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A50.0 ml solution of a weak acid HA (0.100 mol dm^-3) is titrated with a NaOH solution (0.100 mol dm^-3). What is the pH of the solution when 10.0 ml of the base solution has been added? K = 1 * 10^-5 mol dm^-3.

A50.0 ml solution of a weak acid HA (0.100 mol dm^-3) is titrated with a NaOH solution (0.100 mol dm^-3). What is the pH of the solution when 10.0 ml of the base solution has been added? K = 1 * 10^-5 mol dm^-3.

Chemistry
Chemistry
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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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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A 50.0 ml solution of a weak acid HA (0.100 mol dm^-3) is titrated with a NaOH solution (0.100 mol dm^-3). What is the pH of the solution when 10.0 ml of the base solution has been added? K = 1 * 10^-5 mol dm^-3.
Transcribed Image Text:A 50.0 ml solution of a weak acid HA (0.100 mol dm^-3) is titrated with a NaOH solution (0.100 mol dm^-3). What is the pH of the solution when 10.0 ml of the base solution has been added? K = 1 * 10^-5 mol dm^-3.
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V o l u m e space o f space t h e space w e a k space a c i d space equals space 50.0 space m L thin space M o l a r i t y space o f space t h e space w e a k space a c i d comma space H A thin space equals space 0.100 space m o l space d m to the power of blank subscript negative 3 end subscript end exponent W e space h a v e space t o space c a l c u l a t e space t h e space p H thin space o f space t h e space s o l u t i o n space a f t e r space a d d i t i o n space o f space 10.0 space m L thin space o f space N a O H thin space left parenthesis 0.100 space m o l space d m to the power of negative 3 end exponent right parenthesis K thin space equals space 1 space cross times 10 to the power of negative 5 end exponent space m o l space d m to the power of negative 3 end exponent

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