Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 17

(a) Calculate the percent ionization of 0.0075\(M\) butanoic acid \(\left(K_{a}=1.5 \times 10^{-5}\right) .\) (b) Calculate the percent ionization of 0.0075\(M\) butanoic acid in a solution containing 0.085\(M\) sodium butanoate.

Short Answer

Expert verified
For part (a), the percent ionization of 0.0075 M butanoic acid is given by: \[ \% \text{ionization} = \frac{\sqrt{1.5 \times 10^{-5} \times 0.0075}}{0.0075} \times 100 \approx 2.87 \% \] For part (b), the percent ionization of 0.0075 M butanoic acid in a solution containing 0.085 M sodium butanoate is given by: \[ \% \text{ionization} = \frac{\frac{1.5 \times 10^{-5} \times 0.0075}{0.085}}{0.0075} \times 100 \approx 0.294 \% \]

Step by step solution

01

Write the chemical equilibrium equation

Since butanoic acid is a weak acid, it dissociates into the conjugate base butanoate ion and proton as follows: \[ \ce{CH3CH2CH2COOH <=> CH3CH2CH2COO- + H+} \]
02

Set up the ICE table for the initial problem

An ICE table represents the initial concentration, the change in concentration, and the equilibrium concentration for the reactants and products: Initial: \[ [\ce{CH3CH2CH2COOH}] = 0.0075\ M \] \[ [\ce{CH3CH2CH2COO^-}] = 0 \] \[ [\ce{H^+}] = 0 \] Change: \[ [\ce{CH3CH2CH2COOH}] = -x \] \[ [\ce{CH3CH2CH2COO^-}] = +x \] \[ [\ce{H^+}] = +x \] Equilibrium: \[ [\ce{CH3CH2CH2COOH}] = 0.0075 - x\ M \] \[ [\ce{CH3CH2CH2COO^-}] = x \] \[ [\ce{H^+}] = x \] In which \(x\) is the concentration of the ions at equilibrium.
03

Use the equilibrium expression to solve for x

The equilibrium expression is given by the acid dissociation constant, \(K_a\): \[ K_{a} = \frac{[\ce{CH3CH2CH2COO^-}][\ce{H^+}]}{[\ce{CH3CH2CH2COOH}]} \] Since \(K_a = 1.5 \times 10^{-5} \), \[ 1.5 \times 10^{-5} = \frac{x^2}{0.0075 - x} \] Considering the weak dissociation, we can assume that \(x \ll 0.0075\), making the equilibrium simplification of \[ 1.5 \times 10^{-5} \approx \frac{x^2}{0.0075} \] Now, we can solve for \(x\): \[ x = \sqrt{1.5 \times 10^{-5} \times 0.0075} \]
04

Calculate the percent ionization of butanoic acid

The percent ionization is given by the ratio of the concentration of dissociated protons and the initial concentration, multiplied by 100: \[ \% \text{ionization} = \frac{x}{0.0075} \times 100 \] Plug in the value of \(x\) to get the percent ionization of butanoic acid. For part (b), since we have 0.075 M butanoic acid and 0.085 M sodium butanoate: Initial: \[ [\ce{CH3CH2CH2COOH}] = 0.0075\ M \] \[ [\ce{CH3CH2CH2COO^-}] = 0.085 \] \[ [\ce{H^+}] = 0 \] Change: \[ [\ce{CH3CH2CH2COOH}] = -x \] \[ [\ce{CH3CH2CH2COO^-}] = +x \] \[ [\ce{H^+}] = +x \] Equilibrium: \[ [\ce{CH3CH2CH2COOH}] = 0.0075 - x\ M \] \[ [\ce{CH3CH2CH2COO^-}] = 0.085 + x \] \[ [\ce{H^+}] = x \]
05

Calculate the percent ionization of butanoic acid in the sodium butanoate solution

The new equilibrium expression with sodium butanoate present is as follows: \[ K_{a} = \frac{(x)(0.085 + x)}{(0.0075 - x)} \] Again, we can assume that \(x \ll 0.0075\), making the equilibrium simplification of \[ 1.5 \times 10^{-5} \approx \frac{x \times 0.085}{0.0075} \] Now, we can solve for \(x\): \[ x = \frac{1.5 \times 10^{-5} \times 0.0075}{0.085} \] Calculate the percent ionization as we did in Step 4: \[ \% \text{ionization} = \frac{x}{0.0075} \times 100 \] Plug in the value of \(x\) to get the percent ionization of butanoic acid in the sodium butanoate solution.

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

Consider a beaker containing a saturated solution of \(\mathrm{PbI}_{2}\) in equilibrium with undissolved \(\mathrm{PbI}_{2}(s)\). Now solid \(\mathrm{K} \mathrm{I}\) is added to this solution. (a) Will the amount of solid \(\mathrm{PbI}_{2}\) at the bottom of the beaker increase, decrease, or remain the same? (b) Will the concentration of \(\mathrm{Pb}^{2+}\) ions in solution increase or decrease? (c) Will the concentration of \(I\) ions in solution increase or decrease?

How many milliliters of 0.0850\(M \mathrm{NaOH}\) are required to titrate each of the following solutions to the equivalence point: (a) 40.0 \(\mathrm{mL}\) of \(0.0900 \mathrm{M} \mathrm{HNO}_{3},\) (\mathbf{b} ) 35.0 \(\mathrm{mL}\) of \(0.0850 M \mathrm{CH}_{3} \mathrm{COOH},(\mathbf{c}) 50.0 \mathrm{mL}\) of a solution that contains 1.85 \(\mathrm{g}\) of \(\mathrm{HCl}\) per liter?

The beaker on the right contains 0.1 Macetic acid solution with methyl orange as an indicator. The beaker on the left contains a mixture of 0.1\(M\) acetic acid and 0.1\(M\) sodium acetate with methyl orange. (a) Using Figures 16.8 and 16.9, which solution has a higher pH? (b) Which solution is better able to maintain its pH when small amounts of NaOH are added? Explain. [Sections 17.1 and 17.2]

Mathematically prove that the \(\mathrm{pH}\) at the halfway point of a titration of a weak acid with a strong base (where the volume of added base is half of that needed to reach the equivalence point) is equal to p \(K_{a}\) for the acid.

A 20.0 -mL sample of 0.150 \(\mathrm{M} \mathrm{KOH}\) is titrated with 0.125 \(\mathrm{M}\) \(\mathrm{HClO}_{4}\) solution. Calculate the \(\mathrm{pH}\) after the following volumes of acid have been added: (a) \(20.0 \mathrm{mL},\) (b) 23.0 \(\mathrm{mL}\) \((\mathbf{c}) 24.0 \mathrm{mL},(\mathbf{d}) 25.0 \mathrm{mL},(\mathbf{e}) 30.0 \mathrm{mL}\)
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

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