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Chemistry Question on Buffer Solutions

Solubility product of BaCl2 BaCl_{2} is 4×10−9 4\times 10^{-9} . Its solubility would be:

A

1×10−27 1\times 10^{-27}

B

1×10−3 1\times 10^{-3}

C

1×10−7 1\times 10^{-7}

D

1×10−2 1\times 10^{-2}

Answer

1×10−3 1\times 10^{-3}

Explanation

Solution

BaCl2BaCl _{2} is a ternary electrolyte. BaCl2=Ba2+s+2Cl−2sBaCl _{2}= \underset{s}{Ba^{2+}}+\underset{2s}{2 Cl ^{-}} Ksp=[Ba2+][Cl−]2K_{s p} =\left[ Ba ^{2+}\right]\left[ Cl ^{-}\right]^{2} =[s][2s]2=[s][2 s ]^{2} Ksp=4s3K_{s p} =4 s^{3} Ksp=4s3K_{s p} =4 s^{3} where s=molar solubility 4s3=4×10−94 s^{3} =4 \times 10^{-9}(given) or s=10−3Ms=10^{-3} M