Ph De Hcl 10(-8)

Calcular el pH de 10 -8 M HCl.
If we use the relation, pH = – log [H 3 O + ], we get pH equal to 8. Si usamos la relación, pH = - log [H 3 O +], se obtiene un pH igual a 8. But thisis not correct because an acidic solution cannot have pH greater than 7. Pero esto no es correcto, ya que una solución ácida que no pueden tener un pH mayor que 7. It may be notedthat in very dilute acidic solution, when H + concentrations from acid and water are comparable, the concentration of H + from water cannot be neglected. Cabe señalar que en muy diluidosolución ácida, cuando las concentraciones de H + ácido y el agua son comparables, la concentración de H + desde el agua no se puede despreciar.

Therefore, Por lo tanto,
[H + ]total = [H + ] acid + [H + ] water [H +] total = [H +] de ácido + [H +] del agua
Since HCl is a strong acid and is completely ionized Dado que el HCl es un ácido fuerte y estácompletamente ionizado
[H + ] HCl = 1.0 x 10 -8 [H +] HCl = 1,0 x 10 -8
The concentration of H + from ionization is equal to the [OH - ] from water, La concentración de H + de ionizaciónes igual a la [OH -] a partir de agua,
[H + ] H 2 O = [OH - ] H 2 O [H +] H 2 O = [OH -] H 2 O
= x (say) = X (por ejemplo)
[H + ] total = 1.0 x 10 -8 + x [H +] total = 1,0 x 10 -8+ x
But Pero
[H + ] [OH - ] = 1.0 x 10 -14 [H +] [OH -] = 1,0 x 10 -14
(1.0 x 10 -8 + x) (x) = 1.0 x 10 -14 (1,0 x 10 -8 + x) (x) = 1,0 x 10 -14
X 2 + 10 -8 x – 10 -14 = 0 X 2+ 10 -8 x - 10 -14 = 0
Solving for x, we get x = 9.5 x 10 -8 Despejando x, obtenemos x = 9,5 x 10 -8
Therefore, Por lo tanto,
[H + ] = 1.0 x 10 -8 + 9.5 x 10 -8 [H +] = 1,0 x 10-8 + 9,5 x 10 -8
= 10.5 x 10 -8 = 10,5 x 10 -8
= 1.05 x 10 -7 = 1,05 x 10 -7
pH = – log [H + ] = – log (1.05 x 10 -7 ) = 6.98 pH = - log [H +] = - log (1,05 x 10 -7) = 6,98