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3 years ago

The kw for water at 0 °c is 0.12× 10–14 m2. Calculate the ph of a neutral aqueous solution at 0 °c.

Chemistry

1 answer:

jekas [21]3 years ago

8 0

Answer:

pH = 7.46.

Explanation:

The ionization of water is given by the equation :

H₂O(l) ⇄ H⁺(aq) + OH⁻(aq),

The equilibrium constant (Kw) expression is:

Kw = [H⁺][OH⁻] = 0.12 x 10⁻¹⁴.

in pure water and neutral aqueous solution, [H⁺] = [OH⁻]

So, Kw = [H⁺]²

∴ 0.12 x 10⁻¹⁴ = [H⁺]²

∴ [H⁺] = 3.4 x 10⁻⁸ M.

∵ pH = - log [H⁺]

pH = - log (3.4 x 10⁻⁸) = 7.46.

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(a) The Lewis structure is shown in the image below.

(b) $CaF_2$

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Explanation:

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