calcule a concentração de íons H3O+, em mol.L^-1, em soluções com as concentrações de OH-, A 25°C. Dado: Kw=1,0x10^-14:
a) 0,001 mol/L.
b) 2,0x10^-5 mol/L.
c)5,0x10^-9 mol/L.
Respostas
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Kw = [H₃O⁻] . [OH]
a) Kw = [H₃O⁻] . [OH]
1,0 x 10⁻¹⁴ = [H₃O⁻] . [10⁻³]
[H₃O⁻] = 1,0 x 10⁻¹⁴ / [10⁻³]
[H₃O⁻] = 1,0 x 10⁻¹⁴⁺³
[H₃O⁻] = 1,0 x 10⁻¹¹
b) Kw = [H₃O⁻] . [OH]
1,0 x 10⁻¹⁴ = [H₃O⁻] . 2,0 x 10⁻⁵
[H₃O⁻] = 1,0 x 10⁻¹⁴ / 2,0 x 10⁻⁵
[H₃O⁻] = 1,0 x 10⁻¹⁴⁺⁵
[H₃O⁻] = 1,0 x 10⁻⁹
c) Kw = [H₃O⁻] . [OH]
1,0 x 10⁻¹⁴ = [H₃O⁻] . 5,0 x 10⁻⁹
[H₃O⁻] = 1,0 x 10⁻¹⁴ / 5,0 x 10⁻⁹
[H₃O⁻] = 1,0 x 10⁻¹⁴⁺⁹
[H₃O⁻] = 1,0 x 10⁻⁵
Kw = [H₃O⁻] . [OH]
a) Kw = [H₃O⁻] . [OH]
1,0 x 10⁻¹⁴ = [H₃O⁻] . [10⁻³]
[H₃O⁻] = 1,0 x 10⁻¹⁴ / [10⁻³]
[H₃O⁻] = 1,0 x 10⁻¹⁴⁺³
[H₃O⁻] = 1,0 x 10⁻¹¹
b) Kw = [H₃O⁻] . [OH]
1,0 x 10⁻¹⁴ = [H₃O⁻] . 2,0 x 10⁻⁵
[H₃O⁻] = 1,0 x 10⁻¹⁴ / 2,0 x 10⁻⁵
[H₃O⁻] = 1,0 x 10⁻¹⁴⁺⁵
[H₃O⁻] = 1,0 x 10⁻⁹
c) Kw = [H₃O⁻] . [OH]
1,0 x 10⁻¹⁴ = [H₃O⁻] . 5,0 x 10⁻⁹
[H₃O⁻] = 1,0 x 10⁻¹⁴ / 5,0 x 10⁻⁹
[H₃O⁻] = 1,0 x 10⁻¹⁴⁺⁹
[H₃O⁻] = 1,0 x 10⁻⁵
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