Question
Question: What is the molarity of \(Fe\left( {CN} \right)_6^{4 - }\)in a saturated solution of \(A{g_4}\left[ ...
What is the molarity of Fe(CN)64−in a saturated solution of Ag4[Fe(CN)6]? (Ksp=1.6×10−41)
A.1.6×108
B.5.2×10−8
C.2.0×10−8
D.2.3×10−9
Solution
When we have to calculate the molarity of an ion and its solubility product is known, we find the solubility of the ion in the solution as it is equal to the concentration of the ion in the given solution. The solubility product or Ksp represents the equilibrium between a salt and its dissociated ions in a solution.
Complete step by step answer:
The solubility product constant of a salt is the equilibrium constant for the dissolving of a solid salt into water making an aqueous solution. When the solid substance is in equilibrium with its saturated solution, the product of the concentrations of the constituent ions present in the solution gives the solubility product. The solubility product can also be expressed in terms of the solubility of ions formed by dissociation in the solution.
The dissolution of Ag4[Fe(CN)6] can be shown as,
Ag4[Fe(CN)6](s)⇌Fe(CN)64−(aq)+4Ag+(aq)
Solubility at equilibrium | s | s | 4s |
---|
Considering the solubility as s for the salt, the solubilities of the ions is given as s and 4s.
We know that solubility products are given as the product of the concentrations or solubilities of the constituent ions. Hence,
Ksp=s(4s)4
The solubility product is given in the question as Ksp=1.6×10−41.
Substituting the value, we get, Ksp=1.6×10−41=256s5
Solving for solubility s, we get,
s=(2561.6×10−41)1/5
∴s=2.3×10−9 mol/L
Since we know that the concentration and solubility s equal, thus the molarity of this solution is also equal to 2.3×10−9 mol/L
Thus, the correct answer is D.
Note:
While writing the solubility for the constituent ions, care should be taken to multiply them by their respective stoichiometric coefficients and also to raise them to the power of the same when calculating the solubility product.