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

pH = 7.46.

Explanation:

H₂O(l) ⇄ H⁺(aq) + OH⁻(aq),

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.

i think its 100.

hope this !