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Given that Kw for water is 2.4× 10–14 at 37 °C, compute the pH of a neutral aqueous solution at 37 °C, which is the normal human body temperature.

Respuesta :

superman1987

Answer:

pH = 6.81.

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⁻] = 2.4 x 10⁻¹⁴.  

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

So, Kw = [H⁺]²

∴ 2.4 x 10⁻¹⁴ = [H⁺]²

 

∴ [H⁺] = 1.55 x 10⁻⁷ M.

∵ pH = - log [H⁺]  

pH = - log (1.55  x 10⁻⁷) = 6.81.

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