Question

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

Short answer

In conclusion, this color change becomes visible at a pH of approximately 1.66.

Step-by-step solution

Relating pKa to Ka

First, we should express Ka in terms of the given pKa. The relationship between pKa and Ka is given by: \[ \text{pKa} = -\log(\text{Ka}) \] Given that pKa is 3.00, let's solve for Ka: \[ \text{Ka} = 10^{-\text{pKa}} = 10^{-3.00} = 1 \times 10^{-3} \]

Set up equilibrium expression

Let's consider the equilibrium process for the dissociation of the indicator HIn: \[ \text{HIn} \leftrightharpoons \text{In}^{-} + \text{H}^{+} \] As we know, the acid dissociation constant, Ka, is given by the ratio of the product concentrations to the reactant concentration at equilibrium: \[ \text{Ka} = \frac{[\text{In}^-][\text{H}^+]}{[\text{HIn}]} \]

Find the relationship between concentrations

We are given that 7% of the indicator has been converted to the In- form. Therefore, we have the following relationship between the concentrations: - For \(\text{HIn}\ + 0.07\text{HIn} = 1\text{HIn}\) (Initial concentrations) - Conversion: \(0.07\text{HIn} \rightarrow 0.07\text{In}^{-} + 0.07\text{H}^{+}\) - Equilibrium concentrations: \(0.93\text{HIn},\ 0.07\text{In}^{-}, \text{ and }\ 0.07\text{H}^{+}\)

Use Ka expression to find pH

Now, plug these concentrations into the Ka expression and solve for the concentration of \(\text{H}^{+}\) ions: \[ \text{Ka} = \frac{[\text{In}^-][\text{H}^+]}{[\text{HIn}]} \Rightarrow 1 \times 10^{-3} = \frac{0.07[\text{H}^+]}{0.93} \] Now, solving for \([\text{H}^{+}]\) yields: \[ [\text{H}^{+}] = \frac{0.93 \times 10^{-3}}{0.07} \] Then, we can find the pH using the following relationship: \[ \text{pH} = -\log([\text{H}^{+}]) \] Calculating the pH gives: \[ \text{pH} = -\log\left(\frac{0.93 \times 10^{-3}}{0.07}\right) \approx 1.66 \]

Provide the final answer

In conclusion, this color change becomes visible at a pH of approximately 1.66.

Key concepts

pKa Calculation

Understanding the concept of pKa is crucial for students studying chemistry, as it represents the strength of an acid in solution. The pKa value is derived from the acid dissociation constant (Ka), which is a quantitative measure of how well an acid releases its protons into solution. A lower pKa value indicates a stronger acid, as it more readily gives up its protons.

The calculation of pKa is very straightforward: it is the negative base-10 logarithm of the Ka value. Mathematically, this is represented as \[ \text{pKa} = -\log(\text{Ka}) \]. When given a pKa, you can find Ka by taking the antilogarithm or the power of 10 of the negative pKa. Specifically, \[ \text{Ka} = 10^{-\text{pKa}} \].

For example, with a pKa of 3.00, the Ka can be calculated by \[ \text{Ka} = 10^{-3.00} = 1 \times 10^{-3} \]. Knowing how to interconvert these two quantities is essential when analyzing chemical equilibrium, especially in acid-base reactions.

Chemical Equilibrium

Chemical equilibrium is a dynamic state where the rate of the forward chemical reaction equals the rate of the reverse reaction. At equilibrium, the concentrations of reactants and products remain constant but are not necessarily equal.

To quantify the position of equilibrium, we use the equilibrium constant (K). For acid-base reactions, we specifically talk about the acid dissociation constant (Ka), which is the equilibrium constant for the dissociation of an acid to its conjugate base and a proton. The general form of the equation is \[ \text{Ka} = \frac{[\text{Product 1}][\text{Product 2}]}{[\text{Reactant}]} \].

In the context of the exercise, the equilibrium expression for the dissociation of the indicator HIn is given by \[ \text{Ka} = \frac{[\text{In}^-][\text{H}^+]}{[\text{HIn}]} \]. By setting up this expression and knowing the Ka, we can solve for the unknown concentrations, assuming we have some initial concentration inputs, like the percentage of dissociation in this case.

pH Determination

Determining the pH of a solution is a fundamental aspect of chemistry, as it indicates the acidity or basicity of that solution. pH is a scale used to specify the hydrogen ion concentration (\( [\text{H}^+] \) ) in a solution. The pH scale ranges from 0 to 14, with 7 being neutral. A pH less than 7 indicates acidity, while a pH greater than 7 indicates alkalinity.

The pH can be calculated by taking the negative logarithm (base 10) of the hydrogen ion concentration: \[ \text{pH} = -\log([\text{H}^+]) \]. This formula shows the inverse logarithmic relationship between hydrogen ion concentration and pH.

For instance, if the concentration of \( [\text{H}^+] \) is determined to be \( \frac{0.93 \times 10^{-3}}{0.07} \) from the equilibrium calculation, then the pH is calculated as follows: \[ \text{pH} = -\log\left(\frac{0.93 \times 10^{-3}}{0.07}\right) \approx 1.66 \]. This shows that at the point where 7% of the indicator HIn has dissociated, the solution will have a pH of approximately 1.66, making it highly acidic. Thus, pH determination helps in understanding the extent of acid dissociation and visualizing the acid-base nature of a solution.