Question

A certain indicator HIn has a \(\mathrm{p} K_{2}\) of \(3.00\) and a color change becomes visible when \(7.00 \%\) of the indicator has been converted to \(\mathrm{In}^{-}\). At what \(\mathrm{pH}\) is this color change visible?

Short answer

The pH at which the color change becomes visible is approximately 1.88.

Step-by-step solution

Relationship between pH and pOH

In any aqueous solution, the relationship between the acidity (pH) and basicity (pOH) can be expressed as: \(pH + pOH = 14\) For our problem, we will focus on the pH, which is representative of the acidity of the solution.

Calculate the relative amounts of HIn and In⁻ in the solution

We are given that 7.00% of the indicator has been converted to In⁻. If we denote the full amount of HIn as 1, then the amounts of HIn and In⁻ will be as follows: Amount of HIn = 0.93 (since 100% - 7% = 93%) Amount of In⁻ = 0.07 (from information given)

Derive the equilibrium constant from pK₂

We are given that the pK₂ of HIn is 3.00. The pK₂ represents the acidity constant for the second dissociation step. It is related to the equilibrium constant (K₂) by the equation: \[ pK_{2} = -\log K_{2} \] We can now calculate K₂: \[ K_{2} = 10^{-pK_{2}} = 10^{-3.00} = 1.00 \times 10^{-3} \]

Use the equilibrium expression to find the pH

The equilibrium expression for the dissociation of HIn to In⁻ is given by: \[ K_{2} = \frac{[\mathrm{In}^{-}] [\mathrm{H}^{+}]}{[\mathrm{HIn}]} \] Substituting K₂, the amounts of In⁻ and HIn, and denoting the concentration of H⁺ ions as [H⁺], we have: \[ 1.00 \times 10^{-3} = \frac{(0.07) ([\mathrm{H}^{+}])}{(0.93)} \] Now we can solve for the concentration of H⁺ ions: \[ [\mathrm{H}^{+}] = \frac{1.00 \times 10^{-3} \times 0.93}{0.07} \] \[ [\mathrm{H}^{+}] = 1.33 \times 10^{-2} \] Finally, we can convert the concentration of H⁺ ions to pH using the equation: \[ pH = -\log [\mathrm{H}^{+}] \] \[ pH = -\log (1.33 \times 10^{-2}) \] Therefore, the pH at which the color change becomes visible is approximately 1.88.

Key concepts

Acid-Base Equilibrium

Acid-base equilibrium is a fundamental concept in chemistry that looks at how acids and bases balance in a solution. In simple terms, it is the state where the concentrations of acids and their conjugate bases, or bases and their conjugate acids, remain constant over time. This equilibrium is dynamic, meaning that the conversion of acid to base and base to acid occurs continuously.

• In aqueous solutions, water itself plays a role, as it can both donate protons (H⁺) and accept them, acting as either an acid or a base.
• This exchange of protons is central to the behavior of acids and bases in a solution.

Understanding acid-base equilibrium helps predict the behavior of acids and bases, such as when and where an indicator like HIn will change color. This phenomenon is crucial for practical applications in chemistry, such as titrations or pH testing.

Dissociation Constant

The dissociation constant, often represented as K_a for acids, is a measure of the strength of an acid in solution. It quantifies how much an acid dissociates into its ions when dissolved.

• A larger dissociation constant indicates a stronger acid, which dissociates more in solution.
• Conversely, a smaller value suggests a weaker acid, which dissociates less.

In the provided problem, K₂ is the dissociation constant for the second ionization step of the indicator HIn. Knowing the dissociation constant helps us understand the equilibrium position of the reaction and predict the pH at which the indicator will visibly change color. Calculating it involves transforming the pK₂ value (3.00 in this case) using the formula: \[ K_{2} = 10^{-pK_{2}} \] which results in \( 1.00 \times 10^{-3} \).

Equilibrium Expression

The equilibrium expression is a mathematical representation of the balance in a chemical reaction at equilibrium. It relates the concentrations of the reactants and products in a balanced chemical equation.

• For acids, the expression is derived from their dissociation in water, providing insight into where the reaction equilibrium lies.
• It helps in determining unknown concentrations in the system, such as the amount of H⁺ needed to achieve a particular pH.

In our case with the indicator HIn, the equilibrium expression is given by:\[K_{2} = \frac{[\mathrm{In}^{-}] [\mathrm{H}^{+}]}{[\mathrm{HIn}]}\]This equation balances the dissociated form In^-, the concentration of hydrogen ions [H⁺], and the non-dissociated form HIn, allowing us to calculate specific values needed for solving pH-related problems.

pKa and pH Relationship

The relationship between pK_a and pH is crucial in understanding how acidic or basic a solution is. It provides a way to determine the pH at which an acid is half dissociated, known as the acid's strength.

• pK_a, which stands for the negative logarithm of the acid dissociation constant, helps gauge the power of an acid.
• The lower the pK_a, the stronger the acid, meaning it dissociates more completely in solution.

In practical terms, if the pH matches the pK_a, the acid and its conjugate base are in equal concentration. This knowledge is vital in scenarios such as determining the point at which a solution's pH will cause an indicator to change color. For example, at pH 1.88, as calculated in the problem, the balance is such that enough of the indicator has dissociated to make its color change visible, illustrating the interplay between pK₂ and pH.