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snow_tiger [21]
4 years ago
15

The K w for water at 0 ∘ C is 0.12 × 10 − 14 M 2 . Calculate the pH of a neutral aqueous solution at 0 ∘ C . p H = Is a pH = 7.2

5 solution acidic, basic, or neutral at 0 ∘ C ?
Chemistry
1 answer:
labwork [276]4 years ago
5 0

Answer:

pH → 7.46

Explanation:

We begin with the autoionization of water. This equilibrium reaction is:

2H₂O  ⇄   H₃O⁺  +  OH⁻            Kw = 1×10⁻¹⁴         at 25°C

Kw = [H₃O⁺] . [OH⁻]

We do not consider [H₂O] in the expression for the constant.

[H₃O⁺] = [OH⁻] = √1×10⁻¹⁴   →  1×10⁻⁷ M

Kw depends on the temperature

0.12×10⁻¹⁴ = [H₃O⁺] . [OH⁻]  → [H₃O⁺] = [OH⁻]         at 0°C

√0.12×10⁻¹⁴ = [H₃O⁺] → 3.46×10⁻⁸ M

- log [H₃O⁺] = pH

pH = - log 3.46×10⁻⁸ → 7.46

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<h3>Further explanation</h3>

Given

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Solution

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Cathode(reduction-negative pole) : 2H₂O(l)+2e⁻ ⇒ H₂(g)+2OH⁻(aq)

Anode(oxidation-positive pole) : 2H₂O(l)⇒O₂(g)+4H⁻(aq)+4e⁻

Total reaction : 2H₂O(l)⇒2H₂(g)+O₂(g)

So at the cathode H₂ gas is produced

Faraday : 1 mole of electrons (e⁻) contains a charge of 96,500 C

\tt mol~e^-=\dfrac{Q}{96500}

Q = i.t

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mol e⁻ = 8100 : 96500 = 0.084

From equation at cathode , mol ratio e⁻ : H₂ = 2 : 1, so mol H₂ = 0.042

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3 years ago
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