1.

The solubility of AgCl will be minimum in……….

Answer»

`0.001 M AgNO_3`
PURE water
`0.01 M CaCl_2`
`0.01 M NACL`

Solution :`0.01M CaCl_2` gives maximum `Cl^-` ions to keep `K_(sp)` of AGCL constant, decreases in `[Ag^+]` will be maximum.


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