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Calculate the solubility of AgCl , Fe(OH)_3 Ag_2SO_4 and Hg_2Br_2 from their solubility products. K_(sp)(AgCl)=1.8xx10^(-10) ,K_(sp)[Fe(OH)_3]=1.0xx10^(-38), K_(sp)[Ag_2SO_4]=1.4xx10^(-5), K_(sp)=(Hg_2Br_2)=5.6xx10^(-23) .Also calculate the solubilities of salts in gL^(-1)and molarities of the ions. |
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Answer» SOLUTION :(i)`AgCl(s) hArr Ag^(+)(aq)+Cl^(-)(aq)` `K_(sp)=1.8xx10^(-10)` Let solubility be s moles per litre. Then the CONCENTRATION of various species at equilibrium are : `[Ag^+]=s, [Cl^-]=s` `K_(sp)=[Ag^+][Cl^-]=sxxs` `1.8xx10^(-10)=s^2` `therefore s=(1.8xx10^(-10))^(1//2) =1.34xx10^(-5) "mol L"^(-1)` Solubility in g `L^(-1)=1.34xx10^(-5)xx143.5` `=1.92xx10^(-3) gL^(-1)` `[Ag^+]=s=1.34xx10^(-5) "mol L"^(-1)` `[Cl^-]=s=1.34xx10^(-5) "mol L"^(-1)` (ii)For `Fe(OH)_3 : K_(sp)=1.0xx10^(-38)` `Fe(OH)_3 hArr Fe^(3+) + 3OH^-` If solubility is s , then `[Fe^(3+)]=s, [OH^-]=3s` `K_(sp)=[Fe^(3+)] [OH^-]^3` `1.0xx10^(-38)=sxx(3s)^3=27s^4` `27s^4=1.0xx10^(-38)` `s^4=(1.0xx10^(-38))/27` or `s=((1.0xx10^(-38))/27)^(1//4)` `=1.39xx10^(-10) "mol L"^(-1)` Solubility in g `L^(-1) =1.39xx10^(-10)xx106.5` `=1.48xx10^(-8) g L^(-1)` `[Fe^(3+)]=s=1.39xx10^(-10) "mol L"^(-1)` `[OH^-]=3s=3xx1.39xx10^(-10)` `=4.17xx10^(-10) "mol L"^(-1)` (iii)For `Ag_2SO_4 : K_(sp)=1.4xx10^(-5)` `Ag_2SO_4 hArr 2Ag^(+) + SO_4^(2-)` If solubility is s , then `[Ag^+]=2s, [SO_4^(2-)]=s` `K_(sp)=[Ag^+]^2 [SO_4^(2-)]` `1.4xx10^(-5) =(2s)^2 (s)` `1.4xx10^(-5)=4s^3` or `s^3=(1.4xx10^(-5))/4` `s=((1.4xx10^(-5))/4)^(1//3)` `=1.52xx10^(-2) "mol L"^(-1)` Solubility in `gL^(-1) =1.52xx10^(-2) xx312` `=4.74 g L^(-1)` `[Ag^+]=2s=2 xx 1.52xx10^(-2)` `=3.04xx10^(-2) "mol L"^(-1)` `[SO_4^(2-)]=s=1.52xx10^(-2) "mol L"^(-1)` (iv)For `Hg_2Br_2 , K_(sp) = 5.6xx10^(-23)` `Hg_2Br_2 hArr Hg_2^(2+)+ 2Br^-` If solubility is s, then `[Hg_2^(2-)]=s, Br^(-)=2s` `K_(sp)=(s)(2s)^2` `5.6xx10^(-23)=4s^3` or `4s^3 =5.6xx10^(-23)` `s=((5.6xx10^(-23))/4)^(1//3)` =`2.41xx10^(-8) "mol L"^(-1)` Solubility in g `L^(-1) = 2.41xx10^(-8) xx360.4` `=8.68xx10^(-6) gL^(-1)` `[Hg_2^(2+)]=2.41xx10^(-8)` `[Br^-]=2xx2.41xx10^(-8)` `=4.82xx10^(-8)` M |
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