The K sp (25ºC) of sparingly soluble salt of XY 2 is 3.56 × 10 –5 M and at 30ºC the vapour pressure of the saturated solution in water is 31.78 mm of Hg.
At 30ºC, 
= 31.82 mm.
Determine the enthalpy change for the following
reaction in kJ/mol in nearest possible integers
XY 2 (s) 
X 2+ (aq) + 2Y – (aq)
Consider, molarity ~ molality ; log 
= 0.1527
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
53
Sol. At 25ºC, K sp = 3.56 × 10 –5 
= 
(1 – α + x α + y α )
α = 1 , x + y = 3
= 
× 3
× 1000 = 
= 2.23 × 10
–2
= solubility
K sp = 4s 3 = 4 × (2.23 × 10 –2 ) 3
= 5.06 × 10 –5 at 30ºC
∴ 2.303 log 
= 
× 
Δ H = 52.79 kJ ~ 53 kJ/mol
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