Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 28

Derive an expression for the relationship between \(\mathrm{p} K_{\mathrm{a}}\) and \(\mathrm{p} K_{\mathrm{b}}\) for a conjugate acid-base pair. \((p K=-\log K .)\)

Short Answer

Expert verified
The relationship between \(\mathrm{p}K_a\) and \(\mathrm{p}K_b\) for a conjugate acid-base pair can be expressed as: \[\mathrm{p}K_w = \mathrm{p}K_a+\mathrm{p}K_b\]

Step by step solution

01

Write the dissociation equations for the acid and base

Begin by writing the general dissociation equations for the acid (HX) and base (Y-) in water: For acid (HX): \[HX + H_2O \rightleftharpoons H_3O^+ + Y^-\] For base (Y-): \[Y^- + H_2O \rightleftharpoons OH^- + HX\]
02

Find the equilibrium expressions for \(K_a\) and \(K_b\)

Next, write the equilibrium expressions for \(K_a\) (for acid dissociation) and \(K_b\) (for base dissociation): For \(K_a\) (based on the acid dissociation equation): \[K_a = \frac{[H_3O^+][Y^-]}{[HX]}\] For \(K_b\) (based on the base dissociation equation): \[K_b = \frac{[OH^-][HX]}{[Y^-]}\]
03

Relate the two expressions by using the \(K_w\) of water

We will now relate the \(K_a\) and \(K_b\) expressions using the ion product of water, \(K_w\). The ion product of water is defined as: \[K_w = [H_3O^+][OH^-]\] Re-arrange the expression for \(K_b\) to solve for \([OH^-]\): \[[OH^-] = \frac{K_b[ Y^-]}{[HX]}\] Now substitute this expression for \([OH^-]\) into the \(K_w\) equation: \[K_w = [H_3O^+]\left(\frac{K_b[ Y^-]}{[HX]}\right)\] Since we have \([H_3O^+] = \frac{K_a[HX]}{[Y^-]}\) from the \(K_a\) expression, we can substitute this into the above equation: \[K_w =\left(\frac{K_a [HX]}{[Y^-]}\right)\left(\frac{K_b[Y^-]}{[HX]}\right)\]
04

Derive the relationship between \(\mathrm{p}K_a\) and \(\mathrm{p}K_b\)

By simplifying the last equation, we can obtain the relationship between \(K_a\) and \(K_b\): \[K_w = K_a K_b\] -\(pK=-\log K\) \[K_w=10^{-\mathrm{p}K_w}\] \[K_a=10^{-\mathrm{p}K_a}\] \[K_b=10^{-\mathrm{p}K_b}\] By substituting the last three equations into \(K_w = K_a K_b\), we will have: \[10^{-\mathrm{p}K_w} = 10^{-\mathrm{p}K_a} \times 10^{-\mathrm{p}K_b}\] Now we can use the logarithmic property: \(a^{(\log_ax+\log_ay)} = a^{(\log_ax)}\cdot a^{(\log_ay)}\). So, \[10^{-\mathrm{p}K_w} = 10^{-(\mathrm{p}K_a+\mathrm{p}K_b)}\] Hence, the relationship between \(\mathrm{p}K_a\) and \(\mathrm{p}K_b\) is: \[\mathrm{p}K_w = \mathrm{p}K_a+\mathrm{p}K_b\]

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

The equilibrium constant \(K_{\mathrm{a}}\) for the reaction $$\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons{\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}}(\mathrm{OH})^{2+}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)$$ is \(6.0 \times 10^{-3}\) a. Calculate the \(\mathrm{pH}\) of a \(0.10-M\) solution of \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) b. Will a \(1.0-M\) solution of iron(II) nitrate have a higher or lower \(\mathrm{pH}\) than a \(1.0-M\) solution of iron(III) nitrate? Explain.

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. \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\)

What are the major species present in 0.250 \(M\) solutions of each of the following acids? Calculate the pH of each of these solutions. a. \(\mathrm{HNO}_{2}\) b. \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\)

Zinc hydroxide is an amphoteric substance. Write equations that describe \(\mathrm{Zn}(\mathrm{OH})_{2}\) acting as a Brönsted-Lowry base toward \(\mathrm{H}^{+}\) and as a Lewis acid toward \(\mathrm{OH}^{-}\).

Calculate the \(\mathrm{pH}\) of a \(0.200-M\) solution of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\). Hint: \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\) is a salt composed of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\) and \(\mathrm{F}^{-}\) ions. The principal equilibrium in this solution is the best acid reacting with the best base; the reaction for the principal equilibrium is $$\begin{aligned} \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(a q)+\mathrm{F}^{-}(a q) & \rightleftharpoons \\ \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}(a q) &+\mathrm{HF}(a q) \quad K=8.2 \times 10^{-3}\end{aligned}$$
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

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