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31.08.2020 •
Chemistry
How many moles of the precipitate would one expect to be produced from 0.030 moles of calcium nitrate
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Ответ:
The molar solubility of NiS is 7.7 * 10⁻⁷ M
Explanation:
To answer this question, we need to keep in mind two equilibriums.
First, we have the solubilization of NiS:
NiS ⇄ Ni²⁺ + S²⁻ ksp= 3.0 * 10⁻²¹ (we know this from standard tables)
Second, we have the formation of the complex:
Ni²⁺ + 6NH₃ ⇄ [Ni(NH₃)₆]²⁺ kf=2.0 * 10⁻⁸
Combine the two equilibriums and we have
NiS + 6NH₃ ⇄ [Ni(NH₃)₆]²⁺ + S²⁻ K= ksp * kf =6.0* 10⁻¹³=
The molar solubility s is equal to both [Ni(NH₃)₆²⁺] and [S²]
At equilibrium, [NH₃]= 3,1 M - 6s
Thus, if we replace those terms in the formula for K, we're left with:
Using an approximation we can ignore the denominator and we have
s²=6.0 * 10⁻¹³s=7.7 * 10⁻⁷