Imagine you have a light bulb and battery conductivity tester. You are given a dilute solution of copper sulfate and a dilute solution of vinegar, acetic acid. Coppersulfate is an ionic compound. You may recall acetic acid reacts with water to produce ions in solution according to the equation:

Question

H2O+CH3COOH---><---CH3COO-+H3O+
so here is the questions:
#1. What would be the result if you tested each of these solutions for conductivity?
#2. Suppose you added more copper sulfate and acetic acid to the solutions and retested. Would that change the results? Explain.
And #3. If you continued to add copper sulfate abd acetic acid until the solutions held as much solute as possible what would be the result? Expalin.
Please help me this question has totally got me!

Your question isn't exactly clear; however, I can get you started.
CuSO4 is an ionic compound; therefore, the solution, even a dilute one, will contain Cu(II) ions and Sulfate ions. That will conduct electricity. Adding more CuSO4(solid--that is the part that isn't clear) will produce more ions and that will increase the conductivity. And the conductivity will increase until the solutuion is saturated with copper(II) sulfate.

Acetic acid is a weak electrolyte. It will have some ions but not as many as CuSO4; therefore, the conducitivy of the dilute solution of acetic acid will be weak(er) than the CuSO4. Adding more (pure acetic acid--again this is the part that isn't clear) to the solution may increase the conductivity somewhat but only slightly. Changing the concentration of acetic acid from 0.001 M to 0.1M (a hundred fold) changes the conductivity by only 10 times. Changing to 10M (another 100 fold) changes the conductivity by only another 10 times. Adding more pure acetic acid will come to the point that no increase in conductivity is noticed because when all of the water molecules have been used up to ionize the pure acetic acid, no more ions will be formed and no more conductivity will be attained.