In the laboratory, a general chemistry student measured the pH of a 0.591 M aqueous solution ofbenzoic acid, C₂H5COOH to be 2.199.Use the information she obtained to determine the K₂ for this acid.K₂(experiment) =

Answered: In the laboratory, a general chemistry…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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In the laboratory, a general chemistry student measured the pH of a 0.591 M aqueous solution of
benzoic acid, C₂H5COOH to be 2.199.
Use the information she obtained to determine the K₂ for this acid.
K₂(experiment) =

Expert Solution

Step 1: Formula to calculate Ka:

From the provided $p H$ and concentration, we can use the following formula to get the Ka (acid dissociation constant) of benzoic acid ( $C subscript 6 H subscript 5 C O O H$ ):

$K a equals 10 to the power of negative p H end exponent space cross times space fraction numerator left square bracket H A right square bracket over denominator left square bracket A to the power of minus right square bracket end fraction$

$C subscript 6 H subscript 5 C O O H left parenthesis a q right parenthesis space minus negative negative greater than space C subscript 6 H subscript 5 C O O to the power of minus left parenthesis a q right parenthesis space plus space H to the power of plus left parenthesis a q right parenthesis$

The acid dissociation constant we're looking for is represented as $K a$ .
$p H$ is the solution's measured $p H$ , which in this case is $2.199$
[ $H A$ ] is the undissociated acid's starting concentration, which is $0.591 space M$ for benzoic acid ( $C subscript 6 H subscript 5 C O O H$ ).
[ $A to the power of minus$ ] is the concentration of the conjugate base, which is created when the acid dissociates and forms the benzoate ion, $C subscript 6 H subscript 5 C O O to the power of minus$ .

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