Respuesta :
Answer:
a) before base added:
⇒ pH = 6.64
b) after 10 mL:
⇒ pH = 12.23
c) after 20 mL:
⇒ pH = 11.62
Explanation:
a) before base is added:
- HNO + H20 ↔ H3O+ + NO-
∴ pKa = 11.4....from literature
⇒ Ka = 3.98 E-12 = ( [ NO- ] * [ H3O+ ] ) / [ HNO ]
mass balance:
⇒ C HNO = [ HNO ] + [ NO- ] = 0.013 M............(1)
charge balance:
⇒ [ H3O+ ] = [ NO- ] + [ OH- ]...........where [ OH- ] is neglect, comes from water
⇒ [ H3O+ ] = [ NO- ]..........(2)
(2) and (1) in Ka:
⇒ 3.98 E-12 = [ H3O+ ]² / ( 0.013 - [ H3O+ ] )
⇒ [ H3O+ ]² + 3.98 E-12 [ H3O+ ] - 5.17 E-14 = 0
⇒ [ H3O+ ] = 2.273 E-7 M
⇒ pH = 6.64
b) after adding 10.0 mL:
C2H5NH2 + H2O ↔ C2H5NH3+ + OH-
∴ C HNO = ((15*0.013) - (10*0.017)) / (15+10) = 1 E-3 M
∴ C C2H5NH2 = (10*0.017) / (15+10) = 6.8 E-3 M
mass balance:
⇒ 1 E-3 + 6.8 E-3 = [ HNO ] + [ NO- ]
⇒ [ HNO ] = 7.8 E-3 - [ NO- ].........(3)
charge balance:
⇒ [ H3O+ ] + [ C2H5NH3+ ] = [ NO- ] + [ OH-]
∴ [ C2H5NH3+ ] ≅ C2H5NH2 = 6.8 E-3 M
⇒ [ H3O+ ] + 6.8 E-3 = [NO-] + [ OH- ]....the value of [ OH- ] is neglected
⇒ [ NO- ] = [ H3O+ ] + 6.8 E-3..........(4)
(3) and (4) in Ka:
⇒ 3.98 E-12 = (( H3O+ ] + 6.8 E-3) * [ H3O+ ]) / ( 7.8 E-3 - ( [H3O+ ] + 6.8 E-3 ))
⇒ 3.98 E-12 = ([ H3O+]² + 6.8 E-3 [ H3O+]) / ( 1 E-3 - [ H3O+ ] )
⇒ [ H3O+ ]² + 6.8 E-3 [ H3O+ ] - 3.98 E-15 = 0
⇒[ H3O+ ] = 5.85 E-13 M
⇒ pH = 12.23
c) after adding 20,0 mL of C2H5NH2 (excess of base):
∴ C C2H5NH2 = ((20*0.017) - (15*0.013)) / ( 15+20 ) = 4.143 E-3 M E-3 M
⇒ [ OH- ] ≅ C C2H5NH2 = 4.143 E-3 M
⇒ pOH = 2.382
⇒ pH = 11.62
