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Chemistry Question on Equilibrium

The solubility product of BaCl2BaC{{l}_{2}} is 4×10−9.4\times {{10}^{-9}}. Its solubility in mol/L is

A

4×10−34\times {{10}^{-3}}

B

4×10−94\times {{10}^{-9}}

C

1×10−31\times {{10}^{-3}}

D

1×10−91\times {{10}^{-9}}

Answer

1×10−31\times {{10}^{-3}}

Explanation

Solution

Find relationship between solubility product and solubility of BaCl2B a C l_{2} and then solve problem (solubility product of BaCl2BaCl _{2} is 4×10−94 \times 10^{-9} )
BaCl2→Ba2++2Cl−BaCl _{2} \rightarrow Ba ^{2+}+2 Cl ^{-}
Let the solubility of BaCl2xBaCl _{2} x mol / L
∴Ksp=[Ba2+][Cl−]2\therefore K_{s p}=\left[B a^{2+}\right]\left[C l^{-}\right]^{2}
=(x)×(2x)2=(x) \times(2 x)^{2}
=x×4x2−4x3=x \times 4 x^{2}-4 x^{3}
or solubility of BaCl2B a C l_{2}
=( solubility product of BaCl2)134=\frac{\left(\text { solubility product of } BaCl _{2}\right)^{\frac{1}{3}}}{4}
=(4×10−94)1/3=\left(\frac{4 \times 10^{-9}}{4}\right)^{1 / 3}
=10−3 mol/L=10^{-3}\, mol / L