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 \(0.100 M\) solution of chloroacetic acid \(\left(\mathrm{ClCH}_{2} \mathrm{COOH}\right)\) is \(11.0 \%\) ionized. Using this information, calculate \(\left[\mathrm{ClCH}_{2} \mathrm{COO}^{-}\right],\left[\mathrm{H}^{+}\right],\left[\mathrm{ClCH}_{2} \mathrm{COOH}\right]\), and \(K_{a}\) for chloroacetic acid.

Short Answer

Expert verified
The concentrations of the dissociated and undissociated chloroacetic acid and the hydrogen ion concentration are: \[\left[\mathrm{ClCH}_{2}\mathrm{COO}^{-}\right] = \left[\mathrm{H}^{+}\right] = 0.011 M\] \[\left[\mathrm{ClCH}_{2}\mathrm{COOH}\right] = 0.089 M\] and the acid dissociation constant, \(K_{a}\) for chloroacetic acid is \(1.36 \times 10^{-3}\).

Step by step solution

01

1. Calculate the concentration of \(\mathrm{ClCH}_{2}\mathrm{COO}^{-}\) and \(\mathrm{H}^{+}\)

Since we know that the \(0.100 M\) solution is \(11.0\%\) ionized, we can calculate the concentration of ions: \[\left[\mathrm{ClCH}_{2}\mathrm{COO}^{-}\right] = \left[\mathrm{H}^{+}\right] = 0.100 M \times 0.110 = 0.011 M\]
02

2. Calculate the concentration of \(\mathrm{ClCH}_{2}\mathrm{COOH}\)

To calculate the concentration of the non-ionized chloroacetic acid, we need to subtract the concentrations of ions from the initial concentration: \[\left[\mathrm{ClCH}_{2}\mathrm{COOH}\right] = 0.100 M - 0.011 M = 0.089 M\]
03

3. Calculate the acid dissociation constant, \(K_{a}\)

The equilibrium expression for the dissociation of chloroacetic acid is given by: \[K_{a} = \frac{[\mathrm{ClCH}_{2}\mathrm{COO}^{-}][\mathrm{H}^{+}]}{[\mathrm{ClCH}_{2}\mathrm{COOH}]}\] Substitute the values obtained in steps 1 and 2 into the expression to calculate \(K_{a}\): \[K_{a} = \frac{(0.011)(0.011)}{0.089} \approx 1.36 \times 10^{-3}\] So, the concentrations of the dissociated and undissociated chloroacetic acid and the hydrogen ion concentration are: \[\left[\mathrm{ClCH}_{2}\mathrm{COO}^{-}\right] = \left[\mathrm{H}^{+}\right] = 0.011 M\] \[\left[\mathrm{ClCH}_{2}\mathrm{COOH}\right] = 0.089 M\] and the acid dissociation constant, \(K_{a}\) for chloroacetic acid is \(1.36 \times 10^{-3}\).

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.

