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The solubility product of AgCl is 2.8xx1...

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.

Text Solution

Verified by Experts

(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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