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Question

Chemistry Question on Equilibrium

Solubility of MX2M X_{2}-type electrolytes is 0.5×10−40.5 \times 10^{-4} mol/Lmol / L Then find out KspK_{s p} of electrolytes

A

5×10−12 5\times 10^{-12}

B

25×10−10 25\times 10^{-10}

C

1×10−13 1\times 10^{-13}

D

5×10−13 5\times 10^{-13}

Answer

5×10−13 5\times 10^{-13}

Explanation

Solution

(on 100%100 \% ionisation) ∴Ksp of MX2=[M2+][X−]2\therefore K_{s p} \text { of } M X_{2}=\left[M^{2+}\right]\left[X^{-}\right]^{2} =(0.5×10−4)(1.0×10−4)2=\left(0.5 \times 10^{-4}\right)\left(1.0 \times 10^{-4}\right)^{2} =0.5×10−12=5×10−13=0.5 \times 10^{-12}=5 \times 10^{-13}