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Determine the solubilities of silver chromate, barium chromate, ferric, hydroxide, lead, chloride and mercurous iodide at 298 K. Given `K_(sp)` values : `Ag_(2)CrO_(4)=1.1xx10^(-12), BaCrO_(4)=1.2xx10^(-10), Fe(OH)_(3)=1.0xx10^(-38), PbCl_(2)=1.6xx10^(-5), Hg_(2)I_(2)=4.5xx10^(-29)`
Determine also the molarities of individual ions.

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Correct Answer - `S[BaCrO_(4)]=1.1xx10^(-5)M=[Ba^(2+)]=[CrO_(4)^(2-)]`
`S(Fe(OH)_(3))=1.39xx10^(-10)M=[Fe^(3+)], [OH^(-)]=3xx1.39xx10^(-10) M = 4.17xx10^(-10)M`
`S(PbCl_(2))=1.59xx10^(-2)M=[Pb^(2+)],[Cl^(-)]=2xx1.59xx10^(-2)M=3.18xx10^(-2)M`
`S(Hg_(2)I_(2))=2.24xx10^(-10)M=[Hg_(2)^(2+)], [I^(-)]=2xx2.24xx10^(-10)M=4.48xx10^(-10)M`
`{:(Ag_(2)CrO_(4) hArr 2"Ag"^(+)+CrO_(4)^(2-),K_(sp)=(2s)^(2)(s)=4s^(3)"," :. s^(3)=K_(sp)//4,,),(s" " 2s" " s" "=(1.1xx10^(-12))//4=0.275xx106(-12)=2.75xx10^(-13),,):}`
`:. 3 log s = log (2.75xx 10^(-13))=-13+0.4393=-12.5607`
or ` log s = - 4.1869 = bar(5).8131`
`:. s= 6.5xx10^(-5) ` mol `L^(-1)`
`[Ag^(+)]=2xx6.5xx10(-5)=13.0xx10^(-5)M = 1.30 xx 10^(-4)M`
`[CrO_(4)^(2-)] = 6.5xx10^(-5)M`
For other salts, proceed exactly in similar manner .

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