Question

Identify the acid and conjugate base in each reaction. Calculate the pK_A for each acid. List them in order from the strongest to weakest acid.

HC₂H₃O₂ + H₂O --> H₃O⁺ + C₂H₃O₂^-, K_A = 1.7 x 10^-5

HC₇H₅O₂ + H₂O --> H₃O⁺ + C₇H₅O₂^-, K_A = 6.3 x 10^-5

HC₆H₄NO₂+H₂O --> H₃O⁺ + C₆H₄NO₂^-, K_A = 1.4 x 10^-5

Answer 1

1)

HC₂H₃O₂ + H₂O ------------> H₃O⁺ + C₂H₃O₂^-, K_A = 1.7 x 10^-5

acid=HC₂H₃O₂                                    

conjugate base= C₂H₃O₂^-,

pKa = -log Ka

pKa = -log (1.7 x 10^-5)

pKa = 4.77

2)

HC₇H₅O₂ + H₂O --> H₃O⁺ + C₇H₅O₂^-, K_A = 6.3 x 10^-5

acid = HC₇H₅O₂

conjugate base= C₇H₅O₂^-,

pKa = -log Ka

pKa = -log ( 6.3 x 10^-5)

pKa = 4.20

3)

HC₆H₄NO₂+H₂O --> H₃O⁺ + C₆H₄NO₂^-, K_A = 1.4 x 10^-5

acid = HC₆H₄NO₂

conjugate base= C₆H₄NO₂^-,

pKa = -log Ka

pKa = -log ( 1.4 x 10^-5)

pKa = 4.85

acidity order : HC₇H₅O₂ > HC₂H₃O₂ >HC₆H₄NO₂

Ka value is increases acidity also increases