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Determine the solubilities of silver chromate , barium chromate , ferric hydroxide , lead chloride and mercurous constants given in Table 7.9 . Determine also the molarities of individual ions. (i)K_(sp)(Ag_2CrO_4)=1.1xx10^(-12) (ii)K_(sp)(BaCrO_4)=1.2xx10^(-10) (iii)K_(sp)(Fe(OH)_3) = 1.0xx10^(-38) (iv)K_(sp) (PbCl_2) = 1.6xx10^(-5) (v) K_(sp) (Hg_2Cl_2)=1.3xx10^(-18) (vi)K_(sp) (Hg_2I_2)=4.5xx10^(-29) |
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Answer» Solution :(i)Solubilities of `Ag_2CrO_4` : `{:(,Ag_2CrO_(4(s)) to , 2Ag_((aq))^(+) + , CrO_4^(2-)),("Molarity of equili. :",S,2S,S):}` `K_(sp)=[Ag^+]^2[CrO_4^(2-)]` `therefore 1.1xx10^(-12) = (2S)^2 (S) =4S^3` `therefore S=((1.1xx10^(-12))/4)^(1/3)` `=0.6503xx10^4 "MOL L"^(-1)` `=6.503xx10^(-5) "mol L"^(-1)` So, `[Ag^+]=2S=2 (6.503xx10^(-5))=1.3xx10^(-4)` M `[CrO_4^(2-)]=S=6.503xx10^(-5) "mol L"^(-1)` (ii)Solubilities of `BaCrO_4` : `{:(BaCrO_(4(s)) to , Ba_((aq))^(2+)+, CrO_(4(aq))^(2-)),("Molarity at equilibrium",S,S):}` `K_(sp)=[Ba^(2+)]=[CrO_4^(2-)]=(S)(S)=S^2` `therefore S=sqrtK_(sp)=sqrt(1.2xx10^(-10))=1.095xx10^(-5)` M `therefore [Ba^(2+)]=[CrO_4^(2-)] =S=1.095xx10^(-5)` M (iii) Solubilities of `Fe(OH)_3` : `{:("Equilibrium", Fe(OH)_(3(s)) hArr, Fe_((aq))^(3+) + , 3OH_((aq))^(-)),("Solubility/ At equilibrium M", "S M","S M","3 S M"):}` `K_(sp)=[Fe^(3+)][OH^-]^3 = (S)(3S)^3` `therefore 1.0xx10^(-38) = 27S^4 = 100xx10^(-40)` `therefore S=((100xx10^(-40))/27)^(1/4)` `=(3.704xx10^(-40))^(1/4)` `=1.3872xx10^(-10) approx 1.39xx10^(-10) "mol L"^(-1)` So, `[Fe^(3+)]=S=1.39xx10^(-10) "mol L"^(-1)` `[OH^-]=3S=3(1.39xx10^(-10))` `=4.17xx10^(-10) "mol L"^(-1)` (iv) Solubilities of `PbCl_2` : `{:(PbCl_2 hArr, Pb_((aq))^(2+) + , 2Cl_((aq))^(-)),("S M", "S M" , "2S M"):}` `K_(sp)=[Pb^(2+)][Cl^-]^2 = (S) (2S^2)=4S^3` `therefore S=(K_(sp)/4)^(1/3) = ((1.6xx10^(-5))/4)^(1/3)= (16/4 xx 10^(-6))^(1/3)` `therefore S=1.5874xx10^(-2)` M So, `[Pb^(2+)]=S=1.5874xx10^(-2) "mol L"^(-1)` `[Cl^-] = 2S=2 (1.5874xx10^(-2))` `= 3.1748xx10^(-2) "mol L"^(-1)` (v)Solubilities of `Hg_2Cl_2` : `{:(Hg_2Cl_(2(s)) hArr, Hg_(2(aq))^(2+) + , 2Cl_((aq))^(-)),(S, "S M", "2S M"):}` `therefore S=(K_(sp)/4)^(1/3) = (1.3/4 xx 10^(-18))^(1/3)` `=0.6876xx10^(-6) M = 6.876xx10^(-5)` So, `[Hg_2^(2+)]=S=6.876xx10^(-5)` M `[Cl^-]=2S=2(6.876xx10^(-5))` `=1.375xx10^(-4)` M (vi) Solubilities of `Hg_2I_2` : `{:(Hg_2I_(2(s)) hArr , Hg_(2(aq))^(2+) + , 2I_((aq))^(-)),(S, "S M", "2S M"):}` `K_(sp)=[Hg_2^(2+)][I^-]^2 = (S)(2S)^2 =4S^3` `therefore S=root3(K_(sp)/4)=((4.5xx10^(-29))/4)^(1/3)=(11.25xx10^(-30))^(1/3)` `=2.241xx10^(-10) "mol L"^(-1)` `therefore [Hg_2^(2+)]=(S)=2.241xx10^(-10) "mol L"^(-1)` `[I^-]^2 =2(S)=2(2.41xx10^(-10))` `= 4.82xx10^(-10) "mol L"^(-1)` |
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