2. The reversible reaction for the [tex]\(pH\)[/tex] indicator bromophenol blue is:

[tex]\[
\left( \text{C}_{19} \text{H}_{8} \text{Br}_{4} \text{O}_{5} \text{S} \right) \text{H}^{-} \leftrightarrow \left( \text{C}_{19} \text{H}_{8} \text{Br}_{4} \text{O}_{5} \text{S} \right)^{2-} + \text{H}^{+}
\][/tex]

Write the equilibrium expression for this reaction.



Answer :

Certainly! Let's go through the steps to write the equilibrium expression for the given reaction.

### Step 1: Identify the balanced chemical equation

The given reaction for the [tex]$pH$[/tex] indicator bromophenol blue is:
[tex]\[ (C_{19}H_8Br_4O_5S)H^{-} \leftrightarrow (C_{19}H_8Br_4O_5S)^{2-} + H^{+} \][/tex]

### Step 2: Write the general form of the equilibrium constant expression

The general formula for the equilibrium constant [tex]\(K_{eq}\)[/tex] for a reaction of the type:
[tex]\[ aA + bB \leftrightarrow cC + dD \][/tex]
is given by:
[tex]\[ K_{eq} = \frac{[C]^c [D]^d}{[A]^a [B]^b} \][/tex]

Where:
- [tex]\( [C] \)[/tex] and [tex]\( [D] \)[/tex] are the molar concentrations of the products,
- [tex]\( [A] \)[/tex] and [tex]\( [B] \)[/tex] are the molar concentrations of the reactants,
- [tex]\( a, b, c, d \)[/tex] are the stoichiometric coefficients of the reactants and products.

### Step 3: Apply the specific reaction to the general form

For the reaction:
[tex]\[ (C_{19}H_8Br_4O_5S)H^{-} \leftrightarrow (C_{19}H_8Br_4O_5S)^{2-} + H^{+} \][/tex]

The products are:
- [tex]\((C_{19}H_8Br_4O_5S)^{2-}\)[/tex]
- [tex]\(H^+\)[/tex]

The reactant is:
- [tex]\((C_{19}H_8Br_4O_5S)H^{-}\)[/tex]

### Step 4: Substitute the concentrations and coefficients into the equilibrium expression

In this reaction, the stoichiometric coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], [tex]\(c\)[/tex], and [tex]\(d\)[/tex] are all equal to 1, so the equilibrium constant expression becomes:
[tex]\[ K_{eq} = \frac{[(C_{19}H_8Br_4O_5S)^{2-}][H^+]}{[(C_{19}H_8Br_4O_5S)H^{-}]} \][/tex]

### Conclusion
The equilibrium expression for the given reaction is:
[tex]\[ K_{eq} = \frac{[(C_{19}H_8Br_4O_5S)^{2-}][H^+]}{[(C_{19}H_8Br_4O_5S)H^{-}]} \][/tex]