Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 52

Phenylacetic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{COOH}\right)\) is one of the substances that accumulates in the blood of people with phenylketonuria, an inherited disorder that can cause mental retardation or even death. A \(0.085 \mathrm{M}\) solution of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{COOH}\) has a \(\mathrm{pH}\) of \(2.68 .\) Calculate the \(K_{a}\) value for this acid.

Short Answer

Expert verified
The $K_{a}$ value for phenylacetic acid is approximately \(5.1 \times 10^{-5}\).

Step by step solution

01

Convert pH to H3O+ concentration

Using the formula for pH, which is: pH = -log10[H3O+] we will first calculate the concentration of hydronium ions (H3O+): 2.68 = -log10[H3O+]
02

Solve for [H3O+]

Now, we will isolate the [H3O+] by using the inverse of the log function, which is 10^: [H3O+] = 10^(-pH) = 10^(-2.68) [H3O+] ≈ 2.09 × 10^(-3) M
03

Write the equilibrium expression for Ka

The dissociation of phenylacetic acid (C6H5CH₂COOH) can be represented as: C6H5CH₂COOH ⇄ C6H5CH₂COO⁻ + H3O⁺ The equilibrium constant expression for Ka can be written as: Ka = ([C6H5CH₂COO⁻][H3O⁺])/([C6H5CH₂COOH])
04

Set up the ICE table

ICE stands for Initial, Change, and Equilibrium concentrations. The table will look like this: | C6H5CH₂COOH | C6H5CH₂COO⁻ | H3O⁺ ------------------------------------------------ Initial | 0.085 M | 0 M | 0 M Change | -x M | +x M | +x M Equilibrium| 0.085-x M | x M | x M As we know the concentration of H3O+ at equilibrium which is 2.09 × 10^(-3) M, so x = 2.09 × 10^(-3) M Now we will substitute the values in the Ka expression: Ka = ([C6H5CH₂COO⁻][H3O⁺])/([C6H5CH₂COOH])
05

Calculate Ka

Substituting the equilibrium concentrations into the Ka expression: Ka = ((2.09 × 10^(-3))(2.09 × 10^(-3)))/ (0.085 - 2.09 × 10^(-3)) And solving for Ka: Ka ≈ 5.1 × 10^(-5) The Ka value for this acid is approximately 5.1 × 10^(-5).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Calculate \(\left[\mathrm{OH}^{-}\right]\) for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: (a) \(\left[\mathrm{H}^{+}\right]=0.0505 \mathrm{M} ;\) (b) \(\left[\mathrm{H}^{+}\right]=2.5 \times 10^{-10} \mathrm{M} ;(\mathrm{c})\) a solution in which \(\left[\mathrm{H}^{+}\right]\) is 1000 times greater than \(\left[\mathrm{OH}^{-}\right] .\)

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 \(\mathrm{H}_{2} \mathrm{O} ;(\mathrm{c})\) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) 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 \(\mathrm{CH}_{3} \mathrm{COOH}\)

Label each of the following as being a strong base, a weak base, or a species with negligible basicity. In each case write the formula of its conjugate acid, and indicate whether the conjugate acid is a strong acid, a weak acid, or a species with negligible acidity: (a) \(\mathrm{CH}_{3} \mathrm{COO}^{-},\) (b) \(\mathrm{HCO}_{3}^{-}, (\mathrm{c}) \mathrm{O}^{2-}, (\mathrm{d}) \mathrm{Cl}^{-} ,(\mathrm{e}) \mathrm{NH}_{3}\)

At the freezing point of water \(\left(0^{\circ} \mathrm{C}\right), K_{w}=1.2 \times 10^{-15}\) Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for a neutral solution at this temperature.

Calculate \(\left[\mathrm{H}^{+}\right]\) for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: (a) \(\left[\mathrm{OH}^{-}\right]=0.00045 \mathrm{M} ;\) (b) \(\left[\mathrm{OH}^{-}\right]=8.8 \times 10^{-9} \mathrm{M} ;(\mathrm{c})\) a so- lution in which \(\left[\mathrm{OH}^{-}\right]\) is 100 times greater than \(\left[\mathrm{H}^{+}\right]\).
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation
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