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;

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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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