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Consider the following chemical reaction of bromothymol blue indicator. It appears yellow in undissociated form and blue in its dissociated aqueous solution.

[tex]\[
\begin{array}{l}
HC_2H_3O_2(aq) \Leftrightarrow H^+(aq) + C_2H_3O_2^-(aq) \\
\text{yellow} \\
\text{blue} \\
\end{array}
\][/tex]

What will be the color of the solution if a large amount of [tex]\(H_2CO_3\)[/tex] is added?

A. The solution will remain yellow.
B. The solution will turn blue.
C. The solution will turn pink.
D. The solution will turn green.

Sagot :

GinnyAnsweridn
Let's analyze the given problem step-by-step based on the provided chemical equilibrium and the influence of adding [tex]\(H_2CO_3\)[/tex].

1. Understand the Chemical Equilibrium:
The chemical equilibrium provided involves acetic acid (HC[tex]\(_2\)[/tex]H[tex]\(_3\)[/tex]O[tex]\(_2\)[/tex]) dissociating in water:
[tex]\[ \text{HC}_2\text{H}_3\text{O}_2(aq) \Leftrightarrow \text{H}^+(aq) + \text{C}_2\text{H}_3\text{O}_2^-(aq) \][/tex]
- The undissociated form of acetic acid (HC[tex]\(_2\)[/tex]H[tex]\(_3\)[/tex]O[tex]\(_2\)[/tex]) is yellow.
- The dissociated form, with its ions [tex]\( \text{H}^+ \)[/tex] and [tex]\( \text{C}_2\text{H}_3\text{O}_2^- \)[/tex], appears blue.

2. Adding [tex]\( \text{H}_2\text{CO}_3 \)[/tex] (Carbonic Acid):
- When [tex]\( \text{H}_2\text{CO}_3 \)[/tex] is added to the solution, it dissociates to release [tex]\( \text{H}^+ \)[/tex] ions.
[tex]\[ \text{H}_2\text{CO}_3(aq) \rightarrow \text{H}^+(aq) + \text{HCO}_3^-(aq) \][/tex]
- The dissociation of carbonic acid increases the concentration of [tex]\( \text{H}^+ \)[/tex] ions in the solution.

3. Le Chatelier’s Principle:
- According to Le Chatelier's Principle, the system will respond to the increase in [tex]\( \text{H}^+ \)[/tex] ions by shifting the equilibrium position to the left in order to reduce the disturbance (excess [tex]\( \text{H}^+ \)[/tex] ions).
- This shift to the left results in the formation of more undissociated HC[tex]\(_2\)[/tex]H[tex]\(_3\)[/tex]O[tex]\(_2\)[/tex]:
[tex]\[ \text{HC}_2\text{H}_3\text{O}_2(aq) \leftarrow \text{H}^+(aq) + \text{C}_2\text{H}_3\text{O}_2^-(aq) \][/tex]

4. Color Change:
- More HC[tex]\(_2\)[/tex]H[tex]\(_3\)[/tex]O[tex]\(_2\)[/tex] being formed means that the solution will contain more of the yellow undissociated form, and less of the blue ionized form.
- Consequently, the solution will maintain or possibly increase in the yellow hue.

5. Conclusion:
The effect of adding [tex]\( \text{H}_2\text{CO}_3 \)[/tex] is to maintain the solution's yellow color as the equilibrium shifts to favor the formation of the yellow undissociated form of acetic acid.

Therefore, the correct answer is:

The solution will remain yellow.
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