In a bromothymol blue CO2 indicator lab, a green aquatic plant is placed in two corked test tubes after the solution turns yellow from exhaled CO2; why does the tube kept in direct sunlight turn back to blue while the tube kept in the dark stays yellow? (A) Carbon dioxide was removed by photosynthesis. (B) Carbon dioxide was produced by photosynthesis. (C) Oxygen was removed by respiration. (D) Carbon dioxide was produced by respiration.

What the indicator color is really telling you

Bromothymol blue shows changes tied to dissolved $CO_2$ in water. When $CO_2$ dissolves, it forms carbonic acid, which makes the solution more acidic and turns the indicator yellow. With little $CO_2$, the solution is less acidic and the indicator appears blue.

What happened right after the student exhaled

Exhaling into each tube added a lot of $CO_2$ to the water, so this equilibrium shifted toward acid:

$$CO_2 + H_2O \rightleftharpoons H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$$

More $H^+$ means lower pH, so both tubes turned yellow.

Why the sunlight tube turned blue again

In direct sunlight, the aquatic plant carries out photosynthesis, which consumes $CO_2$:

$$6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2$$

As $CO_2$ is removed, the reaction above shifts left, $H^+$ decreases, pH rises, and bromothymol blue returns to blue. That matches choice A.

Why the dark tube stayed yellow

In the dark, photosynthesis stops, but the plant still respires, producing $CO_2$:

$$C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O$$

So $CO_2$ stays high (or increases), keeping the solution acidic and yellow.

Eliminating the other answer choices

• B is wrong because photosynthesis uses $CO_2$ rather than making it.
• C does not explain the color change because bromothymol blue is responding mainly to $CO_2$/pH, not oxygen removal.
• D is true for respiration in general, but it would keep the solution yellow, not turn it blue, so it cannot explain the sunlight result.