Ionization
Ionization is a crucial concept in chemistry, especially when dealing with acids and their behavior in solutions. When an acid like chloroacetic acid dissolves in water, it partially dissociates into ions. In this case, the ionization occurs as the chloroacetic acid separates into chloroacetate anion (\( \mathrm{ClCH}_{2}\mathrm{COO}^{-} \)) and hydrogen cation (\( \mathrm{H}^{+} \)). The extent to which this ionization happens is important because it influences the concentration of ions and ultimately the acid's properties in solution.For the given chloroacetic acid solution, it is noted that the ionization is only 11%. This means that out of the total acid molecules present, only a small fraction has converted into ions.
  • The concentration of ions formed in the solution is directly related to the percentage ionization and the initial concentration of the acid solution.
  • By multiplying the initial molarity by the percentage ionized, one can determine the concentration of each ion formed. In this case, 0.100 M of acid and 11% ionization result in 0.011 M of chloroacetate and hydrogen ions.
Understanding how ionization works can help predict how acidic or conductive a solution will be.
Chloroacetic Acid
Chloroacetic acid, known chemically as \( \mathrm{ClCH}_{2}\mathrm{COOH} \), is an organochlorine compound and a type of carboxylic acid. This acid is a key compound in organic synthesis due to its high reactivity and ability to form derivatives. Due to the presence of a chlorine atom attached to the alpha carbon, chloroacetic acid exhibits higher acidity compared to acetic acid.
  • The added chlorine atom increases its acidity by stabilizing the negative charge on the chloroacetate ion that forms during dissociation.
  • This stabilization occurs through an effect known as "inductive effect," wherein the electron-withdrawing nature of the chlorine atom helps spread out the charge.
With a relatively significant ionization percentage, chloroacetic acid is a moderately strong acid in water. This increased acidity makes it valuable in reactions where strong acid catalysts are required.
Equilibrium Expression
The equilibrium expression is a fundamental tool in understanding how chemical reactions behave in a state of balance. For the dissociation reaction of chloroacetic acid, the equilibrium expression provides insights into how ion concentrations relate to each other and the undissociated acid at equilibrium.The expression for the equilibrium constant (\( K_a \)) of chloroacetic acid is given by:\[ K_a = \frac{[\mathrm{ClCH}_{2}\mathrm{COO}^{-}][\mathrm{H}^{+}]}{[\mathrm{ClCH}_{2}\mathrm{COOH}]} \]
  • \([\mathrm{ClCH}_{2}\mathrm{COO}^{-}]\) and \([\mathrm{H}^{+}]\) represent the concentrations of the dissociated ions.
  • \([\mathrm{ClCH}_{2}\mathrm{COOH}]\) is the concentration of the acid that remains undissociated.
Inserting the given concentrations into this formula allows the calculation of \( K_a \) for chloroacetic acid, which quantifies its strength. A stronger acid will have a larger \( K_a \) value, indicating a greater tendency to donate protons. Here, the \( K_a \) is calculated as \( 1.36 \times 10^{-3} \), marking chloroacetic acid as a relatively strong acid compared to many other carboxyl acids.

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

Predict whether aqueous solutions of the following compounds are acidic, basic, or neutral: (a) \(\mathrm{NH}_{4} \mathrm{Br}\), (b) \(\mathrm{FeCl}_{3 \text {, }}\) (c) \(\mathrm{Na}_{2} \mathrm{CO}_{3}\), (d) \(\mathrm{KClO}_{4}\), (e) \(\mathrm{NaHC}_{2} \mathrm{O}_{4}\).

Addition of the indicator methyl orange to an unknown solution leads to a yellow color. The addition of bromthymol blue to the same solution also leads to a yellow color. (a) Is the solution acidic, neutral, or basic? (b) What is the range (in whole numbers) of possible pH values for the solution? (c) Is there another indicator you could use to narrow the range of possible \(\mathrm{pH}\) values for the solution?

In many reactions the addition of \(\mathrm{AlCl}_{3}\) produces the same effect as the addition of \(\mathrm{H}^{+}\). (a) Draw a Lewis structure for \(\mathrm{AlCl}_{3}\) in which no atoms carry formal charges, and determine its structure using the VSEPR method. (b) What characteristic is notable about the structure in part (a) that helps us understand the acidic character of \(\mathrm{AlCl}_{3}\) ? (c) Predict the result of the reaction between \(\mathrm{AlCl}_{3}\) and \(\mathrm{NH}_{3}\) in a solvent that does not participate as a reactant. (d) Which acid-base theory is most suitable for discussing the similarities between \(\mathrm{AlCl}_{3}\) and \(\mathrm{H}^{+}\)?

Explain the following observations: (a) \(\mathrm{HNO}_{3}\) is a stronger acid than \(\mathrm{HNO}_{2} ;\) (b) \(\mathrm{H}_{2} \mathrm{~S}\) is a stronger acid than \(\mathrm{H}_{2} \mathrm{O}\); (c) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is a stronger acid than \(\mathrm{HSO}_{4}^{-}\); (d) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is a stronger acid than \(\mathrm{H}_{2} \mathrm{SeO}_{4} ;\) (e) \(\mathrm{CCl}_{3} \mathrm{COOH}\) is a stronger acid than \(\mathrm{CCl}_{3} \mathrm{COOH}\).

\mathrm{~A} 0.100 \mathrm{M}\( solution of bromoacetic acid \)\left(\mathrm{BrCH}_{2} \mathrm{COOH}\right)\( is \)13.2 \%\( ionized. Calculate \)\left[\mathrm{H}^{+}\right],\left[\mathrm{BrCH}_{2} \mathrm{COO}^{-}\right],\left[\mathrm{BrCH}_{2} \mathrm{COOH}\right]\( and \)K_{a}$ for bromoacetic acid.

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