Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A student carried out an acid-base titration by adding \(\mathrm{NaOH}\) solution from a burette to an Erlenmeyer flask containing an HCl solution and using phenolphthalein as the indicator. At the equivalence point, she observed a faint reddish-pink color. However, after a few minutes, the solution gradually turned colorless. What do you suppose happened?

Short Answer

Expert verified
Exposure to \(\mathrm{CO_2}\) from the air likely turned the solution colorless over time by forming carbonic acid.

Step by step solution

01

Understand the Problem

In this titration, the student added a base (\(\mathrm{NaOH}\)) to an acid (\(\mathrm{HCl}\)) to reach the equivalence point, where moles of acid are equal to moles of base. Phenolphthalein changes color around this point, changing from colorless to a faint pink as it progresses from acidic to basic.
02

Analyze the Equivalence Point

At the equivalence point in a strong acid-strong base titration, the solution is neutral, and initially, a faint reddish-pink color is observed. This indicates that the phenolphthalein has detected a slight excess of base, turning the solution slightly basic.
03

Consider the Change Over Time

Over time, the solution turned colorless, which suggests that the conditions in the flask returned to a less basic, or neutral-pH environment. This might happen if the solution was exposed to \(\mathrm{CO_2}\) from the air, forming carbonic acid and thus lowering the pH.

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 Vaia!

Key Concepts

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

Equivalence Point
In an acid-base titration, the equivalence point is a crucial concept. It is the stage in the titration process where the amount of acid is exactly neutralized by the base, meaning the moles of acid equal the moles of base.
The chemical equation for such a reaction involving hydrochloric acid (HCl) and sodium hydroxide (NaOH) is:
\[ \text{HCl} (aq) + \text{NaOH} (aq) \rightarrow \text{NaCl} (aq) + \text{H}_2\text{O} (l) \]At this point, the pH of the solution is typically around 7, which is considered neutral for strong acid-strong base titrations.
The equivalence point can be detected using indicators that change color at certain pH levels or via pH meters for greater accuracy. This is why the student observed a faint reddish-pink color at this stage in the experiment, as the indicator signaled that the equivalence had been nearly reached.
Phenolphthalein Indicator
Phenolphthalein is a popular indicator used in titrations. It is very sensitive to pH changes, which makes it an excellent choice for identifying the equivalence point in a titration.
  • In acidic solutions (pH < 7), phenolphthalein is colorless.
  • As the pH approaches neutrality, it begins to change, first to faint pink, and becomes magenta in increasingly basic solutions (pH > 8.2).
The color change occurs because phenolphthalein reacts with the ions in the solution, changing its chemical structure, which alters the way it absorbs and transmits light.
This change signals that the solution has shifted from acidic to slightly basic at the equivalence point, hence the observed faint pink color during the titration process.
Neutralization Reaction
During a titration, a neutralization reaction occurs between the acid and the base. This specific reaction involves hydrochloric acid (HCl) and sodium hydroxide (NaOH).
The primary goal is to determine the concentration of one of the two reactants, often the analyte. Adding NaOH to HCl results in sodium chloride (NaCl) and water (H\(_2\)O), both of which are neutral products.
This reaction is quantitatively predictable; for every mole of HCl, one mole of NaOH is needed to completely neutralize it. As the reaction progresses towards the equivalence point, the acid and base reactants diminish, and the solution contains only the neutral products.
pH Change
The pH change in a solution during a titration is an important observation. It helps in understanding when the equivalence point is reached and past.
  • Initially, in an acidic solution, the pH is low, below 7.
  • As the base is added, the pH gradually increases.
  • At the equivalence point, in a strong acid-strong base titration, the pH approaches a neutral value of 7.
After reaching the equivalence point, any further addition of the base makes the solution increasingly basic, causing a sharp rise in pH.
The reason the solution returned to a colorless state after some time is that the exposure to carbon dioxide (CO\(_2\)) in the air led to the formation of carbonic acid, which slightly increased the acidity, lowering the pH and turning the phenolphthalein colorless once again.

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

The \(\mathrm{pH}\) of a saturated solution of a metal hydroxide MOH is 9.68 . Calculate the \(K_{s p}\) for this compound.

What reagents would you employ to separate the following pairs of ions in solution: (a) \(\mathrm{Na}^{+}\) and \(\mathrm{Ba}^{2+}\) (b) \(\mathrm{K}^{+}\) and \(\mathrm{Pb}^{2+}\) (c) \(\mathrm{Zn}^{2+}\) and \(\mathrm{Hg}^{2+} ?\)

Water containing \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) ions is called hard water and is unsuitable for some household and industrial use because these ions react with soap to form insoluble salts, or curds. One way to remove the \(\mathrm{Ca}^{2+}\) ions from hard water is by adding washing soda \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot 10 \mathrm{H}_{2} \mathrm{O}\right)\). (a) The molar solubility of \(\mathrm{CaCO}_{3}\) is \(9.3 \times 10^{-5} \mathrm{M}\). What is its molar solubility in a \(0.050 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) solution? (b) Why are \(\mathrm{Mg}^{2+}\) ions not removed by this procedure? (c) The \(\mathrm{Mg}^{2+}\) ions are removed as \(\mathrm{Mg}(\mathrm{OH})_{2}\) by adding slaked lime \(\left[\mathrm{Ca}(\mathrm{OH})_{2}\right]\) to the water to produce a saturated solution. Calculate the \(\mathrm{pH}\) of a saturated \(\mathrm{Ca}(\mathrm{OH})_{2}\) solution. (d) What is the concentration of \(\mathrm{Mg}^{2+}\) ions at this \(\mathrm{pH}\) ? (e) In general, which ion \(\left(\mathrm{Ca}^{2+}\right.\) or \(\mathrm{Mg}^{2+}\) ) would you remove first? Why?

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 \(\mathrm{pH}\) ?

If \(\mathrm{NaOH}\) is added to \(0.010 \mathrm{M} \mathrm{Al}^{3+},\) which will be the predominant species at equilibrium: \(\mathrm{Al}(\mathrm{OH})_{3}\) or \(\mathrm{Al}(\mathrm{OH})_{4}^{-} ?\) The \(\mathrm{pH}\) of the solution is \(14.00 .\left[K_{\mathrm{f}}\right.\) for \(\mathrm{Al}(\mathrm{OH})_{4}^{-}=2.0 \times 10^{33}\)

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.

Sign-up for free