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inessss [21]
3 years ago
9

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 :

<em>H₂O(l) ⇄ H⁺(aq) + OH⁻(aq),</em>

<em></em>

  • The equilibrium constant (Kw) expression is:

<em>Kw = [H⁺][OH⁻] = 0.12 x 10⁻¹⁴.  </em>

<em></em>

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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<h3>What is effusion?</h3>

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