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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 \(\mathrm{In}^{-}\). At what \(\mathrm{pH}\) is this color change visible?

Short Answer

Expert verified
The color change becomes visible at a pH of approximately 2.54.

Step by step solution

01

Understand the given information and what needs to be found

We know that pKa = 3.00, and the color change becomes visible when 7.00% of the indicator has been converted to In-. We need to find the pH at which this color change occurs.
02

Write the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is given by: pH = pKa + log (\(\frac{[\mathrm{In}^-]}{[\mathrm{HIn}]}\))
03

Use the percentage of conversion to find the ratio of [In-] to [HIn].

Since the color change occurs when 7.00% of the indicator has been converted to In-, we can set up the following relationship: \(\frac{[\mathrm{In}^-]}{[\mathrm{HIn}] + [\mathrm{In}^-]} = 0.07\)
04

Solve the equation from step 3 for the ratio [\(\mathrm{In}^-\)]/[\(\mathrm{HIn}\)].

To solve for the ratio, we can rewrite the equation as follows: \([\mathrm{In}^-] = 0.07 * [\mathrm{HIn}] + 0.07 * [\mathrm{In}^-]\) Now, we can simplify the equation: \([\mathrm{In}^-](1 - 0.07) = 0.07 * [\mathrm{HIn}]\) Finally, dividing both sides by 0.07 and (1-0.07) to find the ratio: \(\frac{[\mathrm{In}^-]}{[\mathrm{HIn}]} = \frac{0.07}{1 - 0.07} = \frac{0.07}{0.93}\)
05

Plug the ratio into the Henderson-Hasselbalch equation and find the pH

Using the ratio from step 4, we can plug it into the Henderson-Hasselbalch equation: pH = pKa + log(\(\frac{[\mathrm{In}^-]}{[\mathrm{HIn}]}\)) = 3.00 + log(\(\frac{0.07}{0.93}\)) Now, we can calculate the value of the pH: pH = 3.00 + log(\(\frac{0.07}{0.93}\)) ≈ 2.54 The color change becomes visible at a pH of approximately 2.54.

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Key Concepts

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

Understanding pKa
The concept of pKa is fundamental in the study of chemistry, particularly when exploring acid-base properties. It represents the negative logarithm of the acid dissociation constant, designated as Ka, which measures the strength of an acid in a solution. The lower the pKa value, the stronger the acid, as it indicates a greater tendency for the acid to donate protons to the solution.

For instance, an acid with a pKa of 3.00, like the indicator HIn in the exercise, is considered a weak acid since it has a moderate tendency to donate protons. Additionally, understanding pKa aids in predicting the behavior of molecules in different pH environments, which is crucial for determining at what pH a color change of a pH indicator, like HIn, will occur.
Role of pH Indicators
pH indicators are substances that display different colors depending on the acidity or basicity of the environment they are in. These are especially useful in determining the pH of a solution. Each indicator has a specific pH range over which it changes color, known as the transition range. This is largely determined by the pKa value of the indicator.

In our example, the indicator HIn undergoes a visible color change when a certain fraction, in this case, 7%, has been converted to its deprotonated form, In-. Calculating at what pH this color change will be visible requires using the Henderson-Hasselbalch equation, taking into account the percentage of the indicator that has been converted. pH indicators are crucial in various applications, such as laboratory experiments, titrations, and even in medical diagnostics.
Acid-Base Titration and pH Change
Acid-base titration is an analytical procedure used to determine the concentration of an acid or base in a solution. It involves the gradual addition of an acid to a base, or vice versa, and the subsequent monitoring of pH changes that occur during the reaction.

As the titration progresses, the pH of the solution changes, which is often monitored using a pH indicator, like HIn. The indicator's distinct color change points to the equivalence point of the titration—the point at which the number of moles of acid equals the number of moles of base in the solution.

Understanding how the pH changes during a titration and knowing the pKa of the indicator used can be instrumental in accurately interpreting titration curves and determining the end point of a titration, which should ideally coincide with the equivalence point.

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Most popular questions from this chapter

What volumes of \(0.50 \mathrm{M} \mathrm{HNO}_{2}\) and \(0.50 \mathrm{M} \mathrm{NaNO}_{2}\) must be mixed to prepare \(1.00 \mathrm{~L}\) of a solution buffered at \(\mathrm{pH}=3.55\) ?

Malonic acid \(\left(\mathrm{HO}_{2} \mathrm{CCH}_{2} \mathrm{CO}_{2} \mathrm{H}\right)\) is a diprotic acid. In the titration of malonic acid with \(\mathrm{NaOH}\), stoichiometric points occur at \(\mathrm{pH}=3.9\) and \(8.8\). A \(25.00-\mathrm{mL}\) sample of malonic acid of unknown concentration is titrated with \(0.0984 \mathrm{M} \mathrm{NaOH}\), requiring \(31.50 \mathrm{~mL}\) of the \(\mathrm{NaOH}\) solution to reach the phenolphthalein end point. Calculate the concentration of the initial malonic acid solution. (See Exercise 113.)

A \(0.400-M\) solution of ammonia was titrated with hydrochloric acid to the equivalence point, where the total volume was 1.50 times the original volume. At what pH does the equivalence point occur?

Calculate the \(\mathrm{pH}\) of each of the following solutions. a. \(0.100 M\) propanoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{\mathrm{a}}=1.3 \times 10^{-5}\right)\) b. \(0.100 M\) sodium propanoate \(\left(\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\right)\) c. pure \(\mathrm{H}_{2} \mathrm{O}\) d. a mixture containing \(0.100 \mathrm{M} \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\) and \(0.100 \mathrm{M}\) \(\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\)

Calculate the \(\mathrm{pH}\) at the halfway point and at the equivalence point for each of the following titrations. a. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) titrated by \(0.10 \mathrm{M} \mathrm{NaOH}\) b. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)\) titrated by \(0.20 \mathrm{M} \mathrm{HNO}_{3}\) c. \(100.0 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{HCl}\) titrated by \(0.25 \mathrm{M} \mathrm{NaOH}\)

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