Odpowiedź:
Wyjaśnienie:
Dane: Szukane:
Ir = 10⁻⁸ [I⁻]=?
W roztworze jest nasyconym:
PbI₂ ⇄ Pb²⁺ + 2I⁻
Ir = [Pb²⁺] * [I⁻]²
[Pb²⁺]=x
[I⁻]= 2x
Ir = x*(2x)²
Ir = 4x³
x³ = Ir/4
x = ∛Ir/4
x = ∛10⁻⁸/4
x = 1,3572*10⁻³mol/dm³
[I⁻] = 2*1,3572*10⁻³
[I⁻]= 2,7144*10⁻³mol/dm³ ≅ 2,71*10⁻³mol/dm³
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Odpowiedź:
Wyjaśnienie:
Dane: Szukane:
Ir = 10⁻⁸ [I⁻]=?
W roztworze jest nasyconym:
PbI₂ ⇄ Pb²⁺ + 2I⁻
Ir = [Pb²⁺] * [I⁻]²
[Pb²⁺]=x
[I⁻]= 2x
Ir = x*(2x)²
Ir = 4x³
x³ = Ir/4
x = ∛Ir/4
x = ∛10⁻⁸/4
x = 1,3572*10⁻³mol/dm³
[I⁻] = 2*1,3572*10⁻³
[I⁻]= 2,7144*10⁻³mol/dm³ ≅ 2,71*10⁻³mol/dm³