Question

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 =

Answer 1

Answer:

equivalence point:

mL KOH = 15.01 mL

pH = 2.401

Explanation:

• KOH ↔ K+  + OH-

      0.0608 M

• HNO2 + H2O ↔ H3O+  +  NO2-
• Ka = ( [ H3O+ ] * [ NO2- ] ) / [ HNO2 ] = 4.5 E-4....from literature
• 2 H2O ↔ H3O+  +  OH-
• Kw = 1 E-14 = [ H3O+ ] * [ OH- ]

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