consider the following chemical reaction of bromothymol blue indicator. it appears yellow in undissociated form and blue in its dissociated aqueous solution.
\\ce{hc_{2}h_{3}o_{2}(aq) \
ightleftharpoons h^{+}(aq) + c_{2}h_{3}o_{2}^{-}(aq)} \
\quad\quad\quad\quad\quad yellow \quad\quad\quad\quad\quad\quad\quad\quad blue

what will be the color of the solution if a large amount of $\ce{h_{2}co_{3}}$ is added?
$\bigcirc$ the solution will remain yellow.
$\bigcirc$ the solution will turn blue.
$\bigcirc$ the solution will turn pink.
$\bigcirc$ the solution will turn green.

Question

consider the following chemical reaction of bromothymol blue indicator. it appears yellow in undissociated form and blue in its dissociated aqueous solution.
\\ce{hc_{2}h_{3}o_{2}(aq) \
ightleftharpoons h^{+}(aq) + c_{2}h_{3}o_{2}^{-}(aq)} \
\quad\quad\quad\quad\quad yellow \quad\quad\quad\quad\quad\quad\quad\quad blue

what will be the color of the solution if a large amount of $\ce{h_{2}co_{3}}$ is added?
$\bigcirc$ the solution will remain yellow.
$\bigcirc$ the solution will turn blue.
$\bigcirc$ the solution will turn pink.
$\bigcirc$ the solution will turn green.

Accepted Answer

# Brief Explanations:
1. First, recall the properties of $ \ce{H2CO3} $: Carbonic acid ($ \ce{H2CO3} $) is an acid, and in aqueous solution, it dissociates to produce $ \ce{H+} $ ions ($ \ce{H2CO3(aq)
ightleftharpoons H+(aq) + HCO3-(aq)} $, and further dissociation of $ \ce{HCO3-} $ also produces $ \ce{H+} $).
2. Now, look at the given equilibrium for bromothymol blue: $ \ce{HC2H3O2(aq)
ightleftharpoons H+(aq) + C2H3O2-(aq)} $ (yellow form on the left, blue form on the right).
3. According to Le Chatelier's principle, when we add a large amount of $ \ce{H2CO3} $, the concentration of $ \ce{H+} $ ions in the solution increases significantly.
4. An increase in $ \ce{[H+]} $ will shift the equilibrium of the bromothymol blue reaction to the left (because the system tries to counteract the increase in $ \ce{H+} $ by consuming it, so the reverse reaction is favored).
5. Shifting the equilibrium to the left means more of the undissociated (yellow) form $ \ce{HC2H3O2} $ is present, so the solution will remain yellow (or become more yellow - dominated, but since it was likely in some equilibrium before, adding $ \ce{H+} $ pushes it left). The options of pink or green are not related to the given indicator's color change (bromothymol blue is yellow or blue, not pink/green), and turning blue would require the equilibrium to shift right (which would happen with a base, not an acid adding $ \ce{H+} $).

# Answer:
A. The solution will remain yellow.

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