Asaturated 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

Equation :

PbF2 > Pb2+ + 2F-

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

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

answer: i dont know the answer but this might

explanation:

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