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
The answer ia to d to some up what they said.
Adding CuSO4 is adding a common ion (the Cu^2+) so the reaction will shift to the right producing more CuS. Adding Na2SO4 will do nothing. There is no common ion in Na2SO4 and there is nothing to react. So the answer is ........
The answer is D.
To determine the effect of adding CuSO4(aq) and Na2SO4(aq) to the reaction, we need to consider the common ions and the solubility of the product.
(a) If CuSO4(aq) is added, it will introduce additional Cu2+ ions into the solution. Since Cu2+ is already a reactant in the reaction, adding more Cu2+ ions will not have any significant effect on the reaction. Therefore, the answer is: (a) Nothing.
(b) If Na2SO4(aq) is added, it will introduce additional S2- ions into the solution. Since S2- is also a reactant in the reaction, adding more S2- ions will increase the concentration of the reactant and therefore favor the formation of the product. This means that adding Na2SO4(aq) will shift the equilibrium towards the formation of more CuS(s). Therefore, the answer is: (b) More product will form.
So, the correct answer is: (a) Nothing; (b) More product will form (Choice C).