Question

The vapour pressure of a saturated solution of a sparingly soluble salt (A2B3) is 31.8 mm Hg at 40°C. If the vapour pressure of pure water is 31.9 mm Hg, the solubility product (Ksp) of A2B3, at the same temperature is ​

Answer 1

Answer:

5.47*10^(-5)

Explanation:

Let stability of A2B2 be 5 moles/litre

A2B2⇌2A++2B−

                    2s      2s

Ksp=(2s)2×(2s)2=16×s4

Now, PPo−P=Nn=55.564

⇒31.831.9−31.8=55.564s⇒S=0.043 moles/litre

∴Ksp=16×(0.043)4=5.47×10−5