Question

Use the thermodynamic data at 298 k below to determine the ksp for barium carbonate, baco3 at this temperature. substance: ba2+(aq) co32–(aq) baco3(s) δh°f (kj/mol): –538.36 –676.26 –1219 δg°f (kj/mol): –560.7 –528.1 –1139 s°(j/k·mol): 13 –53.1 112

Answer 1

Answer : the correct answer for ksp = 1.59 * 10⁻⁹

Following are the steps to calculate the ksp of reaction

BaCO₃ →Ba ²⁺ + CO₃²⁻ :

Step 1 : To find ΔG° of reaction :

ΔG° of reaction can be calculates by taking difference between ΔG° of products and reactants as :

ΔG° reaction =Sum of ΔG° ( products ) - Sum of Δ G° ( reactants ) .

Given : ΔG° for Ba²⁺ ( product )= -560.7
`(KJ)/(Mol)`

ΔG° for CO₃²⁻ (product ) =- 528.1
`(KJ)/(Mol)`

ΔG° BaCO₃ ( reactant) = –1139
`(KJ)/(Mol)`

Plugging value in formula :

ΔG° for reaction = ( ΔG° of Ba ²⁺ + ΔG° of CO₃²⁻ ) - (ΔG° of BaCO₃ )

⁻ = ( -560.7
`(KJ)/(Mol)` + 528.1
`(KJ)/(Mol)` ) - ( -1139
`(KJ)/(Mol)` )

= ( -1088.8
`(KJ)/(Mol)`) - (-1139
`(KJ)/(Mol)` )

= - 1088.8
`(KJ)/(Mol)` + 1139
`(KJ)/(Mol)`

ΔG° of reaction = 50.2
`(KJ)/(Mol)`

Step 2: To calculate ksp from ΔG° of reaction .

The relation between Ksp and ΔG° is given as :

ΔG° = -RT ln ksp

Where ΔG° = Gibb's Free energy R = gas constant T = Temperature

Ksp = Solubility constant product .

Given : ΔG° of reaction = 50.2
`(KJ)/(Mol)`

T = 298 K R = 8.314
`(J)/(Mol * K)`

Plugging values in formula

`image`

`50.2 (KJ)/(mol) = - 2477.572 (J)/(mol) * ln K`

((Converting 2477
`(J)/(mol) to (KJ)/(mol)`

Since , 1 KJ = 1000 J So ,
`2477 (J)/(mol) * (1 KJ)/(1000J) = 2.477 (KJ)/(mol)` ))

Dividing both side by
`- 2.477 (KJ)/(mol)`

`(50.2(KJ)/(mol))/(-2.477 (KJ)/(mol)) = (-2.477 (KJ)/(mol))/(-2.477 (KJ)/(mol)) * ln ksp`

ln ksp =
`ln ksp = -20.27 (KJ)/(mol)`

Removing ln :

ksp = 1. 59 * 10⁻⁹