Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 98

Calculate the \(\mathrm{pH}\) of a \(0.050 \mathrm{M}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\) solution \(\left(K_{\mathrm{b}}=\right.\) \(\left.1.3 \times 10^{-3}\right)\)

Short Answer

Expert verified
The pH of the \(0.050 \mathrm{M}\) \(\left(\mathrm{C}_{2}\mathrm{H}_{5}\right)_{2}\mathrm{NH}\) solution is approximately 11.33.

Step by step solution

01

Write down the ionization equation

Ethylene diamine acts as a base in the solution, where its ionization equation can be represented as follows: \[\left(\mathrm{C}_{2}\mathrm{H}_{5}\right)_{2}\mathrm{NH} + \mathrm{H}_{2}\mathrm{O} \leftrightarrow \left(\mathrm{C}_{2}\mathrm{H}_{5}\right)_{2}\mathrm{NH}_{2}^{+} + \mathrm{OH}^{-}\]
02

Set up the ICE table

It's essential to use an ICE (initial, change, equilibrium) table to organize the changing concentrations during the ionization process: \[ \begin{array}{c|c|c|c} & (\mathrm{C}_{2}\mathrm{H}_{5})_{2}\mathrm{NH} & (\mathrm{C}_{2}\mathrm{H}_{5})_{2}\mathrm{NH}_{2}^{+} & \mathrm{OH}^{-} \\ \hline I & 0.050 & 0 & 0 \\ C & -x & +x & +x \\ E & 0.050-x & x & x \\ \end{array} \] Where \(I\) indicates initial concentrations, \(C\) is the change in concentrations, and \(E\) represents equilibrium concentrations.
03

Write the expression for \(K_{b}\) and substitute the equilibrium concentrations

The base dissociation constant (\(K_{b}\)) expression can be written as follows: \[K_{b} = \frac{[\left(\mathrm{C}_{2}\mathrm{H}_{5}\right)_{2}\mathrm{NH}_{2}^{+}][\mathrm{OH}^{-}]}{[\left(\mathrm{C}_{2}\mathrm{H}_{5}\right)_{2}\mathrm{NH}]}\] Plug in the equilibrium concentrations from the ICE table: \[K_{b} = \frac{x \cdot x}{0.050 - x} = \frac{x^{2}}{0.050 - x}\]
04

Solve for \(x\) using the given \(K_{b}\) value

Now, we need to insert the given value of \(K_{b}\) and solve for \(x\): \[1.3 \times 10^{-3} = \frac{x^{2}}{0.050 - x}\] This equation can be simplified further: \[x^{2} + 1.3 \times 10^{-3}x - 0.050 \times 1.3 \times 10^{-3} = 0\] The quadratic equation is difficult to solve by hand, so a calculator can be used to find the value of \(x\): \[x \approx 0.00214\] This value, \(x\), represents the concentration of \(\mathrm{OH}^{-}\) at equilibrium.
05

Calculate the \(\mathrm{pOH}\) and then the \(\mathrm{pH}\)

Now, we can calculate the pOH using the \(\mathrm{OH}^{-}\) concentration: \[\mathrm{pOH} = -\log_{10}[\mathrm{OH}^{-}] = -\log_{10}(0.00214) \approx 2.67\] Finally, we can find the pH of the solution by subtracting the pOH from 14: \[\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 2.67 \approx 11.33\] So, the \(\mathrm{pH}\) of the \(0.050 \mathrm{M}\) \(\left(\mathrm{C}_{2}\mathrm{H}_{5}\right)_{2}\mathrm{NH}\) solution is approximately 11.33.

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

What are the major species present in \(0.250 M\) solutions of each of the following acids? Calculate the \(\mathrm{pH}\) of each of these solutions. a. \(\mathrm{HClO}_{4}\) b. \(\mathrm{HNO}_{3}\)

Calculate the percentage of pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\) that forms pyridinium ion, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\), in a \(0.10 \mathrm{M}\) aqueous solution of pyridine \(\left(K_{\mathrm{b}}=1.7 \times 10^{-9}\right)\)

Calculate the \(\mathrm{pH}\) of each of the following solutions. a. \(0.12 \mathrm{M} \mathrm{KNO}_{2}\) c. \(0.40 \mathrm{M} \mathrm{NH}_{4} \mathrm{ClO}_{4}\) b. \(0.45 \mathrm{M} \mathrm{NaOCl}\)

Given that the \(K_{\mathrm{a}}\) value for acetic acid is \(1.8 \times 10^{-5}\) and the \(K_{\mathrm{a}}\) value for hypochlorous acid is \(3.5 \times 10^{-8}\), which is the stronger base, \(\mathrm{OCl}^{-}\) or \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\) ?

Write the reaction and the corresponding \(K_{\mathrm{b}}\) equilibrium expression for each of the following substances acting as bases in water. a. \(\mathrm{NH}_{3}\) b. \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

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