Consider the titration of 100.0 mL of .200 M acetic acid(Ka = 1.8 X 10-5 ) by .100 M KOH. Calculate the pH of the resulting solution after the followingvolumes of KOH have been added: a. 0.0 mL b. 50.0 mL c. 100.0 mL d. 150.0 mL e. 200.0 mL f. 250.0 mL Consider the titration of 100.0 mL of .200 M acetic acid(Ka = 1.8 X 10-5 ) by .100 M KOH. Calculate the pH of the resulting solution after the followingvolumes of KOH have been added: a. 0.0 mL b. 50.0 mL c. 100.0 mL d. 150.0 mL e. 200.0 mL f. 250.0 mL
Explanation / Answer
Balanced chemical equation : CH3COOH + KOH -----------> CH3COOK + H2O a ) When there is no base [H+] = 1.8 X 10-5 * 0.2 M = 1.89 * 10 ^ -3 M pH = 2.72 b) CH3COOH + KOH -----------> CH3COOK + H2O Beforerxn 0.1 L * 0.2 M 0.1M * 0.05 L = 0.02 mols 0.005 mols After rxn 0.02 mols - 0.005 mols 0 0.005 = 0.015 mols [H+] = 0.015 mols / 0.15 L = 0.1 M pH = 1 c) CH3COOH + KOH -----------> CH3COOK + H2O Beforerxn 0.1 L * 0.2 M 0.1M * 0.1 L = 0.02 mols 0.01 mols After rxn 0.02 mols - 0.01 mols 0 0.01 = 0.01 mols [H+] = 0.01 mols / 0.2 L = 0.05 M pH = 1.30 Similarly we can calculate for the remaining . Beforerxn 0.1 L * 0.2 M 0.1M * 0.1 L = 0.02 mols 0.01 mols After rxn 0.02 mols - 0.01 mols 0 0.01 = 0.01 mols [H+] = 0.01 mols / 0.2 L = 0.05 M pH = 1.30 Similarly we can calculate for the remaining .