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The solubility product of AgCl is 2.8xx10^(-10) at 298 K. Calculate the solubility of AgCl in (i) pure water (ii) 0.1M AgNO_(3) solution, and (iii) 0.1M HCl solution. |
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Answer» Solution :(i) `K_(sp)" of "AGCL=underset(s)([Ag^(+)])underset(s)([Cl^(-)])` `therefore` Solubility `S=sqrt(K_(sp))=sqrt(2.8xx10^(-10))=1.67xx10^(-5)"mol "DM^(-3)` (ii) `AgCl_((s))hArrAg^(+)(aq)+Cl^(-)(aq)` `underset(0.1M)(AgNO_(3))tounderset(0.1M)(Ag^(+))+underset(0.1M)(NO_(3)^(-))` `therefore` Solubility of AgCl in the pressure of 0.1M `AgNO_(3)=K_(sp)/c=(2.8xx10^(-10))/0.1` `=2.8xx10^(-9)"mol "dm^(-3)` (iii) `AgCl_((s))hArrAg^(+)(aq)+Cl^(-)(aq)` `underset(0.1M)(HCL)toH^(+)+underset(0.1M)(Cl^(-))` `therefore` Solubility of AgCl in the pressure of 0.1M HCl = `K_(sp)/c` = `(2.8xx10^(-10))/0.1` `=2.8xx10^(-9)" mol "dm^(-3)` |
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