Chemistry, 29.08.2019 16:30 nessamayjuniorp2b9yc

Be sure to answer all parts. find the ph during the titration of 20.00 ml of 0.1000 m butanoic acid, ch3ch2ch2cooh (ka = 1.54 10 ), with 0.1000 m naoh solution after the following additions of titrant. (a) 14.00 ml: ph = (b) 20.50 ml: ph = (c) 30.00 ml: ph =

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Answer from: Ashley606hernandez

a) 14.00 mL NaOH; pH = 5.184

b) 20.50 mL NaOH; pH = 6.444

c) 30.00 mL NaOH; pH = 5.289

Explanation:

CH3CH2CH2COOH + H2O ↔ CH3CH2CH2COO- + H3O+

∴ Ka = ( [ H3O+ ] * [ CH3CH2CH2COO- ] ) / [ CH3CH2CH2COOH ] = 1.54 E-5

CH3CH2CH2COOH + NaOH ↔ CH3CH2CH2COONa + H2O

a) 14,00 mL NaOH:

∴ C CH3CH2CH2COOH = ( moles of acid - moles of base ) / total volumen

⇒ C CH3CH2CH2COOH = ( ( 20 * 0.1 ) - ( 0.1 * 14 ) ) / ( 20 + 14 ) = 0.0176 M

∴ C CH3CH2CH2COONa = moles of base (L.R) / total volume

⇒C CH3CH2CH2COONa = ( 0.1 * 14 ) / ( 20 + 14 ) = 0.0412 M

mass balance:

⇒ 0.0176 + 0.0412 = [ CH3CH2CH2COOH ] + [ CH3CH2CH2COO- ]

⇒ 0.0587 = [ CH3CH2CH2COOH ] + [ CH3CH2CH2COO- ](1)

charge balance:

⇒[ H3O+ ] + [ Na+ ] = [ CH3CH2CH2COO- ]

∴ [ Na+ ] ≅ C CH3CH2CH2COONa = 0.0412 M

⇒ [ H3O+ ] + 0.0412 = [ CH3CH2CH2COO- ](2)

(2) in (1):

⇒ [ CH3CH2CH2COOH ] = 0.0587 - ( [ H3O+ ] + 0.0412 )

⇒ [ CH3CH2CH2COOH ] = 0.0175 - [ H3O+ ](3)

(2) and (3) in Ka:

⇒ Ka = ( [ H3O+ ] * ( [ H3O+ ] + 0.0412 ) ) / ( 0.0175 - [ H3O+ ] ) = 1.54 E-5

⇒ [ H3O+ ]² + 0.0412 [ H3O+ } = 2.695 E-7 - 1.54 E-5 [ H3O+ ]

⇒ [ H3O+ ]² + 0.0412 [ H3O+ ] - 2.695 E-7 = 0

⇒[ H3O+ ] = 6.54 E-6 M

∴ pH = - Log [ H3O+ ]

⇒ pH = 5.184

b) 20.50 mL NaOH:

⇒ C CH3CH2CH2COOH = 1.235 E-3 M

⇒ C CH3CH2CH2COONa = 0.0506 M

mass balance:

⇒ [ CH3CH2CH2COOH ] = 0.0518 - [ CH3CH2CH2COO- ](4)

charge balance:

⇒ [ H3O+ ] + 0.0506 = [ CH3CH2CH2COO- ](5)

(5) in (4):

⇒[ CH3CH2CH2COOH ] = 0.0518 - ( [ H3O+ ] + 0.0506 )(6)

(5) and (6) in Ka:

⇒1.54 E-5 = ( [ H3O+ ] * ( [ H3O+ ] + 0.0506 ) ) / ( 1.183 E-3 - [ H3O+ ] )

⇒ [ H3O+ ]² + 0.0506 [ H3O+ } = 1.821 E-8 - 1.54 E-5 [ H3O+ ]

⇒ [ H3O+ ]² + 0.0506 [ H3O+ ] - 1.821 E-8 = 0

⇒ [ H3O+ ] = 3.599 E-7 M

⇒ pH = 6.444

c) 30.0 mL NaOH:

⇒ C CH3CH2CH2COOH = 0.02 M

⇒ C CH3CH2CH2COONa = 0.06 M

mass balance:

⇒ [ CH3CH2CH2COOH ] = 0.08 - [ CH3CH2CH2COO- ](7)

charge balance:

[ H3O+ ] + 0.06 = [ CH3CH2CH2COO- ](8)

(8) in (7):

⇒[ CH3CH2CH2COOH ] = 0.08 - ( [ H3O+ ] + 0.06 )

⇒ [ CH3CH2CH2COOH ] = 0.02 - [ H3O+ ](9)

(7) and (9) in Ka:

⇒ 1.54 E-5 = ( [ H3O+ ] * ( [ H3O+ ] + 0.06 ) ) / ( 0.02 - [ H3O+ ] )

⇒ [ H3O+ ]² + 0.06 [ H3O+ ] = 3.08 E-7 - 1,54 E-5 [ H3O+ ]

⇒ [ H3O+ ]² + 0.06 [ HJ3O+ ] - 3.08 E-7 = 0

⇒ [ H3O+ ] = 5.133 E-6 M

⇒ pH = 5.289

Answer from: Quest

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Answer from: Quest

possibly, yes.

explanation:

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