Question

Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at `298K` from theor solubility product constants given in Table `7.9`. Determine also the molarities of individual ions.

Answer 1

Silver chromate
`Ag_(2)CrO_(4), K_(sp)` of `Ag_(2)CrO_(4)`
`1.1xx10^(-12)`
Let the solution of `Ag_(2)CrO_(4)` be eauals to `S`
`AgCr_(2)O_(4) hArr 2Ag^(o+)+CrO_(4)^(2-)`
`1.1xx10^(-12)=(2S)^(2)(S)`
or `1.1xx10^(-12)=4S^(3)`
or `S^(3)=(1.1xx10^(-12))/4`
or `S=((1.1xx10^(-12))/4)^(1//3)`
`=0.65xx10^(-4) M`
`[Ag^(o+)]=2xx0.65xx10^(-4)=1.30xx10^(-4) M`
`[CrO_(4)^(2-)]=1.3xx10^(-4)//2=6.5xx10^(-5)M`
ii. Barium chromate= `BaCrO_(4)`:
`BaCrO_(4) hArr Ba^(2+)+CrO_(4)^(2-)`
`K_(sp)` of `BaCrO_(4)=1.2xx10^(-10)`
or `1.2xx10^(-10)=SxxS`
or `1.2xx10^(-10)=S^(2)`
or `S=sqrt(1.2xx10^(-10))=1.1xx10^(-5) M`
iii. Ferric hydroxide: `Fe(OH)_(3)`
`Fe(OH)_(3) hArr Fe^(3+)+3overset(Θ)(O)H`
`K_(sp) of Fe(OH)(3)=1.0xx10^(-38)`
or `1.0xx10^(-38)=Sxx(3S)^(3)`
or `1.0xx10^(-38)=27 S^(4)`
or `S^(4)=(1.0xx10^(-38))/27`
or `S=((1.0xx10^(-38))/27)^(1//4)=1.39xx10^(-10)M`
`[Fe^(3+)]=1.39xx10^(-10)M`
`[overset(Θ)(O)H]=3xx1.39xx10^(-10)=4.17xx10^(-10)M`
iv.Lead chloride `= PbCl_(2)`
`PbCl_(2) hArr Pb^(2+)+2Cl^(Θ)`
`K_(sp)` of `PbCl_(2)=1.6xx10^(-5)`
`1.65xx10^(-5)=Sxx(2S)^(2)`
or `1.6xx10^(-5)=4S^(3)`
or `S^(3)=(1.6xx10^(-5))/4=0.4xx10^(-5)`
`S=(0.4xx10^(-5))^(1//3)=(4xx10^(-6))^(1//3)`
`=1.59xx10^(-2)M=[Pb^(2+)]`
`[Cl^(Θ)]=2xx1.59xx10^(-2)=3.18xx10^(-2)M`
v. Mercuous iodide: `Hg_(2)I_(2)`
`Hg_(2)I_(2) hArr Hg_(2)^(2+)+2I^(Θ)`
`K_(sp)` of `Hg_(2)I_(2)=4/5xx10^(-29)`
`4.5xx10^(-29)=(S) (2S)^(2)=4S^(3)`
or `4.5xx10^(-29)=4S^(3)`
or `S^(3)=4.5/4xx10^(-29)`
or `S=(4.5/4xx10^(-29))^(1//3)`
`=2.24xx10^(-10)M=[Hg_(2)^(2+)]`
`[I^(Θ)]=2xx2.24xx10^(-10)=4.48xx10^(-10) M`.