The kw for water at 0 °c is 0.12× 10–14 m2. Calculate the ph of a neutral aqueous solution at 0 °c.
1 answer:
Answer:
pH = 7.46.
Explanation:
- The ionization of water is given by the equation :
<em>H₂O(l) ⇄ H⁺(aq) + OH⁻(aq),</em>
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- 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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