an indicator HIn is blue and its anion In- is red, Ka of indicator is 0.0001 at what pH value will a solution containing HIn be blue?
To determine the pH value at which a solution containing HIn will be blue, we need to consider the equilibrium of the indicator HIn in water. The equilibrium can be represented as:
HIn ⇌ H+ + In-
In this reaction, HIn represents the blue protonated form of the indicator, and In- represents the red deprotonated form of the indicator.
The equilibrium constant, Ka, for this reaction is given as 0.0001. Ka is defined as the ratio of the concentration of the products (H+ and In-) to the concentration of the reactant (HIn) at equilibrium. Mathematically, it can be expressed as:
Ka = [H+][In-] / [HIn]
In general, when the ratio of [In-] to [HIn] is greater than 1, the solution will appear red, as the deprotonated form dominates. On the other hand, when the ratio of [HIn] to [In-] is greater than 1, the solution will appear blue, as the protonated form dominates.
Since we know the value of Ka, we can use it to calculate the pH at which the solution will be blue. At this pH, the ratio [HIn] / [In-] will be greater than 1.
Let x be the concentration of HIn at equilibrium. Since the initial concentration of HIn is usually much greater than that of H+ or In-, we can approximate the starting concentration of HIn as the total concentration of the indicator in the solution.
At equilibrium, the concentration of H+ will be equal to the concentration of In-. Therefore, the expression for Ka can be rewritten as:
Ka = x * x / (total concentration - x)
Where total concentration is the sum of [HIn] and [In-].
Simplifying this expression, we get:
Ka = x^2 / (total concentration - x)
Substituting the known value of Ka (0.0001), we can solve this equation to find the value of x, which represents the concentration of HIn at equilibrium.
After finding x, we can calculate the pH using the relation:
pH = -log[H+]
where [H+] is the concentration of H+ in the solution, which is equal to x.
Therefore, by solving for x and substituting its value in the pH equation, we can determine the pH value at which the solution containing HIn will appear blue.