Explanation Explanation The [ OH − ] of given NH 3 is 4 . 1 5 × 1 0 - 4 M _ . The compound NH 3 is a weak base and establishes equilibrium upon dissociation in an aqueous solution. The [ OH − ] value is calculated by the formation of an ICE (Initial Concentration Equilibrium) table. NH 3 + H 2 O ⇌ NH 4 + + OH − Initial ( M ) : 0.10 0 0 Change ( M ) : − x x x Equilibrium ( M ) : 0.10 − x x x The equilibrium constant expression is calculated by the formula, K b = [ Concentration of products ] [ Concentration of reactants ] The equilibrium constant expression for the stated reaction is, K b = [ NH 4 + ] [ OH − ] [ NH 3 ] The value of K b is 1.8 × 10 − 5 . Substitute the value of K b , [ NH 4 + ] , [ OH − ] and [ NH 3 ] from ICE table in the above expression. 1.8 × 10 − 5 = x × x 0.10 − x x 2 + 1.8 × 10 − 5 − 1.8 × 10 − 7 = 0 x = 4 . 1 5 × 1 0 - 4 M _ Thus, the value of [ OH − ] is 4 . 1 5 × 1 0 - 4 M _ . The pOH of given NH 3 solution is 3 . 3 8 _ . The pOH value is calculated by the formula, pOH = − log [ OH − ] Substitute the value of [ OH − ] in the above formula. pOH = − log [ 4.15 × 10 − 4 ] pOH = 3 . 3 8 _ The pH of given NH 3 is 1 0 . 6 2 _ . The sum, pH + pOH = 14 Substitute the calculated value of pOH in the above equation. pH + 3.38 = 14 pH = 14 − 3.38 pH = 1 0 . 6 2 _ The pH of KCl is 7 _ . The strong salts like KCl are formed by the reaction between a strong acid and a strong base. KCl is made up by the combination of a strong acid, that is HCl and a strong base, that is KOH . These are strong salts and dissociate completely in an aqueous. The strong salts like KCl are neutral in nature and have pH value equal to 7 _ . The [ OH − ] value of KCl is. 0 . 1 0 M _ . The given compound KOH is a strong base and dissociates completely and readily in an aqueous solution. The dissociation reaction that takes place is, KOH → K + + OH − Thus, the given concentration of KOH will correspond to the [ OH − ] . Therefore, the value of [ OH − ] is 0 . 1 0 M _ . The pOH of given KCl solution is 1 _ . The pOH value is calculated by the formula, pOH = − log [ OH − ] Substitute the value of [ OH − ] in the above formula

Bundle: Chemistry: An Atoms First Approach, 2nd, Loose-Leaf + OWLv2, 4 terms (24 months) Printed Access Card

2nd Edition

ISBN: 9781305717633

Author: Steven S. Zumdahl, Susan A. Zumdahl

Publisher: Cengage Learning

See similar textbooks

Concept explainers

Ionic Equilibrium

Chemical equilibrium and ionic equilibrium are two major concepts in chemistry. Ionic equilibrium deals with the equilibrium involved in an ionization process while chemical equilibrium deals with the equilibrium during a chemical change. Ionic equilibri…

Arrhenius Acid

Arrhenius acid act as a good electrolyte as it dissociates to its respective ions in the aqueous solutions. Keeping it similar to the general acid properties, Arrhenius acid also neutralizes bases and turns litmus paper into red.

Bronsted Lowry Base In Inorganic Chemistry

Bronsted-Lowry base in inorganic chemistry is any chemical substance that can accept a proton from the other chemical substance it is reacting with.

Videos

Question

Chapter 13, Problem 166CWP

Interpretation Introduction

Interpretation: The given solutions are to be arranged in increasing order of pH.

Concept introduction: Strong acids dissociate completely and rapidly when dissolved in an aqueous solution and have a very large value of Ka and a lower value of pH while the weak acids dissociate partially and slowly in an aqueous solution and have a very low value of Ka.and a higher value of pH.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 13, Problem 165CWP

Chapter 13, Problem 167CWP

Chapter 13 Solutions

Bundle: Chemistry: An Atoms First Approach, 2nd, Loose-Leaf + OWLv2, 4 terms (24 months) Printed Access Card

Ch. 13 - Define each of the following: a. Arrhenius acid b....Ch. 13 - Define or illustrate the meaning of the following...Ch. 13 - Prob. 3RQ Ch. 13 - How is acid strength related to the value of Ka?...Ch. 13 - Two strategies are followed when solving for the...Ch. 13 - Prob. 6RQ Ch. 13 - Prob. 7RQ Ch. 13 - For conjugate acidbase pairs, how are Ka and Kb...Ch. 13 - What is a salt? List some anions that behave as...Ch. 13 - For oxyacids, how does acid strength depend on a....

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Solution Summary: The author explains that the given solutions are to be arranged in increasing order of pH. Strong acids dissociate completely and rapidly when dissolved in an aqueous solution.

Concept explainers

Videos

Want to see the full answer?

Chapter 13 Solutions