Chemistry, 26.03.2020 05:55 sandrahg3

(a) Given that Ka for acetic acid is 1.8 X 10^-5 and that for hypochlorous acid is 3.0 X 10^-8, which is the stronger acid? (b) Which is the stronger base, the acetate ion or the hypochlorite ion? (c) Calculate Kb values for CH3COO- and ClO-.

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Acetic acid is stronger than hypochlorous acid. Hypochlorite ion is stronger than acetate ion. Kb for CH₃COO⁻ is 5.6 × 10⁻¹⁰ and the Kb for ClO⁻ is 3.3 × 10⁻⁷.

(a) Given that Ka for acetic acid is 1.8 × 10⁻⁵ and that for hypochlorous acid is 3.0 × 10⁻⁸, which is the stronger acid?

The acid ionization constant (Ka) is a way to measure the strength of an acid. The larger the Ka, the stronger the acid. Thus, acetic acid is stronger than hypochlorous acid.

(b) Which is the stronger base, the acetate ion or the hypochlorite ion?

Acetate ion is the conjugate base of acetic acid and hypochlorite ion is the conjugate base of hypochlorous acid. There is an inverse relationship between the strength of an acid and its conjugate base. The stronger the acid, the weaker the conjugate base. Then, hypochlorite ion is stronger than acetate ion.

Given the Ka of an acid, we can calculate the Kb of its conjugate base using the following expression.

$Ka\times Kb = Kw$

The Kb for CH₃COO⁻ is:

$Kb = \frac{Kw}{Ka} = \frac{1.0 \times 10^{-14} }{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$

The Kb for ClO⁻ is:

$Kb = \frac{Kw}{Ka} = \frac{1.0 \times 10^{-14} }{3.0 \times 10^{-8}} = 3.3 \times 10^{-7}$

Acetic acid is stronger than hypochlorous acid. Hypochlorite ion is stronger than acetate ion. Kb for CH₃COO⁻ is 5.6 × 10⁻¹⁰ and the Kb for ClO⁻ is 3.3 × 10⁻⁷.

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Answer from: romeroalexis817

HOAc is stronger acid than HClO

ClO⁻ is stronger conjugate base than OAc⁻

Kb(OAc⁻) = 5.5 x 10⁻¹⁰

Kb(ClO⁻) = 3.3 x 10⁻⁷

Explanation:

Assume 0.10M HOAc => H⁺ + OAc⁻ with Ka = 1.8 x 10⁻⁵

=> [H⁺] = √Ka·[Acid] =√(1.8 x 10⁻⁵)(0.10) M = 1.3 x 10⁻³M H⁺

Assume 0.10M HClO => H⁺ + ClO⁻ with Ka = 3 x 10⁻⁸

=> [H⁺] = √(3 x 10⁻⁸)(0.10)M = 5.47 x 10⁻⁵M H⁺

HOAc delivers more H⁺ than HClO and is more acidic.

Kb = Kw/Ka, Kw = 1 x 10⁻¹⁴

Kb(OAc⁻) = 5.5 x 10⁻¹⁰

Kb(ClO⁻) = 3.3 x 10⁻⁷

Answer from: Quest

It shows that the experiment is designed properly.

Answer from: Quest

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