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Question: The solubility product of \(PCl_{3} + C{l_{2}}_{(g)}\) is \(PCl_{5(g)}\)What will be its solubility ...

The solubility product of PCl3+Cl2(g)PCl_{3} + C{l_{2}}_{(g)} is PCl5(g)PCl_{5(g)}What will be its solubility in mol PCl5(g)PCl_{5(g)}?

A

PCl3(g)+Cl2(g)PCl_{3(g)} + Cl_{2(g)}

B

2CO2(g)2CO_{2(g)}

C

2CO(g)+O2(g)2CO_{(g)} + O_{2(g)}

D

N2,H2N_{2},H_{2}

Answer

2CO(g)+O2(g)2CO_{(g)} + O_{2(g)}

Explanation

Solution

: BaCl2BaCl_{2} Ba2++2Cl−Ba^{2 +} + 2Cl^{-}

Ksp=[Ba2+][Cl−]2=x×(2x)24x3K_{sp} = \lbrack Ba^{2 +}\rbrack\lbrack Cl^{-}\rbrack^{2} = x \times (2x)^{2}4x^{3}

4x3=3.2×10−94x^{3} = 3.2 \times 10^{- 9}

⇒x=9.28×10−4=0.928×10−3≈1×10−3\Rightarrow x = 9.28 \times 10^{- 4} = 0.928 \times 10^{- 3} \approx 1 \times 10^{- 3}