Answer:
a) pH = 2.95
b) pH = 8.94
c) pH = 7.00
d) pH = 4.89
Explanation:
a) The reaction is:
CH₃CH₂COOH(aq) + H₂O(l) ⇄ CH₃CH₂COO⁻(aq) + H₃O⁺(aq)(1)
At equlibrium: 0.1 - x x x
The dissociation constant of equation (1), Ka, is:
[tex] K_{a} = \frac{[C_{3}H_{5}O_{2}^{-}][H_{3}O^{+}]}{[C_{3}H_{6}O_{2}]} [/tex]
[tex] 1.3\cdot 10^{-5} = \frac{x^{2}}{(0.1 - x)} [/tex] (2)
Solving equation (2) for x, we have:
x = 0.00113 = [H₃O⁺] = [CH₃CH₂COO⁻]
Hence, the pH is:
[tex] pH = -log [H_{3}O^{+}] = -log (0.00113) = 2.95 [/tex]
b) The reaction is:
CH₃CH₂COO⁻(aq) + H₂O(l) ⇄ CH₃CH₂COOH(aq) + OH⁻(aq)
At equlibrium: 0.1 - x x x
The dissociation constant of the above equation, Kb, is:
[tex] K_{b} = \frac{[C_{3}H_{6}O_{2}][OH^{-}]}{[C_{3}H_{5}O_{2}^{-}]} [/tex]
Also, we have that:
[tex]K_{w} = K_{a}*K_{b} \rightarrow K_{b} = \frac{K_{w}}{K_{a}} = \frac{1 \cdot 10^{-14}}{1.3 \cdot 10^{-5}} = 7.69 \cdot 10^{-10} [/tex]
[tex] 7.69\cdot 10^{-10} = \frac{x^{2}}{0.1 - x} [/tex] (3)
Solving equation (3) for x, we have:
x = 8.77x10⁻⁶ = [OH⁻] = [CH₃CH₂COOH]
Hence, the pH is:
[tex] pOH = -log[OH^{-}] = -log(8.77\cdot 10^{-6}) = 5.06 \rightarrow pH = 14 - pOH = 8.94 [/tex]
c) Pure water:
H₂O ⇄ H⁺ + OH⁻
[tex] K_{w} = [H^{+}][OH^{-}] \rightarrow 1\cdot 10^{-14} = [H^{+}][OH^{-}] [/tex]
Since is pure water: [H⁺] = [OH⁻]
[tex] 1\cdot 10^{-14} = [H^{+}]^{2} \rightarrow [H^{+}] = \sqrt{1\cdot 10^{-14}} = 1 \cdot 10^{-7} [/tex]
Therefore, the pH is:
[tex] pH = -log [H^{+}] = -log (1 \cdot 10^{-7}) = 7.00 [/tex]
d) The reaction is:
CH₃CH₂COOH(aq) + H₂O(l) ⇄ CH₃CH₂COO⁻(aq) + H₃O⁺(aq)
At equlibrium: 0.1 0.1 x
The pH is:
[tex] pH = pKa + log(\frac{[C_{3}H_{5}O_{2}^{-}]}{[C_{3}H_{6}O_{2}]}) = -log(1.3 \cdot 10^{-5}) + log(\frac{0.1}{0.1}) = 4.89 [/tex]
I hope it helps you!
