Answer:
1) pH = 9.37
2) α = 0.97%
Explanation:
1) Pyridine is a weak base that ionizes in water as follows:
C₅H₅N(aq) + H₂O(l) ⇄ C₅H₅NH⁺(aq) + OH⁻(aq)
Since pKb is 8.75, we can find the equilibrium constant Kb using the following expression:
pKb = -log Kb
Kb = antilog (-pKb) = antilog (-8.75) = 1.78 × 10⁻⁹
If we know Kb and the initial concentration of the base (Cb) we can find [OH⁻] using the following expression:
[OH⁻] = √(Kb × Cb)
[OH⁻] = √(1.78 × 10⁻⁹ × 0.305) = 2.33 × 10⁻⁵ M
Then, we can find pOH and pH.
pOH = -log [OH⁻]
pOH = -log (2.33 × 10⁻⁵) = 4.63
pH + pOH = 14
pH = 14 - pOH = 14 - 4.63 = 9.37
2) Benzoic acid is a weak acid that ionizes in water as follows:
C₆H₅COOH(aq) ⇄ C₆H₅COO⁻(aq) + H⁺(aq)
Since pKa is 4.2, we can find the equilibrium constant Ka using the following expression:
pKa = -log Ka
Ka = antilog (-pKa) = antilog (-4.2) = 6.3 × 10⁻⁵
If we know Ka and the initial concentration of the acid (Ca) we can find [H⁺] and [C₆H₅COO⁻] using the following expression:
[H⁺] = [C₆H₅COO⁻] = √(Ka × Ca)
[H⁺] = [C₆H₅COO⁻] = √(6.3 × 10⁻⁵ × 0.66) = 6.4 × 10⁻³ M
The percentage of ionized molecules (α) is:
[tex]\alpha =\frac{[C_{6}H_{5}COO^{-}]_{eq}}{Ca} .100 \%= \frac{6.4 \times 10^{-3} M }{0.66M} .100 \% = 0.97\%[/tex]
