Question

1) Pyridine is a weak base that is used in the manufacture of pesticides and plastic resins. It is also a component of cigarette smoke. Pyridine ionizes in water as follows: C5H5N+H2O<->C5H5NH++OH?The pKb of pyridine is 8.75. What is the pH of a 0.305M solution of pyridine? Express the pH numerically to two decimal places.Benzoic acid is a weak acid that has antimicrobial properties. Its sodium salt, sodium benzoate, is a preservative found in foods, medications, and personal hygiene products. Benzoic acid ionizes in water: C6H5COOH<->C6H5COO?+H+2) The pKa of this reaction is 4.2. In a 0.66M solution of benzoic acid, what percentage of the molecules are ionized?Express your answer to two significant figures and include the appropriate units. --------------.
3 )Ammonia, NH3, is a weak base with a Kb value of 1.8

Answer 1

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:

$\alpha =\frac{[C_{6}H_{5}COO^{-}]_{eq}}{Ca} .100 \%= \frac{6.4 \times 10^{-3} M }{0.66M} .100 \% = 0.97\%$