# A sparingly soluble salt gets precipitated only when the product of the concentration of its ions in the solution   becomes greater than its solubility product. If the solubility of in water is calculate its solubility in 0.01 mol dm-3 of .

As per the information given in the question, the solubility of  in water is

Hence, We can write the equation of disassociation of  as: -

 At t=0 0 0 At t=0 S S

Now, it is known to us that,

Thus, in the presence of 0.01  soln

 Initial 0.01 0 0 Final 0 0.02 0.01

Now we know that,

 Final S S+0.01

The expression for Ksp in the presence of sulphuric acid will be

Ksp = (S) (S + 0.01)

Since value of Ksp will not change in the presence of sulphuric acid,
therefore,

(S) (S + 0.01) = 64 × 10–8
S+ 0.01 S = 64 × 10–8
S+ 0.01 S – 64 × 10–8 = 0

On solving quadratic equation, we get S= $6\times10^{-4}$.

Thus the solubility of  in 0.01 mol dm-3  of  is $6\times10^{-4}$ mol dm-3

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