A 50.0-mL solution of 0.100 M propanoic acid (HC3H5O2; Ka= 1.34 x 10–5) is titrated with 0.200 M NaOH solution. The net ionic equation for the reaction is as follows: HC3H5O2(aq) + OH–(aq) --> C3H5O2–(aq)+ H2O (a) What is the initial pH of the acid solution? (b) Calculate the pH of the resulting solution after each of the following amount of 0.200 M NaOH has been added, and the limiting reactant is completely reacted. (i) 10.0 mL of 0.200 M NaOH; (ii) 15.0 mL of 0.100 M NaOH; (iii) 20.0 mL of 0.100 M NaOH;
A 50.0-mL solution of 0.100 M propanoic acid (HC3H5O2; Ka= 1.34 x 10–5) is titrated with 0.200 M NaOH solution. The net ionic equation for the reaction is as follows: HC3H5O2(aq) + OH–(aq) --> C3H5O2–(aq)+ H2O (a) What is the initial pH of the acid solution? (b) Calculate the pH of the resulting solution after each of the following amount of 0.200 M NaOH has been added, and the limiting reactant is completely reacted. (i) 10.0 mL of 0.200 M NaOH; (ii) 15.0 mL of 0.100 M NaOH; (iii) 20.0 mL of 0.100 M NaOH;
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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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A 50.0-mL solution of 0.100 M propanoic acid (HC3H5O2; Ka= 1.34 x 10–5) is titrated with 0.200 M NaOH solution. The net ionic equation for the reaction is as follows:
HC3H5O2(aq) + OH–(aq) --> C3H5O2–(aq)+ H2O
(a) What is the initial pH of the acid solution?
(b) Calculate the pH of the resulting solution after each of the following amount of 0.200 M NaOH has been added, and the limiting reactant is completely reacted.
(i) 10.0 mL of 0.200 M NaOH;
(ii) 15.0 mL of 0.100 M NaOH;
(iii) 20.0 mL of 0.100 M NaOH;
(iv) 25.0 mL of 0.100 M NaOH;
(v) 30.0 mL of 0.200 M NaOH
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