Classify each phrase according to whether it applies to photophosphorylation, oxidative phosphorylation, or both
1 answer:
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Answer:
a. pH = 2.52
b. pH = 8.67
c. pH = 12.83
Explanation:
The equation of the titration between the benzoic acid and NaOH is:
C₆H₅CO₂H + OH⁻ ⇄ C₆H₅CO₂⁻ + H₂O (1)
a. To find the pH after the addition of 20.0 mL of NaOH we need to find the number of moles of C₆H₅CO₂H and NaOH:
From the reaction between the benzoic acid and NaOH we have the following number of moles of benzoic acid remaining:
The concentration of benzoic acid is:
Now, from the dissociation equilibrium of benzoic acid we have:
C₆H₅CO₂H + H₂O ⇄ C₆H₅CO₂⁻ + H₃O⁺
0.14 - x x x
(2)
By solving equation (2) for x we have:
x = 0.0030 = [C₆H₅CO₂⁻] = [H₃O⁺]
Finally, the pH is:
b. At the equivalence point, the benzoic acid has been converted to its conjugate base for the reaction with NaOH so, the equilibrium equation is:
C₆H₅CO₂⁻ + H₂O ⇄ C₆H₅CO₂H + OH⁻ (3)
The number of moles of C₆H₅CO₂⁻ is:
The volume of NaOH added is:
The concentration of C₆H₅CO₂⁻ is:
From the equilibrium of equation (3) we have:
C₆H₅CO₂⁻ + H₂O ⇄ C₆H₅CO₂H + OH⁻
0.14 - x x x
By solving the equation above for x, we have:
x = 4.64x10⁻⁶ = [C₆H₅CO₂H] = [OH⁻]
The pH is:
c. To find the pH after the addition of 100 mL of NaOH we need to find the number of moles of NaOH:
From the reaction between the benzoic acid and NaOH we have the following number of moles remaining:
The concentration of NaOH is:
Therefore, the pH is given by this excess of NaOH:
I hope it helps you!