Answered

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Encuentre el pH, pOH, [H⁺] y [OH⁻] de 1.7 × 10⁻⁴ M de Sr(OH)₂:

Sr(OH)₂ → Sr²⁺ + 2OH⁻

1.7 × 10⁻⁴ M × 2 / 1 = 3.4 × 10⁻⁴ M = [OH⁻]

pOH = -log(3.4 × 10⁻⁴) = 3.46
pH = 14 - pOH = 14 - 3.46 = 10.54

[H⁺] = 10⁻³.46 = 3.4 × 10⁻⁴ M
[OH⁻] = 10⁻¹⁰.⁵⁴ = 2.9 × 10⁻¹¹ M

Sagot :

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