A15.0 ml sample of 0.013 m hno is titrated with 0.017mm kb= 3.9 x 10-10. determine the ph at these points: 02 m at the beginning (before base is added) ph = -log(h₂o) = log (0.013) a 1-886 ph after adding 10.0 ml of czhsnh2 ph after adding 20.00 ml of c2h5nh2

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:

∴ pKa = 11.4from 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