40g Ba(NO 3 ) 2 is dissolved in 1Kg of water ate 313.15K. a) Compute the equilib

Question

40g Ba(NO3)2 is dissolved in 1Kg of water ate 313.15K.

a) Compute the equilibrium vapor pressure above the solution if none of the Ba(NO3)2 ionizes. Assume the solution behaves ideally in the Roult's law regime. The equilibrium vapor pressure above pure water at 313.15K is 55.324 torr.

b) What would the vapor pressure be if barium nitrate ionizes completely?

c) The actual vapor pressure is found to be 54.909 torr. Calculate the percent ionization for barium nitrate.

d) Calculate Gmixing for the solution in (c). Assume the solution is ideal.

e) Using Debye-Huckel theory, Calculate for NO3- for the solution in (c).

Explanation / Answer

a) MolecularWeight of Ba(NO3)2 = 261.37 g/mol & Molecular Weight of Water = 18g/mol

Weight of Ba(NO3)2 dissolved in 1 Kg of water = 40g

Moles of Ba(NO3)2 dissolved = 40/261.37 = 0.153 moles

No. of Moles of 1Kg water = Weight/Molecular weight = 1000/18 = 55.55 moles

Mole fraction of water (Solvent) = No. of moles of water / (No. of moles of Ba(NO3)2 + No. of moles of Water)

X = 55.55 / (0.153 + 55.55) = 0.9972

According to the Raolut's Law,

Pressure of Solvent, P = Mole fraction of Solvent (X) * Pressure of pure solvent(P0)

P = X * P0

P = 0.9972 * 55.324 = 55.17 torr

Pressure of Water, P = 55.17 torr

b)

Ba(NO3)2 will dissociate in water as follows:

Ba(NO3)2 -----> Ba2+ + 2NO3- ................... (1)

1 mole Ba(NO3)2 will dissociate into 3 moles of ions (1 mole of Ba2+ & 2 moles of NO3-:).

Molecular Weight of Ba(NO3)2 = 261.37 g/mol & Molecular Weight of Water = 18g/mol

Weight of Ba(NO3)2 dissolved in 1 Kg of water = 40g

Moles of Ba(NO3)2 dissolved = 40/261.37 = 0.153 moles

As per equation (1), Total moles of ions in water = 3 * 0.153 = 0.459 moles

No. of Moles of 1Kg water = Weight/Molecular weight = 1000/18 = 55.55 moles

Mole fraction of water (Solvent) = No. of moles of water / (No. of moles of ions + No. of moles of Water)

X = 55.55 / (0.459 + 55.55) = 0.9918

According to the Raolut's Law,

Pressure of Solvent, P = Mole fraction of Solvent (X) * Pressure of pure solvent(P0)

P = X * P0

P = 0.9918 * 55.324 = 54.87 torr

Pressure of Water, P = 54.87 torr

c)

According to the Raolut's Law,

Pressure of Solvent, P = Mole fraction of Solvent (X) * Pressure of pure solvent(P0)

P = X * P0

54.909 = X * 55.324

X = 54.909/55.324 = 0.9925

Mole fraction of water (Solvent) = No. of moles of water / (No. of moles of ions + No. of moles of Water)

X = 55.55 / ( I + 55.55) ( I = No. of moles of ions)

0.9925 = 55.55 / ( I + 55.55)

I = 0.4198 moles

Ba(NO3)2 -----> Ba2+ + 2NO3-   

3 moles of ions will be formed by dissociation of 1 mole of Ba(NO3)2.

Therefore, 0.4198 moles will be formed by dissociation of Ba(NO3)2. moles = 0.4198/3 = 0.14 moles

Total moles of Ba(NO3)2. = 0.153

Moles dissociated = 0.14

Percentage dissociation = 0.14*100/0.153 = 91.45 %

(d)

Ba(NO3)2 -----> Ba2+ + 2NO3-

No. of moles of Ba2+ = 0.14

No. of moles of NO3- = 2 * 0.14 = 0.28

Equilibrium Constant,K = [Concentration of Ba2+] * [Concentration of NO3-]2

K = 0.14 * (0.28)2 = 1.0976 x 10-2

Gibbs Energy of mixing = - R*T*ln K = - 8.314 * 313.15 * ln (0.010976) = 11747.238 J

e) Using Debye-Huckel theory, Calculate for NO3- for the solution in (c).

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