Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 69

The \(\mathrm{p} K_{\mathrm{a}}\) of butyric acid (HBut) is 4.7 . Calculate \(K_{\mathrm{b}}\) for the butyrate ion (But \(^{-}\) ).

Short Answer

Expert verified
After performing the calculation, you find that the value of \(K_{\mathrm{b}}\) for the Butyrate ion (But-\(-\))) is approximately 2 x 10^{-10}.

Step by step solution

01

Understanding pKa and pKb

The the pKa and pKb are the negative logs of the acid dissociation constant (Ka) and base dissociation constant (Kb), respectively. Here, the pKa is given, which can be used to find the Ka for Butyric Acid, which will then be used to find the Kb of Butyrate ion, using the relationship \(K_{\mathrm{a}} * K_{\mathrm{b}} = K_{\mathrm{w}}\), where \(K_{\mathrm{w}}\) is the ion product for water.
02

Find Ka

To find the Ka of butyric acid, use the formula \( K_{\mathrm{a}} = 10^{- pK_{\mathrm{a}}}\). Substituting \( pK_{\mathrm{a}} = 4.7\) into the formula gives \( K_{\mathrm{a}} = 10^{-4.7}\).
03

Calculate Kb

Use the formula \( K_{\mathrm{b}} = \frac{K_{\mathrm{w}}}{K_{\mathrm{a}}}\) to calculate Kb. The value for \( K_{\mathrm{w}}\) is typically \( 1.0 x 10^{-14}\) at 25°C. Substituting these values gives \( K_{\mathrm{b}} = \frac{ 1.0 x 10^{-14}}{10^{-4.7}}\).
04

Simplify Kb

Simplify the expression for Kb to get the final numerical answer.

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 maximum allowable concentration of \(\mathrm{Pb}^{2+}\) ions in drinking water is \(0.05 \mathrm{ppm}\) (that is, \(0.05 \mathrm{~g}\) of \(\mathrm{Pb}^{2+}\) in 1 million g of water). Is this guideline exceeded if an underground water supply is at equilibrium with the mineral anglesite, \(\mathrm{PbSO}_{4}\left(K_{\mathrm{sp}}=1.6 \times 10^{-8}\right) ?\)

A sample of \(0.1276 \mathrm{~g}\) of an unknown monoprotic acid was dissolved in \(25.0 \mathrm{~mL}\) of water and titrated with \(0.0633 \mathrm{M} \mathrm{NaOH}\) solution. The volume of base required to reach the equivalence point was \(18.4 \mathrm{~mL}\). (a) Calculate the molar mass of the acid. (b) After \(10.0 \mathrm{~mL}\) of base had been added to the titration, the \(\mathrm{pH}\) was determined to be 5.87 . What is the \(K_{\mathrm{a}}\) of the unknown acid?

What are the criteria for choosing an indicator for a particular acid-base titration?

Acid-base reactions usually go to completion. Confirm this statement by calculating the equilibrium constant for each of the following cases: (a) a strong acid reacting with a strong base, (b) a strong acid reacting with a weak base \(\left(\mathrm{NH}_{3}\right),\) (c) a weak acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) reacting with a strong base, \((\mathrm{d})\) a weak acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) reacting with a weak base \(\left(\mathrm{NH}_{3}\right)\) (Hint: Strong acids exist as \(\mathrm{H}^{+}\) ions and strong bases exist as \(\mathrm{OH}^{-}\) ions in solution. You need to look up the \(K_{\mathrm{a}}, K_{\mathrm{b}}\), and \(K_{\mathrm{w}}\) values.)

A volume of \(75 \mathrm{~mL}\) of \(0.060 \mathrm{M} \mathrm{NaF}\) is mixed with \(25 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2} .\) Calculate the concentra- tions in the final solution of \(\mathrm{NO}_{3}^{-}, \mathrm{Na}^{+}, \mathrm{Sr}^{2+},\) and \(\mathrm{F}^{-} \cdot\left(K_{\mathrm{sp}}\right.\) for \(\left.\mathrm{SrF}_{2}=2.0 \times 10^{-10} \mathrm{.}\right)\)
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

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