/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 Acid-base indicators mark the en... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Acid-base indicators mark the end point of titrations by "magically" turning a different color. Explain the "magic" behind acid-base indicators.

Short Answer

Expert verified
The "magic" behind acid-base indicators is due to the equilibrium between their protonated (HIn) and deprotonated (In-) forms, each displaying a distinct color. As the pH of the solution changes during titration, the ratio of HIn to In- also changes, causing a visible color change when their concentrations are equal. The exact pH at which this occurs depends on the indicator's pKa value, following the Henderson-Hasselbalch equation. Factors affecting the color change include the specific indicator chosen, its concentration, and the pH range of the titration.

Step by step solution

01

1. Acid-base indicators and their role in titration

Acid-base indicators are weak acids or bases that display different colors in their ionized and unionized forms. In titrations, they are used to determine the end point – the point at which the reaction between the analyte and the titrant is complete. At the end point, the indicator changes its color, signaling the completion of the reaction.
02

2. How color change in indicators occur

Acid-base indicators are composed of molecules that exist in equilibrium between their protonated (HIn) and deprotonated (In-) forms. Each form has a distinct color. When the pH of the solution changes during titration, the ratio of HIn to In- also changes, which eventually causes a visible color change. The equation representing this equilibrium is: \[HIn(aq) \rightleftharpoons H^+(aq) + In^-(aq)\] The color change occurs when the concentrations of the protonated HIn and deprotonated In- forms are equal. The exact pH at which this occurs depends on the indicator's pKa value, as given by the Henderson-Hasselbalch equation: \[\text{pH} = \text{p}K_a + \log_{10}\Big(\frac{[\text{In}^-]}{[\text{HIn}]}\Big)\]
03

3. Factors affecting the color change

The color change of an acid-base indicator depends on: 1. The specific indicator chosen: Different indicators have different pKa values and distinctive color changes that occur at different pH levels. 2. The concentration of the indicator: Higher concentrations can make the color change more intense and easier to see. 3. The pH range of the titration: An appropriate indicator should be chosen based on the pH range of the titration. For instance, if the titration is expected to have a pH range of 4-6, an indicator with a pKa value close to 5 should be chosen. In summary, the "magic" behind acid-base indicators lies in the equilibrium between their protonated and deprotonated forms along with their distinctive colors in different forms and pH-dependent color changes, which allow them to signal the end point of a titration.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider a solution formed by mixing \(50.0 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}, 30.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HOCl}, 25.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{NaOH}\). \(25.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\), and \(10.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{KOH}\). Calculate the \(\mathrm{pH}\) of this solution.

Mixing together solutions of acetic acid and sodium hydroxide can make a buffered solution. Explain. How does the amount of each solution added change the effectiveness of the buffer?

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 \(\mathrm{pH}=3.55\) ?

Calculate the \(\mathrm{pH}\) of a solution that is \(1.00 \mathrm{M} \mathrm{HNO}_{2}\) and \(1.00 \mathrm{M}\) \(\mathrm{NaNO}_{2}\)

Amino acids are the building blocks for all proteins in our bodies. A structure for the amino acid alanine is All amino acids have at least two functional groups with acidic or basic properties. In alanine, the carboxylic acid group has \(K_{\mathrm{a}}=4.5 \times 10^{-3}\) and the amino group has \(K_{\mathrm{b}}=7.4 \times 10^{-5} .\) Because of the two groups with acidic or basic properties, three different charged ions of alanine are possible when alanine is dissolved in water. Which of these ions would predominate in a solution with \(\left[\mathrm{H}^{+}\right]=1.0 M ?\) In a solution with \(\left[\mathrm{OH}^{-}\right]=\) \(1.0 M ?\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.