A sample of AgCl was treated with 5.00 ml of 1.5 M Na_(2)CO_(3) solution to give Ag_(2)CO_(3). The remaining solution contained 0.0026 g of Cl^(-) per litre. Calculate the solubility product of AgCl (K_(sp) "for" Ag_(2)CO_(3)=8.2xx10^(-2))

A sample of AgCl was treated with 5.00 ml of 1.5 M Na_(2)CO_(3) soluti

1.	A sample of AgCl was treated with 5.00 ml of 1.5 M Na_(2)CO_(3) solution to give Ag_(2)CO_(3). The remaining solution contained 0.0026 g of Cl^(-) per litre. Calculate the solubility product of AgCl (K_(sp) "for" Ag_(2)CO_(3)=8.2xx10^(-2))
Answer» <html><body><p></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> :`1.5 M Na_(2)CO_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>) ` gives `[CO_(3)^(2-)]=1.5M` <br/> `K_(sp)` for `Ag_(2)CO_(3)=[Ag^(+)]^(2)[CO_(3)^(2-)]` <br/> `:. [Ag^(+)]=<a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>((K_(sp) "for" Ag_(2)CO_(3))/([CO_(3)^(2-)]))=sqrt((8.2xx10^(-12))/(1.5))=2.34xx106(-6)M` <br/> `K_(sp) "for" AgCl=[Ag^(+)][<a href="https://interviewquestions.tuteehub.com/tag/cl-408888" style="font-weight:bold;" target="_blank" title="Click to know more about CL">CL</a>^(-)]=(2.34xx10^(-6))((0.0026)/(35.5))=1.71xx10^(-10)`</body></html>