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

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:

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