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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Ответ:
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 + H2Oa) 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
Ответ:
i think the 3,4,2
Explanation: