The solubility of AgCl(s) in water at 25 °C is 1.33 * 10 - 5 mol...

Question

The solubility of AgCl(s) in water at 25 °C is 1.33 * 10 - 5 mol>L 
and its ΔH° of solution is 65.7 kJ>mol. What is its solubility at 
50.0 °C?

Accepted Answer

Hi everyone here we have a question asking us to determine the Celje bility of lithium fluoride solid at 60 degrees Celsius. If it's changed an entropy of solution and Celje bility at 25 degrees Celsius is 2.18 kg per mole and 0.616 most per liter respectively. So our step one is to calculate RKSP at 25°C. So with um fluoride is going to disassociate into lithium and flooring. So our K sp equals lithium times. Flooring equals S two Equals 0. squared Equals 3. 46 times 10 To the negative 3rd. And now we're going to calculate our K. S. P at 60 degrees Celsius. So we're going to use the natural log of K two Equals the natural log of K one plus our change in entropy over our gas constant times one over our first temperature -1 over our second temperature. So our T one equals 25 Plus 0.15 Which equals 298.15. Because we're changing it to Kelvin. R T two equal Plus 273. Equals 333 0.15. Our natural log of K two equals the natural log of three 0.79 Times 10 to the -3 plus 2.18 Times to the 3rd, divided by 8. times one Over 298. minus one over 333 0.15. So our K two Equals 4. times 10 to the negative third. And now we need to calculate our Selya bility at 40°C. So our K sp equals S two S equals the square root. F k sp equals the square root of 4.161, 9 Times 10 to the negative 3rd Equals 0. most per leader. And that is our final answer. Thank you for watching. Bye.

The solubility of AgCl(s) in water at 25 °C is 1.33 * 10 - 5 mol>L and its ΔH° of solution is 65.7 kJ>mol. What is its solubility at 50.0 °C?

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