Respuesta :

kenmyna
 The  solubility  product   expression  for  mercury (i)  iodide  Hg2 i2

  write  the  equation  for  dissociation of   Hg2i2 in   water

that  is  Hg2i2    ---.>  Hg2^2+    + 2 i ^-


therefore  the  solubility  product  is  expressed  as follows    (Hg2^2+)  (I-)^2

  I-   is  squared   since there  is two  iodine  ions   in  the  equation

Answer : The solubility product expression is,

[tex]K_{sp}=[Hg^+]^2[I^-]^2[/tex]

Explanation :

Solubility product constant : It is defined as the product of the concentration of the ions present in a solution raised to the power by its stoichiometric coefficient in a solution of a salt. This takes place at equilibrium only. The solubility product constant is represented as, [tex]K__{sp}[/tex].

The dissociation of mercury(i) iodide is written as:

[tex]Hg_2I_2\rightleftharpoons 2Hg^++2I^-[/tex]

The expression for solubility constant for this reaction will be,

[tex]K_{sp}=[Hg^+]^2[I^-]^2[/tex]

Therefore, the solubility product expression is [tex]K_{sp}=[Hg^+]^2[I^-]^2[/tex]

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