Respuesta :
Here is the full question
Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (K a = 1.54 × 10 − 5), with 0.1000 M NaOH solution after the following additions of titrant (total volume of added base given):
a) 10.00 mL
pH =
b) 20.10 mL
pH =
c) 25.00 mL
pH =
Answer:
pH = 4.81
pH = 10.40
pH = 12.04
Explanation:
a)
Number of moles of butanoic acid
= [tex]20.00 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}[/tex]
= 0.002000 mol
Number of moles of NaOH added
= [tex]10.00 \ mL * \frac{L}{1000 \ mL }* \frac{0.1000 \ mol }{L}[/tex]
= 0.001000 mol
pKa of butanoic acid = - log Ka
= - log ( 1.54 × 10⁻⁵)
= 4.81
Equation for the reaction is expressed as follows:
CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O
The ICE Table is expressed as follows:
CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O
Initial 0.002000 0.001000 0
Change - 0.001000 - 0.001000 + 0.001000
Equilibrium 0.001000 0 0.001000
Total Volume = (20.00 + 10.00 ) mL
= 30.00 mL = 0.03000 L
Concentration of [CH₃CH₂CH₂COOH] = [tex]\frac{0.001000 \ mol}{ 0.03000 \ L }[/tex]
= 0.03333 M
Concentration of [CH₃CH₂COO⁻] = [tex]\frac{0.001000 \ mol}{ 0.03000 \ L}[/tex]
= 0.03333 M
By Henderson- Hasselbalch equation
pH = pKa + log [tex]\frac{conjugate \ base}{acid }[/tex]
pH = pKa + log [tex]\frac{CH_3CH_2CH_2COO^-}{CH_3CH_2CH_2COOH}[/tex]
PH = 4.81 + log [tex]\frac{0.03333}{0.03333}[/tex]
pH = 4.81
Thus; the pH of the resulted buffer solution after 10.00 mL of NaOH was added = 4.81
b )
After the equivalence point, we all know that the pH of the solution will now definitely be determined by the excess H⁺
Number of moles of butanoic acid
= [tex]20.00 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}[/tex]
= 0.002000 mol
Number of moles of NaOH added
= [tex]20.10 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}[/tex]
= 0.002010 mol
Following the previous equation of reaction , The ICE Table for this process is as follows:
CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O
Initial 0.002000 0.002010 0
Change - 0.002000 -0.002000 + 0.002000
Equilibrium 0 0.000010 0.002000
We can see here that the base is present in excess;
NOW, number of moles of base present in excess
= ( 0.002010 - 0.002000) mol
= 0.000010 mol
Total Volume = (20.00 + 20.10 ) mL
= 40.10 mL × [tex]\frac{1 \ L}{1000 \ mL }[/tex]
= 0.04010 L
Concentration of acid [OH⁻] = [tex]\frac{0.000010 \ mol}{0.04010 \ L }[/tex]
= [tex]2.494*10^{-4} M[/tex]
Using the ionic product of water:
[tex][H_3O^+] = \frac{K \omega }{[OH^-]}[/tex]
where
[tex]K \omega = 10^{-14}[/tex]
[tex][H_3O^+] = \frac{1.0*10^{-14}}{2.494*10^{-14}}[/tex]
= [tex]4.0*10^{-11}M[/tex]
pH = - log [tex][H_3O^+}][/tex]
pH = - log [[tex]4.0*10^{-11}M[/tex]]
pH = 10.40
Thus, the pH of the solution after the equivalence point = 10.40
c)
After the equivalence point, pH of the solution is determined by the excess H⁺.
Number of moles of butanoic acid
= [tex]20.00 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}[/tex]
= 0.002000 mol
Number of moles of NaOH added
= [tex]25.00 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}[/tex]
= 0.002500 mol
From our chemical equation; The ICE Table can be illustrated as follows:
CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O
Initial 0.002000 0.002500 0
Change - 0.002000 -0.002000 +0.002000
Equilibrium 0 0.000500 0.002000
Base is present in excess
Number of moles of base present in excess = [ 0.002500 - 0.002000] mol
= 0.000500 mol
Total Volume = ( 20.00 + 25.00 ) mL
= 45.00 mL
= 45.00 × [tex]\frac{1 \ L}{1000 \ mL }[/tex]
= 0.04500 L
Concentration of acid [OH⁻] = [tex]\frac{0.0005000 \ mol}{ 0.04500 \ L }[/tex]
= 0.01111 M
Using the ionic product of water [tex][H_3O^+] = \frac{K \omega }{[OH^+]}[/tex]
= [tex]\frac{1.0*10^{-14}}{0.01111}[/tex]
= [tex]9.0*10^{-13}[/tex] M
pH = - log [tex][H_3O^+}][/tex]
pH = - log [[tex]9.0*10^{-13}M[/tex]]
pH = 12.04
Thus, the pH of the solution after the equivalence point = 12.04
