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What is the percent ionization for each of the following acids?

a. 0.022 M HClO2 solution of pH= 3.88
Answer obtained: .60%

pH = 10^(-3.88) = 1.32 x 10^-4

% ionization= [H+] formed divided by
MHA (original acid concent) x 100
= (1.32x10^-4)/0.022 x 100
= .60%

b. 0.0027 M HClO2 solution of pH=4.70
Answer obtained from solving: .44%

pH = 10^(-4.70) = 1.20 x 10^-5

% ionization
= (1.20x1-^-5)/0.0027 x 100
= .44%

Question ID
525078

Created
April 4, 2011 3:11am UTC

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0

URL
https://questions.llc/questions/525078

Answers
2

Views
2206

2 answers

  1. I can't check your answers because you didn't provide the Ka for HClO2. I don't have it listed in any of my references.

    Answer ID
    525099

    Created
    April 4, 2011 4:28am UTC

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    0

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  2. B is wrong pH is .00001953 =2.0x10^-5
    2.0x10^-5/.0027 = .74%
    6 1/2 yrs and nobody saw this?

    Answer ID
    1634457

    Created
    October 26, 2017 9:45pm UTC

    Rating
    1

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