Be sure to answer all parts. find the ph of the equivalence point and the volume (ml) of 0.0608 m koh needed to reach the equivalence point in the titration of 23.4 ml of 0.0390 m hno2. volume: ml koh ph =

equivalence point:

mL KOH = 15.01 mL

pH = 2.401

Explanation:

      0.0608 M

equivalence point:

⇒  mn HNO2 = mn KOH

⇒ ( 0.039 mn/mL ) * ( 23.4 mL ) = ( 0.0608 mn/mL ) * VKOH

⇒ VKOH = 15.01 mL

mass balance:

⇒ 0.039 M = [ HNO2 ] + [ NO2- ] (1)

charge balance:

⇒ [ H3O+ ] = [ NO2- ] + [ OH- ]where [ OH- ] comes from water is negligible

⇒ [ H3O+ ] = [ NO2- ] (2)

(2) in (1):

[ HNO2 ] = 0.039 - [ H3O+ ](3)

(3) and (2) in Ka:

⇒ Ka = [ H3O+ ]² / ( 0.039 - [ H3O+ ] ) = 4.5 E-4

⇒ [ H3O+ ]² + 4.5 E-4 [ H3O+ ] - 1.755 E-5 = 0

⇒ [ H3O+ ] = 3.97 E-3 M

⇒ pH = - Log [ H3O+ ]

⇒ pH = 2.401

it is the most commonly accepted one