Respuesta :
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
Based on the data provided, the volume of KOH required is 15.01 mL and the pH at equivalence point is 2.40
What is the pH at equivalence point?
At equivalence point, equal amounts of KOH and HNO2 will have reacted.
The volume of KOH at equivalence point is determined as follows:
- Moles = molarity × volume
Moles of HNO2 = moles of KOH
0.039 M × 23.4 mL = 0.0608 × Volume of KOH
Volume of KOH = 15.01 mL
Calculating the pH at equivalence point:
- Ka of HNO2 = 4.5 × 10^-4
From the dissociation equation of HNO2;
- Ka = ( [ H3O+ ] * [ NO2- ] ) / [ HNO2 ]
At equilibrium:
0.039 M = [ HNO2 ] + [ NO2- ]
Also;
[ H3O+ ] = [NO2- ] + [OH-]
where [ OH- ] is negligible
Thus:
[ H3O+ ] = [NO2-]
Substituting [NO2-] above;
[ HNO2 ] = 0.039 - [ H3O+ ]
Therefore
Ka = [ H3O+ ]² / ( 0.039 - [ H3O+ ] ) = 4.5 × 10^-4
[ H3O+ ]² = (0.039 - [ H3O+ ]) × 4.5 × 10^-4
[ H3O+ ]² + 4.5 × 10^-4 [ H3O+ ] - 1.755 ×10^-5 = 0
[ H3O+ ] = 3.97 ×10^-3 M
Thus:
pH = - log (3.97 ×10^-3)
pH = 2.40
Therefore, the pH of the solution at equivalence point is 2.40
Learn more about pH and equivalence point at: https://brainly.com/question/14501686
