The Handbook of Chemistry and Physics gives the following data for the solubilit

Question

The Handbook of Chemistry and Physics gives the following data for the solubility of lead(ll) bromide in water. What is the concentration of lead (ll) bromide in moles per liter in each of these saturated solutions? Using the values from question 3, calculate the solubility product constants for lead (II) bromide at 0, 20, and 100 degree C. Using the values from question 4, calculate Delta G for dissolving lead (II) bromide at 0, 20, and 100 degree C What do the solubilities given in question 3 imply about the sign of Delta H for the dissolving of lead(Il) bromide? Explain.

Explanation / Answer

4)

Ksp = [Pb+2][Br-]^2

[PB+2] = S = 1.24*10^-2 ; 2.3*10^-2; 1.28*10^-1

[Br-] = 2S = 2*(1.24*10^-2 ; 2.3*10^-2; 1.28*10^-1) ; = 2.48*10^-2; 4.6*10^-2; 1.56*10^-1

Ksp(0ºC) = (1.24*10^-2)(2.48*10^-2)^2 = 0.00000762649

Ksp(20ºC) = (2.3*10^-2)(4.6*10^-2)^2 = 0.000048668

KSp(100ºC) = (1.28*10^-1)(1.56*10^-1)^2 = 0.003115008

5)

G = -RT*lnK

G (0ºC) = -8.314*(273)*ln(0.00000762649) = 26746.1381381 J/mol

G(20ºC) = -8.314*293*ln(0.000048668) = 24190.6906461 J/mol

G(100ºC) = -8.314*373*ln(0.003115008) = 17898.1986809 J/mol

6)

As temperature increases; the salt becomes more soluble

therefore, the Hsolubility must be endothermic, that is, it requires energy in the form of heat ( increase of T)

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