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:
![\eta_{NaOH} = C*V = 0.250 M*0.020 L = 5.00 \cdot 10^{-3} moles](https://tex.z-dn.net/?f=%20%5Ceta_%7BNaOH%7D%20%3D%20C%2AV%20%3D%200.250%20M%2A0.020%20L%20%3D%205.00%20%5Ccdot%2010%5E%7B-3%7D%20moles%20)
From the reaction between the benzoic acid and NaOH we have the following number of moles of benzoic acid remaining:
![\eta_{C_{6}H_{5}CO_{2}H} = \eta_{C_{6}H_{5}CO_{2}H}i - \eta_{NaOH} = 0.015 moles - 5.00 \cdot 10^{-3} moles = 0.01 moles](https://tex.z-dn.net/?f=%5Ceta_%7BC_%7B6%7DH_%7B5%7DCO_%7B2%7DH%7D%20%3D%20%5Ceta_%7BC_%7B6%7DH_%7B5%7DCO_%7B2%7DH%7Di%20-%20%5Ceta_%7BNaOH%7D%20%3D%200.015%20moles%20-%205.00%20%5Ccdot%2010%5E%7B-3%7D%20moles%20%3D%200.01%20moles)
The concentration of benzoic acid is:
![C = \frac{\eta}{V} = \frac{0.01 moles}{(0.020 + 0.050) L} = 0.14 M](https://tex.z-dn.net/?f=%20C%20%3D%20%5Cfrac%7B%5Ceta%7D%7BV%7D%20%3D%20%5Cfrac%7B0.01%20moles%7D%7B%280.020%20%2B%200.050%29%20L%7D%20%3D%200.14%20M%20)
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)
![Ka = \frac{x*x}{0.14 - x}](https://tex.z-dn.net/?f=%20Ka%20%3D%20%5Cfrac%7Bx%2Ax%7D%7B0.14%20-%20x%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:
![\eta_{C_{6}H_{5}CO_{2}^{-}} = \eta_{C_{6}H_{5}CO_{2}H}i = 0.015 moles](https://tex.z-dn.net/?f=%20%5Ceta_%7BC_%7B6%7DH_%7B5%7DCO_%7B2%7D%5E%7B-%7D%7D%20%3D%20%5Ceta_%7BC_%7B6%7DH_%7B5%7DCO_%7B2%7DH%7Di%20%3D%200.015%20moles%20)
The volume of NaOH added is:
The concentration of C₆H₅CO₂⁻ is:
![C = \frac{\eta}{V} = \frac{0.015 moles}{(0.060 L + 0.050 L)} = 0.14 M](https://tex.z-dn.net/?f=%20C%20%3D%20%5Cfrac%7B%5Ceta%7D%7BV%7D%20%3D%20%5Cfrac%7B0.015%20moles%7D%7B%280.060%20L%20%2B%200.050%20L%29%7D%20%3D%200.14%20M%20)
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)
![(\frac{Kw}{Ka})*(0.14 - x) - x^{2} = 0](https://tex.z-dn.net/?f=%20%28%5Cfrac%7BKw%7D%7BKa%7D%29%2A%280.14%20-%20x%29%20-%20x%5E%7B2%7D%20%3D%200%20)
![(\frac{1.00 \cdot 10^{-14}}{6.5 \cdot 10^{-5}})*(0.14 - x) - x^{2} = 0](https://tex.z-dn.net/?f=%20%28%5Cfrac%7B1.00%20%5Ccdot%2010%5E%7B-14%7D%7D%7B6.5%20%5Ccdot%2010%5E%7B-5%7D%7D%29%2A%280.14%20-%20x%29%20-%20x%5E%7B2%7D%20%3D%200%20)
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)
![pH = 14 - pOH = 14 - 5.33 = 8.67](https://tex.z-dn.net/?f=%20pH%20%3D%2014%20-%20pOH%20%3D%2014%20-%205.33%20%3D%208.67%20)
c. To find the pH after the addition of 100 mL of NaOH we need to find the number of moles of NaOH:
![\eta_{NaOH}i = C*V = 0.250 M*0.100 L = 0.025 moles](https://tex.z-dn.net/?f=%20%5Ceta_%7BNaOH%7Di%20%3D%20C%2AV%20%3D%200.250%20M%2A0.100%20L%20%3D%200.025%20moles%20)
From the reaction between the benzoic acid and NaOH we have the following number of moles remaining:
The concentration of NaOH is:
![C = \frac{\eta}{V} = \frac{0.010 moles}{0.100 L + 0.050 L} = 0.067 M](https://tex.z-dn.net/?f=%20C%20%3D%20%5Cfrac%7B%5Ceta%7D%7BV%7D%20%3D%20%5Cfrac%7B0.010%20moles%7D%7B0.100%20L%20%2B%200.050%20L%7D%20%3D%200.067%20M%20)
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)
![pH = 14 - pOH = 12.83](https://tex.z-dn.net/?f=%20pH%20%3D%2014%20-%20pOH%20%3D%2012.83%20)
I hope it helps you!