Respuesta :

Willchemistry
Molarity :

M = n / V

M = 7.00 x 10⁻³ / 12.0

= 0.000583 M  => H+

Kw = [ H+] x [OH-]

1 x 10⁻¹⁴ = 0.000583 x [ OH-]

[ OH-] = 1 x 10⁻¹⁴ / 0.000583

[OH-] = 1.71 x 10⁻¹¹

hope this helps!
to find it in the best way is 
7.00×10^−3 mol of HBr / 18.0 L of solution.= 3.89 e-4 Molar HBr 

3.89 e-4 Molar HBr releases 3.89 e-4 Molar H+ 

the pH = - log of 3.89 e-4 Molar H+ 

since [H+] [OH-] = 1 e-14 the Kof water 

[OH-] = 1 e-14 / 3.89 e-4 = 2.57 e-11 
so we concluded that 
[OH-} = 2.57 X 10^-11 

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