Question

Part E

Find the pH of a 0.140 M solution of a weak monoprotic acid having Ka= 0.12.

Part F

Find the percent ionization of a 0.140 M solution of a weak monoprotic acid having Ka= 0.12.

Answer 1

weak acid = AH molarity = 0.140 M Ka = 0.120 Construct ICE : reaction AH H H + A- Initial(M) 0.140 0 0 Change(M) -x +x +x Equilibrium(M) 0.140-X Equilibrium expression: Ka = [H+][ A- 1/[ AH 0.120 = (x?)/( 0.140 -X) Calculating x Value: x? + 0.120x - 0.016800 1 = 0 By solving the above quadratic equation, we get x = * = 0.08283 M Part-E : Calculating pH: pH = -log[H+] = -log[ 0.08283 ]= 1.08 Calculating Fraction of dissociation: Fraction of dissociation = (Change in concentration /Initial concentration ) Fraction of dissociation = ( 0.08283 / 0.140 ) = 0.5916 Part-F: Calculating percent of dissociation: percent of dissociation = Fraction of dissociation 100 percent of dissociation = 0.5916 x 100 = 59.2 % percent of ionization = 0.5916 x 100 = 59.2 %