/ibiscms/mod/ibis/view.php?id 2533191 WWW sapling learning.com App Attempts Scor

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/ibiscms/mod/ibis/view.php?id 2533191 WWW sapling learning.com App Attempts Score Print Calculator Periodic Table 25 Questi 37 of 39 26 90 calculate the concentrations of all species in a 1.74 M Na2SO3 (sodium s solution. The ionization Map ulfite) constants for sulfurous acid are Kat E 4x 10 and Ka2 E 6.3x 10 8 27 28 M 29 67 30 29 32 33 35 OH 36 Number 37 38 A Previous ® Give Up & V Solution Check Answer 0 Next HExit 9 Hint 39 Copyright 2011 2016 Sapling Learning, Inc. 37 about us careers partners privacy policy terms of use tact us help Ask me anything Available From 20/2016 11:30 PM 5/13/2016 11:30 PM Due Date Points Possible 100 Grade Category: Graded Description: Policies: Homework can check your a You can view solutions when you complete or give up on any question. You can keep trying to answer each question until you get it right or give up You lose 5% of the points available to each answer in your question for each incorrect attempt at that o Help With This Topic OWeb Help & Videos Technical Support and Bug Reports 903 PM 4/23/2016

Explanation / Answer

Solution :-

Na2SO3 is the salt so it dissociates completely

So the Na^+ concentration is twice the concentration of Na2SO3

[Na^+] =1.74 M *2 = 3.48 M

SO3^2- ions acts as base and forms the OH- when reacts with water

SO3^2- + H2O   -------- > HSO3^-   + OH-

1.74 M                                       0           0

-x                                              +x            +x

1.74-x                                        x             x

Kb1 = kw/ka2

     = 1*10^-14 / 6.3*10^-8

       = 1.59*10^-7

Kb1 = [HSO3^-][OH-] /[SO3^2-]

1.59*10^-7 = [x][x]/[1.74-x]

1.59*10^-7 *1.74 = x^2

2.77*10^-7 = x^2

Taking square root of both side we get

5.26*10^-4 M = x = [OH-]=[HSO3^-]

now HSO3^- reacts with water to form H2SO3

HSO3^-   + H2O    ------------- > H2SO3   + OH-

5.26*10^-4                                  0                  0

-x                                                 +x                 +x

5.26*10^-4-x                             x                     x

Kb2=KW/ka1

     = 1*10^-14 / 1.4*10^-2

     = 7.14*10^-13

Kb2 = [H2SO3][OH-]/[HSO3-]

7.14*10^-13 = [x][x]/[5.26*10^-4]

7.14*10^-13 * 5.26*10^-4 = x^2

3.76*10^-16 =x^2

Taking square root of both sides we get

1.94*10^-8 M = x = [H2SO3]

So the concentration of each species are as follows

[Na+] = 3.48 M

[SO3^2-] = 1.74 M – x = 1.74 M – 5.26*10^-4 M = 1.739 M

[HSO3-] = 5.26*10^-4 M

[H2SO3] = 1.94*10^-8 M

[OH-] = 5.26*10^-4 M

[H+] = 1*10^-14 / 5.26*10^-4 = 1.90*10^-11 M

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