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A solution which is $10^{- 3}$ $M$ each in $M n^{2 +}$ , $F e^{2 +}, Z n^{2 +}$ and $H g^{2 +}$ is treated with $10^{- 16} M$ sulphide ion. If $K_{s p} o f$ $M n S, F e S, Z n S$ and $H g S$ are $10^{- 15}$ , $10^{- 23}$ , $10^{- 20}$ and $10^{- 54}$ respectively. Which of the following will be precipitated first?

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$F e S$

$M n S$

$H g S$

$Z n S$

answer is C.

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ease of $P P T α \frac{1}{K_{s p}}$

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