SATHEE: Equilibrium - Result Question 89

Equilibrium - Result Question 89

####93. Solubility of $M X_{2}$ -type eletrolytes is $0.5 \times 10^{- 4}$ $m o l /$ lit, then find out $K_{s p}$ of electrolytes [2002]

(a) $5 \times 10^{- 12}$

(b) $25 \times 10^{- 10}$

(c) $1 \times 10^{- 13}$

(d) $5 \times 10^{- 13}$

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Solution:

(d) Given $s = 0.5 \times 10^{- 4} m o l e s / L$

$[M X_{2} \to M^{2 +} + 2 X]$

$∵$ For $M X_{2}, K_{s p} = s \times (2 s)^{2} = 4 s^{3}$

$K_{s p} = 4 \times (0.5 \times 10^{- 4})^{3} = 4 \times 0.125 \times 10^{- 12}$

$= 0.5 \times 10^{- 12} = 5 \times 10^{- 13}$

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