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The ionization constant \(K_{\mathrm{a}}\) of an indicator HIn is \(1.0 \times 10^{-6} .\) The color of the nonionized form is red and that of the ionized form is yellow. What is the color of this indicator in a solution whose \(\mathrm{pH}\) is \(4.00 ?\)

Short Answer

Expert verified
The color of the indicator is red in a solution with pH 4.00.

Step by step solution

01

Write the Ionization Equation

For the weak acid indicator HIn, the ionization can be represented as:\[\text{HIn} \rightleftharpoons \text{H}^+ + \text{In}^-\]where \(\text{HIn}\) is the nonionized (red) form and \(\text{In}^-\) is the ionized (yellow) form.
02

Express the Ionization Constant

The ionization constant \(K_a\) for the indicator HIn is given by:\[K_a = \frac{[\text{H}^+][\text{In}^-]}{[\text{HIn}]}\]Given that \(K_a = 1.0 \times 10^{-6}\).
03

Calculate the [H⁺] in the Solution

Since the pH of the solution is 4.00, use the relation:\[pH = -\log[\text{H}^+]\]\[4.00 = -\log[\text{H}^+]\]Solving for \([\text{H}^+]\):\[[\text{H}^+] = 10^{-4} \text{ M}\]
04

Determine the Dominant Form of the Indicator

We need to find the ratio of \([\text{In}^-]\) to \([\text{HIn}]\), using \(\[ \text{H}^+\] \):\[ \frac{[\text{In}^-]}{[\text{HIn}]} = \frac{K_a}{[\text{H}^+]} = \frac{1.0 \times 10^{-6}}{10^{-4}} = 0.01\]This indicates that the concentration of \([\text{HIn}]\) is 100 times that of \([\text{In}^-]\), so the non-ionized (red) form is dominant.
05

Conclude the Color of the Solution

Since the non-ionized (red) form is present in higher concentration compared to the ionized (yellow) form, the indicator will appear red in the solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ionization Constant
The ionization constant, often denoted as \(K_a\) for acids, is an essential concept in understanding how indicators work in chemistry. This constant provides insight into the degree of ionization of a weak acid in solution. In the context of acid-base indicators, the ionization constant tells us how readily the indicator dissociates into its ionized and non-ionized forms. For our specific indicator HIn, the given \(K_a\) is \(1.0 \times 10^{-6}\), indicating a relatively low level of ionization. This means that the equilibria strongly favor the non-ionized form (in this case, red) over the ionized form (yellow), unless perturbed by a pH change. Understanding \(K_a\) helps in predicting the behavior of indicators in different pH solutions, crucial for determining the color change point.
pH Calculations
pH calculations are crucial for determining the color of an acid-base indicator in a solution. The pH of a solution is a measure of its acidity or basicity, and it affects the equilibrium of the indicator between its ionized and non-ionized forms.
  • The pH can be calculated using the formula: \(pH = -\log[\text{H}^+]\).
  • In our situation, a solution with a pH of 4.00 corresponds to a hydronium ion concentration of \([\text{H}^+] = 10^{-4} \text{ M}\).
Knowing the pH allows us to calculate the proportions of each form of the indicator present. It tells us how much of each form we have at equilibrium, which directly informs us about the color the solution will display. In the exercise provided, the calculation helps us to see that at pH 4, the majority form is non-ionized, resulting in the red color.
Color Change in Indicators
The color of an acid-base indicator arises from its chemical structure. These structures often change with differing pH, leading to changes in color. A common feature of many indicators is their ability to exist in two different forms: a non-ionized form and an ionized form.
  • At low pH, indicators tend to be in their non-ionized form, while high pH favors the ionized form.
  • For the indicator HIn, red represents the non-ionized form, and yellow represents the ionized form.
In the context of this exercise, the indicator exhibits a reddish color at pH 4 because the concentration of the non-ionized form (red) is significantly higher than that of the ionized form (yellow). When \(\text{HIn}\) outweighs \(\text{In}^-\) by a factor of 100, the red color is dominant. This demonstrates the sensitivity of the indicator to changes in the pH environment, making it a useful tool for detecting changes in acidity or basicity in different solutions.

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Most popular questions from this chapter

Phenolphthalein is the common indicator for the titration of a strong acid with a strong base. (a) If the \(\mathrm{p} K_{\mathrm{a}}\) of phenolphthalein is \(9.10,\) what is the ratio of the nonionized form of the indicator (colorless) to the ionized form (reddish pink) at pH \(8.00 ?\) (b) If 2 drops of \(0.060 M\) phenolphthalein are used in a titration involving a \(50.0-\mathrm{mL}\) volume, what is the concentration of the ionized form at \(\mathrm{pH} 8.00 ?\) (Assume that 1 drop \(=\) \(0.050 \mathrm{~mL}\).)

Using only a pH meter, water, and a graduated cylinder, how would you distinguish between an acid solution and a buffer solution at the same \(\mathrm{pH}\) ?

The \(\mathrm{pH}\) of a sodium acetate-acetic acid buffer is \(4.50 .\) Calculate the ratio \(\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right] /\left[\mathrm{CH}_{3} \mathrm{COOH}\right] .\)

A student mixes \(50.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) with \(86.4 \mathrm{~mL}\) of \(0.494 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4} .\) Calculate the mass of \(\mathrm{BaSO}_{4}\) formed and the \(\mathrm{pH}\) of the mixed solution.

The molar solubility of \(\mathrm{AgCl}\) in \(6.5 \times 10^{-3} \mathrm{M} \mathrm{AgNO}_{3}\) is \(2.5 \times 10^{-8} M\). In deriving \(K_{\mathrm{sp}}\) from these data, which of the following assumptions are reasonable? (a) \(K_{\mathrm{sp}}\) is the same as solubility. (b) \(K_{\mathrm{sp}}\) of \(\mathrm{AgCl}\) is the same in \(6.5 \times 10^{-3} \mathrm{M} \mathrm{AgNO}_{3}\) as in pure water. (c) Solubility of \(\mathrm{AgCl}\) is independent of the concentration of \(\mathrm{AgNO}_{3}\). (d) \(\left[\mathrm{Ag}^{+}\right]\) in solution does not change significantly upon the addition of \(\mathrm{AgCl}\) to \(6.5 \times 10^{-3} \mathrm{M} \mathrm{AgNO}_{3}\). (e) \(\left[\mathrm{Ag}^{+}\right]\) in solution after the addition of \(\mathrm{AgCl}\) to \(6.5 \times 10^{-3} \mathrm{M}\) \(\mathrm{AgNO}_{3}\) is the same as it would be in pure water.

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