A solution is0.01 M Kland 0. 1 M KCl . If solidAgNO_3is added to the solution, what is the [1^(-) ] when AgClbegins to precipitate[K_(sp) (Agl) =1.5 xx 10^(-16) , K_(sp)(AgCl) =1.8 xx 10^(-10)]

A solution is0.01 M Kland 0. 1 M KCl . If solidAgNO_3is added to the s

1.	A solution is0.01 M Kland 0. 1 M KCl . If solidAgNO_3is added to the solution, what is the [1^(-) ] when AgClbegins to precipitate[K_(sp) (Agl) =1.5 xx 10^(-16) , K_(sp)(AgCl) =1.8 xx 10^(-10)]
Answer» <html><body><p>` 3.5 xx 10^(-7)` <br/>` 6.1 xx 10^(-8) ` <br/>` 2.2 xx 10^(-7) ` <br/>` 8.3 xx 10^(-8) ` </p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :` K_(sp)=[<a href="https://interviewquestions.tuteehub.com/tag/ag-362275" style="font-weight:bold;" target="_blank" title="Click to know more about AG">AG</a>^(+) ][Cl^(-) ] ` <br/> ` [Ag^(+) ] =(1.8 xx 10^(-10))/( 10^(-1) )= 1.8 xx 10^(-<a href="https://interviewquestions.tuteehub.com/tag/9-340408" style="font-weight:bold;" target="_blank" title="Click to know more about 9">9</a>) ` <br/> ` [I^(-) ]=(K_(sp) )/( [Ag^(+) ]_(cl^(-) ))=(1.5 xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/16-276476" style="font-weight:bold;" target="_blank" title="Click to know more about 16">16</a>))/( 1.8 xx 10^(-<a href="https://interviewquestions.tuteehub.com/tag/19-280618" style="font-weight:bold;" target="_blank" title="Click to know more about 19">19</a>)) = 8.3 xx 10 ^(-8) M`</body></html>