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Introduction to Optical Isomerism

Which pair of...

Question

Which pair of ions will have highest Ksp values of their sulphides.
(1) ${Z n}^{2 +}, {C u}^{2 +}$
(2) ${P b}^{2 +}, {H g}^{2 +}$
(3) ${Z n}^{2 +}, {M n}^{2 +}$
(4) ${C d}^{2 +}, {B i}^{3 +}$

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Solution

Explanation for correct option:
(3) In case to determine the highest $K_{s p}$ , we have to check its precipitation rate. More easily the compound get precipitate lesser is its $K_{s p}$ .
Group 2 and group 4 are precipitated as their sulphides. Whereas the $K_{s p}$ of group 2 is less than group A ions, thus group 2 ions can be easily precipitated.
Hence, option (3) is correct.

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K_{4} [{F e (C N)}_{6}]

can be used to detect some ions out of

{C d}^{2 +}, {C u}^{2 +}, {Z n}^{2 +}, {F e}^{3 +}, {P b}^{2 +}

{B i}^{3 +}

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Q. The pair(s) of ions where BOTH the ions are precipitated upon passing

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Q. A solution which is

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Q. A solution which is

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

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