Chemistry: An Atoms-Focused Approach (Second Edition)
Chemistry: An Atoms-Focused Approach (Second Edition)
2nd Edition
ISBN: 9780393614053
Author: Thomas R. Gilbert, Rein V. Kirss, Stacey Lowery Bretz, Natalie Foster
Publisher: W. W. Norton & Company
Question
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Chapter 15, Problem 15.70QA
Interpretation Introduction

To find:

The pH of 5.00×10-4M H2SO4.

Expert Solution & Answer
Check Mark

Answer to Problem 15.70QA

Solution:

pH=3.02

Explanation of Solution

1) Concept:

Sulfuric acid is a strong diprotic acid in that one hydrogen atom per molecule ionizes completely (Ka1>>1), but the second hydrogen atom may not completely ionize for every molecule (Ka2>>0.012).

We start with the first ionization reaction that goes to completion.

H2SO4(aq)+ H2O(l)HSO4(aq)-+H3O(aq)+

Therefore, as the second ionization step begins [HSO4-]= [H3O+]=5.00×10-4M.

Ionization of HSO4- then produces additional H3O+ ions

HSO4(aq)-+ H2O(l)SO4(aq)2-+H3O(aq)+

The increase in [H3O+] due to second ionization will be between 0  to  5.00×10-4M, which means the total [H3O+] value will be between 5.00×10-4 and 0.001M and the pH of the solution at equilibrium should be greater than 1.

We have to set up a RICE table for the second ionization reaction. Initially [HSO4-]= [H3O+]=5.00×10-4M. The change in [H3O+] during the second ionization step is +x filling the other cells of the table based on the stoichiometry of the second ionization step.

2) Formula:

i) pH= -log[H3O+]

ii) x=-b±b2-4ac 2a

3) Given:

Concentration of  H2SO4 is 5.00×10-4M

4) Calculations:

Reaction HSO4aq-+ H2Ol   SO4aq2-+ H3Oaq+
[HSO4-] (M) [SO42-] (M) [H3O+](M)
Initial 5.00×10-4 0 5.00×10-4
Change -x +x +x
Equilibrium (5.00×10-4-x) x (5.00×10-4+x)

Inserting the equilibrium concentrations in the equilibrium constant expression for Ka2.

Ka2= [H3O+] [SO42-] [HSO4-]

1.2× 10-2= 5.00×10-4+x(x)(5.00×10-4-x)

6.0×10-6-0.012x= x2+5.0×10-4x

x2+0.0125x-6×10-6=0

x=-b±b2-4ac 2a

Here, a=1, b=0.0125 and c= -6×10-6.

x=-0.0125±(0.0125)2-(4×1×(-6×10-6)) 2×1

x=-0.0125±0.00018 2

Solving this equation for x yields a positive value and a negative value.

x=0.0004628      and x= -0.01296

The negative value for x has no physical meaning because it gives us a negative [SO42-] value. Therefore,

[H3O+]=5.00×10-4+x=5.00×10-4+0.0004628=0.0009628

The corresponding pH  is

pH= -logH3O+= -log0.0009628=3.02

Conclusion:

The total hydrogen ion concentration in an H2SO4  solution is the result of a complete first dissociation and partial second dissociation,

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Chapter 15 Solutions

Chemistry: An Atoms-Focused Approach (Second Edition)

Ch. 15 - Prob. 15.11QACh. 15 - Prob. 15.12QACh. 15 - Prob. 15.13QACh. 15 - Prob. 15.14QACh. 15 - Prob. 15.15QACh. 15 - Prob. 15.16QACh. 15 - Prob. 15.17QACh. 15 - Prob. 15.18QACh. 15 - Prob. 15.19QACh. 15 - Prob. 15.20QACh. 15 - Prob. 15.21QACh. 15 - Prob. 15.22QACh. 15 - Prob. 15.23QACh. 15 - Prob. 15.24QACh. 15 - Prob. 15.25QACh. 15 - Prob. 15.26QACh. 15 - Prob. 15.27QACh. 15 - Prob. 15.28QACh. 15 - Prob. 15.29QACh. 15 - Prob. 15.30QACh. 15 - Prob. 15.31QACh. 15 - Prob. 15.32QACh. 15 - Prob. 15.33QACh. 15 - Prob. 15.34QACh. 15 - Prob. 15.35QACh. 15 - Prob. 15.36QACh. 15 - Prob. 15.37QACh. 15 - Prob. 15.38QACh. 15 - Prob. 15.39QACh. 15 - Prob. 15.40QACh. 15 - Prob. 15.41QACh. 15 - Prob. 15.42QACh. 15 - Prob. 15.43QACh. 15 - Prob. 15.44QACh. 15 - Prob. 15.45QACh. 15 - Prob. 15.46QACh. 15 - Prob. 15.47QACh. 15 - Prob. 15.48QACh. 15 - Prob. 15.49QACh. 15 - Prob. 15.50QACh. 15 - Prob. 15.51QACh. 15 - Prob. 15.52QACh. 15 - Prob. 15.53QACh. 15 - Prob. 15.54QACh. 15 - Prob. 15.55QACh. 15 - Prob. 15.56QACh. 15 - Prob. 15.57QACh. 15 - Prob. 15.58QACh. 15 - Prob. 15.59QACh. 15 - Prob. 15.60QACh. 15 - Prob. 15.61QACh. 15 - Prob. 15.62QACh. 15 - Prob. 15.63QACh. 15 - Prob. 15.64QACh. 15 - Prob. 15.65QACh. 15 - Prob. 15.66QACh. 15 - Prob. 15.67QACh. 15 - Prob. 15.68QACh. 15 - Prob. 15.69QACh. 15 - Prob. 15.70QACh. 15 - Prob. 15.71QACh. 15 - Prob. 15.72QACh. 15 - Prob. 15.73QACh. 15 - Prob. 15.74QACh. 15 - Prob. 15.75QACh. 15 - Prob. 15.76QACh. 15 - Prob. 15.77QACh. 15 - Prob. 15.78QACh. 15 - Prob. 15.79QACh. 15 - Prob. 15.80QACh. 15 - Prob. 15.81QACh. 15 - Prob. 15.82QACh. 15 - Prob. 15.83QACh. 15 - Prob. 15.84QACh. 15 - Prob. 15.85QACh. 15 - Prob. 15.86QACh. 15 - Prob. 15.87QACh. 15 - Prob. 15.88QACh. 15 - Prob. 15.89QACh. 15 - Prob. 15.90QACh. 15 - Prob. 15.91QACh. 15 - Prob. 15.92QACh. 15 - Prob. 15.93QACh. 15 - Prob. 15.94QACh. 15 - Prob. 15.95QACh. 15 - Prob. 15.96QACh. 15 - Prob. 15.97QACh. 15 - Prob. 15.98QACh. 15 - Prob. 15.99QACh. 15 - Prob. 15.100QACh. 15 - Prob. 15.101QACh. 15 - Prob. 15.102QACh. 15 - Prob. 15.103QACh. 15 - Prob. 15.104QACh. 15 - Prob. 15.105QACh. 15 - Prob. 15.106QACh. 15 - Prob. 15.107QACh. 15 - Prob. 15.108QACh. 15 - Prob. 15.109QACh. 15 - Prob. 15.110QA
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