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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. |
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Answer» 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`. |
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