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The concentration of sulphide ion in 0.1 M HCl solution saturated with hydrogen sulphide is 1.0xx10^(-19) M. if 10 mL of this is added to 5 mL of 0.04 M solution of the following : FeSO_4 , MnCl_2 , ZnCl_2 and CdCl_2 |
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Answer» Solution :`K_(SP)(FeS)=6.3xx10^(-18) , K_(sp) (MnS) = 2.5xx10^(-13)` `K_(sp)(ZnS)=1.6xx10^(-24) , K_(sp) (CdS)=8.0xx10^(-27)` CALCULATION `[S^(2-)]` in mixture : `[S^2]=1.0xx10^(-19) = M_1` its 10 mL solution + 5 mL SALT solution Total volume = (10+5)=15 mL `{:("Initial","In mixture"),(M_1V_1,M_2V_2),(1.0xx10^(-19)xx10,M_2xx15):}` `therefore M_2=(1.0xx10^(-19)xx10)/15=6.667xx10^(-20)` M Thus in mixture `[S^(2)]=6.667xx10^(-20)` M Concentration of metal ion in mixture : [metal salt ] =[metal ion] = 5 mol 0.004 M + volume of sulphide ion =10 mL `therefore` Volume of solution in mix two =15 mL So, concentrationof metal ion in mixture `{:("Initial","In mixture"),(M_1V_1,=M_2V_2),(0.04M xx 5"ml",M_2xx15 "ml"):}` `therefore M_2=(0.04M xx 5 mL)/"15 ml"=1.333xx10^(-2)` M Therefore in mixed solution `[Fe^(2+)]=[Mn^(2+)]` `=[Zn^(2+)]=[Cd^(2+)]` `=1.333xx10^(-2)` M Calculate value of `Q_(sp)` For salt M S `Q_(sp)=[M^(2+)][S^(2-)]=(1.333xx10^(-2))(6.667xx10^(-20))` `=8.8891xx10^(-22)` Ionic product Prediction of precipitation : For precipitation , `Q_(sp) gt K_(sp)` so, [`K_(sp)` of ZnS=`1.6xx10^(-24)` ] [`K_(sp)` of CdS=`8.0xx10^(-27)` ] Ionic product of both is more than their solubility product `Q_(sp)` is more. `therefore` Precipitation are formedfrom ZnS and CdS |
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