Question

At 25°C, Ksp for A2B salt is equal to 5 x 10-7. If the salt is 50% dissociated then the solubility of A2B in mol/litre will be
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Answer 1

Given - Ksp for A2B salt = 5 x 10-7 .

Find - The solubility of A2B in mol/litre .

Solution- A2B salt will be dissociated as -

A2B ⇌ 2A+ + B .

• The salt is 50% dissociated so the solubility will be 50/100 s = 0.5 s (s = solubility).

• As, the solubility is 0.5s so for B it is 0.5s while for A it is twice of 0.5s.

A2B ⇌ 2A+ + B-

2*0.5s 0.5s

• As we know , the formula for ksp, so by putting the value of given Ksp we can find the solubility of A2B type salt.

Ksp = [2A]²*[B]

=> 5*10-7 = [2*0.5s]²*[0.5s]

=> 5*10-7 = s² * 0.5 s

=> 5*10-7 = 0.5s³

=> s³ = 10-6

=> s = 10-2 mol/L

• The solubility of A2B in mol/litre will be 10-2 mol/L