do you have any ideas what to do?
I believe the second would increase solubility because it forms a complex, but I do not know the pH of copper (ii) carbonate. Because I think the pH would effect solubility
Right b will increase it for the reason you cite.
Adding HCl certainly will increase the solubility.
CuCO3 ==> Cu^2+ + CO3^2-
You will not that adding HCl reacts with the CO3^2- to form H2CO3 which then decomposes to H2O and CO2. So decreasing the (CO3^2-) forces the equilibrium to the right and means more CuCO3 must dissolve to keep the CO3^2- increasing.
Dissolving in Cu(Ac) [Cu(CH3COO)2] will decreases the solubility because of the common ion effect.
CuCO3 ==> Cu^2+ + CO3^2-
Adding Cu(CH3COO)2 adds the Cu^2+ ion. The reaction will shift to the left so as to reduce the added copper ion and means more CuCO3 will precipitate.
Adding HCl certainly will increase the solubility.
CuCO3 ==> Cu^2+ + CO3^2-
You will not that adding HCl reacts with the CO3^2- to form H2CO3 which then decomposes to H2O and CO2. So decreasing the (CO3^2-) forces the equilibrium to the right and means more CuCO3 must dissolve to keep the CO3^2- increasing.
Dissolving in Cu(Ac) [Cu(CH3COO)2] will decreases the solubility because of the common ion effect.
CuCO3 ==> Cu^2+ + CO3^2-
Adding Cu(CH3COO)2 adds the Cu^2+ ion. The reaction will shift to the left so as to reduce the added copper ion and means more CuCO3 will precipitate.
1 answer