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A saturated solution of lead(II) fluoride, PbF2, was prepared by dissolving solid PbF2 in water. The concentration of Pb2+ ion in the solution was found to be 2.08×10−3 M . Calculate Ksp for PbF2.

Answer :

pstnonsonjoku

Answer:

3.6×10-8 mol2dm-6

Explanation:

The solubility product is the equilibrium product of ions dissolved in water. It is an equilibrium constant obtained as the product of the molar solubility of the ionic species in solution raised to the appropriate power of their stoichiometric coefficients as shown in the image attached. See attached image for detailed solution.

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Answer:

Explanation:

Equation of reaction:

PbF2(s) --> Pb2+(aq) + 2F-(aq)

Solubility product is given as:

Ksp = [Pb2+][F-]^2

Let x = [Pb2+]

Ksp = x × (2x)^2

= 4 × (x)^3

= 4 × (2.08 × 10^-3)^3

Ksp = 3.60 × 10^-8 M^3

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