Given the reaction:
Cu2+(aq) + S2-(aq) CuS(s)
What will happen (a) if CuSO4(aq) is added? And (b) if Na2SO4(aq) is added?
A. (a) Nothing; (b) Less product will form
B. (a) Less product will form; (b) Nothing
C. (a) Nothing; (b) More product will form
D. (a) More product will form; (b) Nothing
I answered this for you yesterday.
To determine the effect of adding CuSO4(aq) and Na2SO4(aq) on the given reaction, we need to consider the possible formation of a precipitate and the solubility of the compounds involved.
(a) If CuSO4(aq) is added:
CuSO4(aq) contains Cu2+ ions, which are already present in the reaction. Adding more Cu2+ ions will not affect the equilibrium of the reaction because the concentration of Cu2+ ions does not change. Therefore, no additional product (CuS) will be formed, and the answer is (a) Nothing.
(b) If Na2SO4(aq) is added:
Na2SO4(aq) contains S2- ions, which react with Cu2+ ions to form the product CuS(s). Adding more S2- ions from Na2SO4(aq) will shift the equilibrium of the reaction towards the formation of more CuS(s). Therefore, more product (CuS) will be formed, and the answer is (c) (a) Nothing; (b) More product will form.
Therefore, the correct answer is (c) (a) Nothing; (b) More product will form.