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
![Ka = \frac{[C_{6}H_{5}CO_{2}^{-}][H_{3}O^{+}]}{[C_{6}H_{5}CO_{2}H]}](https://tex.z-dn.net/?f=%20Ka%20%3D%20%5Cfrac%7B%5BC_%7B6%7DH_%7B5%7DCO_%7B2%7D%5E%7B-%7D%5D%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5BC_%7B6%7DH_%7B5%7DCO_%7B2%7DH%5D%7D%20)

(2)
By solving equation (2) for x we have:
x = 0.0030 = [C₆H₅CO₂⁻] = [H₃O⁺]
Finally, the pH is:
![pH = -log([H_{3}O^{+}]) = -log (0.0030) = 2.52](https://tex.z-dn.net/?f=%20pH%20%3D%20-log%28%5BH_%7B3%7DO%5E%7B%2B%7D%5D%29%20%3D%20-log%20%280.0030%29%20%3D%202.52%20)
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
![Kb = \frac{[C_{6}H_{5}CO_{2}H][OH^{-}]}{[C_{6}H_{5}CO_{2}^{-}]}](https://tex.z-dn.net/?f=Kb%20%3D%20%5Cfrac%7B%5BC_%7B6%7DH_%7B5%7DCO_%7B2%7DH%5D%5BOH%5E%7B-%7D%5D%7D%7B%5BC_%7B6%7DH_%7B5%7DCO_%7B2%7D%5E%7B-%7D%5D%7D)


By solving the equation above for x, we have:
x = 4.64x10⁻⁶ = [C₆H₅CO₂H] = [OH⁻]
The pH is:
![pOH = -log[OH^{-}] = -log(4.64 \cdot 10^{-6}) = 5.33](https://tex.z-dn.net/?f=%20pOH%20%3D%20-log%5BOH%5E%7B-%7D%5D%20%3D%20-log%284.64%20%5Ccdot%2010%5E%7B-6%7D%29%20%3D%205.33%20)

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:
![pOH = -log([OH^{-}]) = -log(0.067) = 1.17](https://tex.z-dn.net/?f=%20pOH%20%3D%20-log%28%5BOH%5E%7B-%7D%5D%29%20%3D%20-log%280.067%29%20%3D%201.17%20)

I hope it helps you!