What Side of the Bromothymol Blue Equilibrium is Favored?

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The equilibrium of bromothymol blue favors the reactants when in its bottle, as indicated by its blue color, suggesting a low concentration of H+ ions. Bromothymol blue is a weak acid with a pKa around 7, meaning it can exist in both acidic and basic forms. The solution's low concentration affects the dissociation of the acid, resulting in a predominance of the undissociated form. The color change associated with bromothymol blue is due to the presence of its conjugate base, which differs in color from the acid. Understanding this equilibrium is crucial for its application in analytical chemistry.
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Homework Statement


Bromothymol blue is an acid-base indicator. When in solution, it forms the following equilibrium:
HInd(aq) ⇌ H+(aq) + Ind-(aq)
What side of the equilibrium is favored while the solution is in the bromothymol blue bottle?

Homework Equations



Keq = [(C)c(D)d]/[(A)a(B)b]

The Attempt at a Solution



I think the equilibrium favors the reactants because since it is blue, then bromothymol blue is a base so there is small amounts of H+ ions so the K expressions is a small number divided by a bigger number which results in a small number.

Can anyone confirm that my thinking is correct? Thanks!
 
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Yes. Bromothymol blue is a weak acid, with pKa around 7.
 
The equation you give is not the relevant one. The right one would also suggest answer.
 
Thank you, but why is Bromothymol Blue an acid even though it is blue?
 
Badgeray said:
Thank you, but why is Bromothymol Blue an acid even though it is blue?

You are mistaking several things at the same time. Being an acid doesn't mean its solutions have low pH. First, it is a weak acid, second, its solutions - as used in the analytical chemistry - have very low concentration.

What is important is that - as a weak acid - it dissociates, it has a conjugate base (which has a different color than the acid itself), and the ratio of concentrations of both forms depends on the solution pH.
 
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