Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298 K form their solubility product constants given below. Determine also the molarities of individual ions.
`K_(SP(Ag_(2)CrO_(4)))=1.1xx10^(-12)`,
`K_(SP(BaCrO_(4)))=1.2xx10^(-10)`,
`K_(SP[Fe(OH)_(3)])=1.0xx10^(-38)`,
`K_(SP(PbCI_(2)))=1.6xx10^(-5)`,
`K_(SP(Hg_(2)I_(2)))=4.5xx10^(-29)`.

Answer

`6.5xx10^(-5)M, 1.09xx10^(-5)M , 1.38xx10^(-10)M, 1.59xx10^(-2) M , 8.18xx10^(-8)`M

Explanation

(i)Silver chromate, `Ag_2CrO_4`
`Ag_2CrO_4 hArr 2Ag^(+) + CrO_4^(2-)`
`[Ag^+]=2s, CrO_4^(2-) =s`
`K_(sp)=(2s)^2 (s) =4s^3`
`therefore 4s^3 =1.1xx10^(-12)`
`s=((1.1xx10^(-12))/4)^(1//3)=6.5xx10^(-5)`
`[Ag^+]=1.30xx10^(-4)` M, `[CrO_4^(2-)]=6.5xx10^(-5)`M .
(ii)Barium chromate, `BaCrO_4`
`BaCrO_4 hArr Ba^(2+) + CrO_4^(2-)`
`[Ba^(2+)]=s, CrO_4^(2-)=s`
`K_(sp)=(s)(s)=s^2`
`s^2=1.2xx10^(-10)`
`s=(1.2xx10^(-10))^(1//2) =1.09xx10^(-5)` M
`[Ba^(2+)]=1.09xx10^(-5)` M , `[CrO_4^(2-)]=1.09xx10^(-5)` M
(iii)Ferric hydroxide , `Fe(OH)_3`
`Fe(OH)_3 hArr Fe^(3+) + 3OH^(-)`
`[Fe^(3+)]=s, [OH^-]=3s`
`K_(sp)=(s) (3s)^3 =27s^4`
`27s^4 =1.0xx10^(-38)`
`s=((1.0xx10^(-38))/27)^(1//4)`
`=1.38xx10^(-10) M`
`[Fe^(3+)]=1.38xx10^(-10) M, [OH^-]=4.14xx10^(-10)` M
(iv) Lead chloride, `PbCl_2`
`PbCl_2 hArr Pb^(2+) + 2Cl^-`
`[Pb^(2+)]=s, [Cl^-]=2s`
`K_(sp)=(s)xx(2s)^2 =4s^2`
`4s^3=1.6xx10^(-5)`
`s=((1.6xx10^(-5))/4)^(1//3) =1.59xx10^(-2)`M
`[Pb^(2+)]=1.59xx10^(-2)`M , `[Cl^-]=3.18xx10^(-2)` M
(v)Mercurous iodide , `Hg_2I_2`
`Hg_2I_2 hArr 2Hg^(+) + 2I^-`
`[Hg^+]=2s , [I^-]=2s`
`K_(sp) =(2s)^2 (2s)^2`
`=16s^4`
`=16s^4 =4.5xx10^(-29))`
`s=((4.5xx10^(-29))/16)^(1//4)=4.09xx10^(-8) ` M
`[Hg^+]=8.18xx10^(-8), [I^-]=8.18xx10^(-8)` M .