Question

Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants. Determine also the molarities of individual ions.

Answer 1

(1) Silver chromate:

\[\ce{Ag_2CrO_4 -> 2Ag^+ + CrO_4^{2-}}\]

then

`"K"_("sp") = ["Ag"^+]^2["CrO"_4^(2-)]`

Let the solubility of \[\ce{Ag_2CrO_4}\] be s.

`=> ["Ag"^+] 2"s" and ["CrO"_4^(2-)] = "s"`

Then

`"K"_("sp") = ("2s")^2 "s" = "4s"^3`

`=> 1.1 xx 10^(-12) = "4s"^3`

`therefore 275 xx 10^(-12) = "s"^3`

`"s" = 0.65 xx 10^(-4) "M"`

Molarity of `"Ag"^+`= 2s = 2 × 0.65 × 10^–4 = 1.30 × 10^–4 M

Molarity of `"CrO"_4^(2-)`= s = 0.65 × 10^–4 M

(2) Barium chromate:

`"BaCr"_4 -> "Ba"^(2+) + "CrO"_4^(2-)`

Then `"K"_("sp") = ["Ba'^(2+)]["CrO"_4^(2-)]`

Let s be the solubility of `"BaCrO"_(4+)`

Thus `["Ba"^(2+)] = "s" and ["CrO"_4^(2-)] = "s"`

`=> "K"_("SP") = "s"^2`

`=> 1.2 xx 10^(-10) = "s"^2`

`=> "s" = 1.09 xx 10^(-5) "M"`

Molarity of `"Ba"^(2+)`= Molarity of `"CrO"_4^(2-) = "s" = 1.09 xx 10^(-5) "M"` 

(3) Ferric hydroxide:

`"Fe"("OH")_3 -> "Fe"^(2+) + 3"OH"^(-)`

`"K"_("sp") = ["Fe"^(2+)] ["OH"^(-)]^3`

Let s be the solubility of  `"Fe"("OH")_3`

Thus `["Fe"^(3+)]` = s and `["OH"^-] = "3s"`

`=> "K"_("sp") = "s".(3"s")^3`

`= "s".27 "s"^3`

`"K"_("sp") =  27"s"^4`    

`1.0 xx 10^(-38 ) = 27 "s"^4`

`.037 xx 10^(-38) = "s"^4`

`.00037 xx 10^(-36) = "s"^4` ` => 1.39 xx  10^(-10) "M" = "S"`

Molarity of `"Fe"^(3+) = "s" = 1.39 xx 10^(-10) "M"`

Molarity of `"OH"^(-) = "3s" = 4.17 xx 10^(-10) "M"`

(4) Lead chloride:

\[\ce{PbCl_2 -> Pb^{2+} + 2Cl-}\]

`"K"_("SP") = ["Pb"^(2+)]["Cl"^-]^2`

Let K_SP be the solubility of `"PbCl"_2`

`["PB"^(2+)] = "s" and ["Cl"^(-)] = "2s"`

Thus `"K"_("sp") = "s".("2s")^2`

`= "4s"^3`

`=> 1.6 xx 10^(-5) = "4s"^3`

`=> 0.4 xx 10^(-5) = "s"^3`

`4 xx 10^(-6) = "s"^3 => 1.58 xx 10^(-2) "M" = "S".1`

Molarity of `"PB"^(2+) = "s" = 1.58 xx 10^(-2) "M"`

Molarity of chloride = 2s = `3.16 xx 10^(-2) "M"`

(5) Mercurous iodide:

`"Hg"_2I_2 -> "Hg"^(2+) + "2I"^(-)`

`"K"_("sp") = ["Hg"_2^(2+)]^(2)[I^(-)]^2`

Let s be the solubility of `"Hg"_2I_2`

`=> ["Hg"_2^(2+)]` = "s" and `["I"^(-)] = "2s"`

Thus `"K"_("sp") = "s"("2s")^2`

`=> "K"_("sp") = 4"s"^3`

`4.5 xx 10^(-29) = 4"s"^3`

`1.125 xx 10^(-29) = "s"^3`

`=> "s" = 2.24 xx 10^(-10) "M"`

Molarity of `"Hg"_2^(2+) = "s" = 2.24 xx  10^(-10)  "M"`

Molarity of `"I"^(-) = "2s" = 4.48 xx 10^(-10) "M"`