Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 179

Consider \(1000 . \mathrm{mL}\) of a \(1.00 \times 10^{-4}-M\) solution of a certain acid HA that has a \(K_{\mathrm{a}}\) value equal to \(1.00 \times 10^{-4}\). How much water was added or removed (by evaporation) so that a solution remains in which \(25.0 \%\) of HA is dissociated at equilibrium? Assume that HA is nonvolatile.

Short Answer

Expert verified
A 1000 mL solution of acid HA with a molarity of \(1.00 \times 10^{-4}M\) and Ka value equal to \(1.00 \times 10^{-4}\), 666.7 mL of water must be removed by evaporation to achieve a solution in which 25% of HA is dissociated at equilibrium.

Step by step solution

01

Write the equilibrium expression for the acid dissociation

\ First, we write the dissociation reaction for the acid HA: \[ \textnormal{HA} \rightleftharpoons \textnormal{H}^+ + \textnormal{A}^-\] Now, let's write the equilibrium expression: \[K_a = \frac{[\textnormal{H}^+][\textnormal{A}^-]}{[\textnormal{HA}]}\] Here, \(K_a = 1.00 \times 10^{-4}\).
02

Set up initial and equilibrium concentrations for the reaction

\ Initial concentrations, in moles per liter: \[ [\textnormal{HA}]_0 = 1.00 \times 10^{-4} M \] \[ [\textnormal{H}^+]_0 = [\textnormal{A}^-]_0 = 0 \] At equilibrium, 25% of HA dissociates, therefore: \[ [\textnormal{HA}]_{eq} = 0.75 \times [\textnormal{HA}]_0 \] \[ [\textnormal{H}^+]_{eq} = [\textnormal{A}^-]_{eq} = 0.25 \times [\textnormal{HA}]_0 \]
03

Use the equilibrium concentrations to set up a Ka equation

\ Now, let's insert the equilibrium concentrations into the Ka equation: \[1.00 \times 10^{-4} = \frac{[\textnormal{H}^+_{eq}][\textnormal{A}^-_{eq}]}{[\textnormal{HA}_{eq}]}\] \[1.00 \times 10^{-4} = \frac{(0.25 \times [\textnormal{HA}_0])^2}{0.75 \times [\textnormal{HA}_0]}\]
04

Solve for the new molar concentration of HA

\ Now, we can solve the equation for the new molar concentration of HA: \[1.00 \times 10^{-4} = \frac{1.00 \times 10^{-4} \times 0.25^2}{0.75}\] New molarity of HA, \( [ \textnormal{HA}_{f}] = 3.00 \times 10^{-4} M\)
05

Calculate the new volume of the solution and the amount of water added or removed

\ Finally, we can calculate the new volume of the solution since the number of moles of HA remains constant in this reaction: \[V_f = \frac{n_0}{[\textnormal{HA}_{f}]} = \frac{[\textnormal{HA_0}] \times V_0}{[\textnormal{HA}_{f}]}\] \[V_f = \frac{1.00 \times 10^{-4} \times 1000}{3.00 \times 10^{-4}} = 333.3 \: \textnormal{mL}\] As the initial volume was 1000 mL and the final volume is 333.3 mL, it means that \(1000 - 333.3 = 666.7 \: \textnormal{mL}\) of water has been removed by evaporation.

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

Sodium azide \(\left(\mathrm{NaN}_{3}\right)\) is sometimes added to water to kill bacteria. Calculate the concentration of all species in a \(0.010-M\) solution of \(\mathrm{NaN}_{3}\). The \(K_{\mathrm{a}}\) value for hydrazoic acid \(\left(\mathrm{HN}_{3}\right)\) is \(1.9 \times 10^{-5}\)

Place the species in each of the following groups in order of increasing base strength. Give your reasoning in each case. a. \(\mathrm{IO}_{3}^{-}, \mathrm{BrO}_{3}^{-}\) b. \(\mathrm{NO}_{2}^{-}, \mathrm{NO}_{3}^{-}\) c. \(\mathrm{OCl}^{-}, \mathrm{OI}^{-}\)

Identify the Lewis acid and the Lewis base in each of the following reactions. a. \(\mathrm{B}(\mathrm{OH})_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{B}(\mathrm{OH})_{4}^{-}(a q)+\mathrm{H}^{+}(a q)\) b. \(\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q)\) c. \(\mathrm{BF}_{3}(g)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BF}_{4}^{-}(a q)\)

A \(0.15-M\) solution of a weak acid is \(3.0 \%\) dissociated. Calculate \(K_{\mathrm{a}}\)

Hemoglobin (abbreviated \(\mathrm{Hb}\) ) is a protein that is responsible for the transport of oxygen in the blood of mammals. Each hemoglobin molecule contains four iron atoms that are the binding sites for \(\mathrm{O}_{2}\) molecules. The oxygen binding is \(\mathrm{pH}\) dependent. The relevant equilibrium reaction is $$\mathrm{HbH}_{4}^{4+}(a q)+4 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)+4 \mathrm{H}^{+}(a q)$$ Use Le Châtelier's principle to answer the following. a. What form of hemoglobin, \(\mathrm{HbH}_{4}{ }^{4+}\) or \(\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}\), is favored in the lungs? What form is favored in the cells? b. When a person hyperventilates, the concentration of \(\mathrm{CO}_{2}\) in the blood is decreased. How does this affect the oxygenbinding equilibrium? How does breathing into a paper bag help to counteract this effect? (See Exercise 146.) c. When a person has suffered a cardiac arrest, injection of a sodium bicarbonate solution is given. Why is this necessary? (Hint: \(\mathrm{CO}_{2}\) blood levels increase during cardiac arrest.)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation

Chemistry Branches

Read Explanation

The Earths Atmosphere

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