Answer the following conditions given they represent a titration of 25.00 ml of 0.150 m ch3co2h with 0.150 m naoh. the reaction can be represented as latex: {\text{ch}}_{3}{\text{co}}_{2}\text {h}+{\text{oh}}^{-}\longrightarrow {\text{ch}}_{3}{\text{co}}_{2}{}^{- }+{\text{h}}_{2}\text{o}
what is the ph of the acid solution before any base is added?
what is the ph of the solution after 14.1ml of base is added?
what is the ph of the solution after 25.00ml of base is added?
what is the ph of the solution after 30.04ml of base is added?
what is the ph of the solution after 30.04ml of base is added?

a) pH = 2.793; 0.0mL of NaOH

b) pH = 4.866; after 14.1mL of NaOH

c) pH = 8.816; after 25mL of NaOH

d) pH = 12.914; after 30.04mL of NaOH

Explanation:

a) CH3COOH ↔ CH3COO- + H3O+

∴ Ka = ( [CH3COO-] * [ H3O+] ) / [ CH3COOH ] = 1.75 E-5from literature

mass balance:

⇒ C CH3COOH = [ CH3COO- ] + [ CH3COOH ] = 0.150 M(1)

charge balance:

⇒ [ H3O+ ] = [ CH3COOH ] + [ OH- ] ...where  [ OH- ] is despised since it comes from water

⇒ [ H3O+ ] = [ CH3COO- ](2)

(2) in (1):

⇒ [ CH3COOH ] = 0.150 - [ H3O+ ](3)

(3) and (2) in Ka:

⇒ Ka = 1.75 E-5 = [ H3O+ ]² / ( 0.150 - [ H3O+ ] )

⇒ [ H3O+ ]² + 1.75 E -5 [ H3O+ ] - 2.625 E-6 = 0

⇒[ H3O+ ] = 1.611 E-3 M

⇒ pH = - Log [ H3O+ ]

⇒ pH = 2.793

b) after 14.1 mL of NaOH:

∴ C CH3COOH = (((25)*(0.150)) - ((14.1)*(0.150))) / (25 + 141.1) = (3.75 - 2.115)/39.1 = 0.0418 M

∴ C CH3COONa = ((14.1)*(0.150)) / 39.1 = 0.054 M

mass balance:

⇒ 0.0418 + 0.054 = [ CH3COOH ] + [ CH3COO- ](4)

charge balance:

⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH-]where  [ OH- ] is despised since it comes from water  and [ Na+ ] ≅  C CH3COONa = 0.054 M

⇒ [ H3O+ ] + 0.054 = [ CH3COO- ](5)

(5) in (4):

⇒ [ CH3COOH ] = 0.0968 - ( [H3O+] + 0.054 )

⇒ [ CH3COOH ] = 0.042 - [H3O+](6)

(5) and (6) in Ka:

⇒ 1.75 E-5 = (( [H3O+] * ( [H3O+] + 0.054 )) / ( 0.042 - [H3O+] )

⇒ 1.75 E-5 * ( 0.042 - [H3O+] ) = ( [H3O+]² + 0.054 [H3O+] )

⇒ [H3O+]² + 0.05402 [H3O+] - 7.35 E-7 = 0

⇒ [H3O+] = 1.3602 E-5 M

⇒ pH = 4.866

c) after 25.00 mL of NaOH:

∴ C CH3COOH =  ( 3,75 - 3.75 ) / 50 mL = 0at the equivalence point there is complete salt formation (CH3COONa); Then the pH is calculated through it.

∴ C CH3COONa = ((0.150) * (25.00)) / 50.00mL = 0.075 M

∴ Kh = Kw/Ka = ( [H3O+] * [OH-] ) / (( [H3O+] * [ CH3COO-] ) / [CH3COOH] ))

⇒ Kh = 1 E-14 / 1.75 E-5 = ( [ CH3COOH ] * [ OH-] ) / [ CH3COO-]

mass balance:

⇒ 0.075 = [ CH3COOH ] + [ CH3COO- ](7)

charge balance:

⇒ [ H3O+ ] + 0.075 = [ CH3COO- ] + [OH-] where [ H3O+ ] = 0

⇒[ CH3COO-] = 0.075 - [OH-](8)

(8) in (7):

⇒ [ CH3COOH ] = 0.075 - ( 0.075 - [ OH-] )

⇒ [ CH3COOH ] = [ OH ](9)

(8) and (9) in Kh:

⇒ Kh = [ OH- ]² / ( 0.075 - [OH-] ) = 5.714 E-10

⇒ [ OH- ]² + 5.714 E-10 [ OH- ] - 4.286 E-11 = 0

⇒ [ OH- ] = 6.546 E-6 M

⇒ pOH = 5.184

⇒ pH = 8.816

d) after 30.04mL of NaOH ( excess of OH ):

∴ C CH3COONa = ((30.04)*(0.150) - (25.0)*(0.150))/(25.0 + 30.04) = 0.0137 M

after equivalence point:

CH3COO- + H2O ↔ CH3COOH + OH-

∴ [ CH3COO- ] ≅ C CH3COONa = 0.082 M ≅ [ OH- ]

⇒ pOH = 1.0861

⇒ pH = 12.914