Chemistry: An Atoms-Focused Approach
Chemistry: An Atoms-Focused Approach
14th Edition
ISBN: 9780393912340
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster
Publisher: W. W. Norton & Company
Question
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Chapter 15, Problem 15.54QA
Interpretation Introduction

To find:

The Ka value of the acid HNO3

Expert Solution & Answer
Check Mark

Answer to Problem 15.54QA

Solution:

The Ka value for 7.5M concentration aqueous solution of nitric acid is 1.2.

Explanation of Solution

1) Concept:

We are asked to determine the Ka value for the nitric acid from known initial concentration of the acid and percent ionization. The value for the NO3-equilibrium can be calculated using the percent dissociation formula. At equilibrium,H3O+=NO3-, and the equilibrium concentration of the acid is calculated from the initial value minus the equilibrium concentration of NO3- ions.

The acid ionization reaction for HNO3 is written as

HNO3aq+H2O lNO3-aq+H3O+(aq)

The dissociation constant expression for the acid is written as

Ka=H3O+NO3-[HNO3]

2) Formula:

i) Percent ionization=A-equilibriumHAinitial ×100

ii) Ka=H3O+NO3-[HNO3]

3) Given:

i) A 50% solution of 7.5M  nitric acid; only 33% of the molecules dissociate into H+ and  NO3-.

ii) Mass percent of solution is given as 50%

iii) Ionization is given as 33%

iv) Initial concentration of the nitric acid is given as 7.5 M

4) Calculations:

From the given percent dissociation, we can find the NO3-equilibrium using the percent dissociation formula as:

Percent ionization=A-equilibriumHAinitial ×100

33%=NO3-equilibrium7.5 M ×100

0.33=NO3-equilibrium7.5 M

NO3-equilibrium=0.33 ×7.5 M=2.475M

The equilibrium concentration of NO3- is calculated as 2.475 M.

We can set up a RICE table to solve this problem.

Reaction HNO3aq+H2O lNO3-aq+H3O+(aq) HNO3 (M) NO3-(M) H3O+ (M)
Initial (I) 7.5 0.00 0.00
Change (C) -x +x +x
Equilibrium (E) 7.5-x x x

At equilibrium, H3O+=NO3-=x=2.475

The equilibrium concentration for the acid is

HNO3equilibrium=HNO3initial-H3O+equilibrium

HNO3equilibrium=7.5-2.475M= 5.025M

Inserting the equilibrium values for all the species in the Ka expression, we get

Ka=H3O+NO3-[HNO3]

Ka=2.4752.4755.025 =1.219

Thus, the acid dissociation constant value for the strong acid HNO3 is 1.2.

Conclusion:

The Ka value is calculated from the initial concentration and percent dissociation by using RICE table.

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Chapter 15 Solutions

Chemistry: An Atoms-Focused Approach

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