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  • A sparingly soluble salt gets precipitated only when the product of the concentration of its ions in the solution (Q_{sp}) becomes greater than its solubility product.

A sparingly soluble salt gets precipitated only when the product of the concentration of its ions in the solution (Q_{sp})  becomes greater than its solubility product. If the solubility of BaSO_{4} in water is 8\times 10^{-4} mol \; dm^{-3} calculate its solubility in 0.01 mol dm-3 of H_{2}SO_{4}..

Answers (1)

As per the information given in the question, the solubility of BaSO_{4} in water is 8 \times 10^{-4} g/L

Hence, We can write the equation of disassociation of BaSO_{4} as: -

BaSO_{4}\rightleftharpoons Ba^{2+}++SO{_{4}}^{2-}

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

Now, it is known to us that,

K_{sp}=S\times S=S^{2}

K_{sp}=(8\times 10^{-8})^{2}

K_{sp}=64\times 10^{-8}

Thus, in the presence of 0.01 H_{2}SO_{4} soln

H_{2}SO_{4}\rightleftharpoons 2H^{+}+SO{_{4}}^{2-}

Initial 0.01 0 0
Final 0 0.02 0.01

Now we know that,

BaSO_{4}\rightleftharpoons Ba^{2+}+SO{_{4}}^{2-}

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
S2 + 0.01 S = 64 × 10–8
S2 + 0.01 S – 64 × 10–8 = 0

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

Thus the solubility of BaSO_{4} in 0.01 mol dm-3  of H_{2}SO_{4} is 6\times10^{-4} mol dm-3

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