To write the equilibrium expression for the given reversible reaction involving bromophenol blue, we need to start by understanding the general form of an equilibrium constant expression. For a reaction of the form:
[tex]\[ aA + bB \leftrightarrow cC + dD \][/tex]
The equilibrium expression (law of mass action) is given by:
[tex]\[ K_{eq} = \frac{[C]^c[D]^d}{[A]^a[B]^b} \][/tex]
where:
- [tex]\( K_{eq} \)[/tex] is the equilibrium constant.
- Square brackets [tex]\([ ]\)[/tex] denote the concentration of the species at equilibrium.
- The letters [tex]\( A, B, C, \)[/tex] and [tex]\( D \)[/tex] represent the chemical species.
- The lowercase letters [tex]\( a, b, c, \)[/tex] and [tex]\( d \)[/tex] are their respective stoichiometric coefficients.
Now, let's apply this to the given reaction for bromophenol blue:
[tex]\[ \left( C_{19}H_8Br_4O_5S \right)H^{-} \leftrightarrow \left( C_{19}H_8Br_4O_5S \right)^{2-} + H^{+} \][/tex]
Here, the reaction shows the dissociation of [tex]\(\left( C_{19}H_8Br_4O_5S \right)H^{-}\)[/tex] into [tex]\(\left( C_{19}H_8Br_4O_5S \right)^{2-}\)[/tex] and [tex]\( H^{+} \)[/tex].
- [tex]\( A \)[/tex] is [tex]\(\left( C_{19}H_8Br_4O_5S \right)H^{-} \)[/tex]
- [tex]\( B \)[/tex] is not present on the reactant side
- [tex]\( C \)[/tex] is [tex]\(\left( C_{19}H_8Br_4O_5S \right)^{2-} \)[/tex]
- [tex]\( D \)[/tex] is [tex]\( H^{+} \)[/tex]
The stoichiometric coefficients [tex]\( a, b, c, \)[/tex] and [tex]\( d \)[/tex] are all [tex]\( 1 \)[/tex].
So, the equilibrium expression becomes:
[tex]\[ K_{eq} = \frac{[\left( C_{19}H_8Br_4O_5S \right)^{2-}][H^{+}]}{[\left( C_{19}H_8Br_4O_5S \right)H^{-}]} \][/tex]
Therefore, the equilibrium expression for the reversible reaction of the pH indicator bromophenol blue is:
[tex]\[ K_{eq} = \frac{[\left( C_{19}H_8Br_4O_5S \right)^{2-}][H^{+}]}{[\left( C_{19}H_8Br_4O_5S \right)H^{-}]} \][/tex]
