The following relationship between solubility of barium bromide and temperature

Question

The following relationship between solubility of barium bromide and temperature was determined by measuring the temperature at which barium bromide solutions of different concentration reached equilibrium and formed crystalline BaBr_2 solid. Using the relationship above, determine if the system is in equilibrium when a 250.50 g sample of BaBr_2 is dissolved in 25.00 mL of water at 20.0 degree C. If not, what spontaneous process will occur? Calculate the Gibbs free energy change (Delta G) to support and explain your answer.

Explanation / Answer

dEtermine if this is in equilibrium

m = 250.50 g of BaBr2 in V =" 250 mL

S = mol/V = (mass/MW) / V = (250.50/297.14) / (25*10^-3) = 33.721 mol per liter

at T = 20°C

find Ksp

1/T = 1/(20+273) = 0.00341296

ln(Ksp) = 2777.6*0.00341296 + 3.5548 = 13.0346

Ksp = exp(13.0346) = 457988.796

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

457988.796 = S*(2S)^2

4*S^3 = 457988.796

S = (457988.796/4)^(1/3)

S = 48.558 mol per liter

we had

= 33.721 mol per liter

therefore, this is unsaturated

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