A student chemist named Hannah Montana set up an electrical conductivity tester similar to that shown on page 141 and conducted an experiment involving two clear, colorless liquids, identified as solution A and solution B. Solution A and Solution B are both 0.050 M. The experiment is described in the steps listed below.


1. When 50.0 mL of Solution A were placed into a beaker and tested, the bulb glowed brightly. Likewise, when Solution B was placed into a different beaker and tested, the bulb also glowed – although slightly less brightly.

2. When Solution B was added to the beaker containing Solution A, a little at a time from a buret (see page 165), the bulb began to dim a little and the mixture of Solution A and Solution B in the beaker became whitish and cloudy. As more and more Solution B was added, the more the bulb dimmed and the cloudier the mixture became.

3. Eventually, when 50.0 mL of Solution B were added, the bulb did not glow at all.

4. When still more Solution B was added, the bulb began to brighten again.

Possible identities of Solution A and Solution B are shown in the table below.
Explain clearly, with appropriate detail, each observation described in the 4 steps. In your answer, you should name each of the two solutions (selected from the list below) and identify the substance causing the white cloudiness in step 2. Table 4.1 on page 147 should be of some assistance to you in identifying this white substance causing the cloudiness.

Solution A Solution B
acetic acid magnesium hydroxide potassium chloride ammonium hydroxide
sulfuric acid barium hydroxide
nitric acid zinc hydroxide
ammonium nitrate ammonium sulfate
hydrochloric acid potassium hydroxide

I'm not about to write a thesis for you but I can get you started. Here is what has happened. ONe of the solutions probably is H2SO4 and the other probably is Ba(OH)2. Both are strong electrolytes but since Ba(OH)2 is not as soluble it probably is responsible for not glowing as brightly as the other one. As B is added to A, the reaction produces BaSO4, a white insoluble precipitate. Since it doesn't dissolve very much (only a few milligrams in a liter), the bulb glows less brightly because fewer and fewer ions are available to conduct the electricity. At the equivalence point, one as BaSO4 and H2O, neither of which is a good electrolyte; therefore, that is essentially no conduction. I assume you can take it from here.

To identify the identity of Solution A and Solution B, let's analyze each of the observations described in the four steps:

1. When 50.0 mL of Solution A were placed into a beaker and tested, the bulb glowed brightly. Likewise, when Solution B was placed into a different beaker and tested, the bulb also glowed, although slightly less brightly.

This observation suggests that both Solution A and Solution B are capable of conducting electricity. Since electrical conductivity occurs when ions are present in a solution, we can eliminate acetic acid, ammonium nitrate, and ammonium sulfate from the list of possible solutions since they do not dissociate into ions when dissolved in water. This leaves us with the following potential solutions that can conduct electricity: sulfuric acid, nitric acid, hydrochloric acid, and potassium hydroxide.

2. When Solution B was added to the beaker containing Solution A, a little at a time, the bulb began to dim, and the mixture of Solution A and Solution B in the beaker became whitish and cloudy. As more Solution B was added, the bulb dimmed further, and the mixture became cloudier.

This cloudiness indicates the formation of a precipitate when Solution B was added to Solution A. Based on Table 4.1 on page 147, which lists common ionic compounds and their solubilities, we can determine that the precipitate formed is likely insoluble in water. The only compound from the list of possible solutions that matches this requirement is barium hydroxide (Ba(OH)2).

3. Eventually, when 50.0 mL of Solution B were added, the bulb did not glow at all.

At this point, we have reached the limit of conductivity, indicating that the solution is saturated with the precipitate formed. This tells us that Solution A must contain a solution that reacts with barium hydroxide to form a precipitate, inhibiting the electrical conductivity. Since barium hydroxide is a strong base, a strong acid is required to react with it and form a precipitate. Among the remaining options from step 1 (sulfuric acid, nitric acid, and hydrochloric acid), sulfuric acid (H2SO4) is the likely candidate. Therefore, Solution A is sulfuric acid (H2SO4).

4. When still more Solution B was added, the bulb began to brighten again.

The fact that the bulb begins to brighten again indicates that the excess Solution B is neutralizing the acid present in Solution A. From the options left in step 1 (nitric acid and hydrochloric acid), nitric acid (HNO3) is known to create a bright glow in the conductivity tester. Therefore, Solution B is nitric acid (HNO3).

In summary:

Solution A: Sulfuric acid (H2SO4)
Solution B: Nitric acid (HNO3)

The substance causing the white cloudiness in step 2 is the precipitate formed from the reaction between barium hydroxide (Ba(OH)2) in Solution B and sulfuric acid (H2SO4) in Solution A.