A solution which is 10^(-3) M each in Mn^(2+),Fe^(2+),Zn^(2+)andHg^(2+) is treated with 10^(-16)M sulphide ion. If K_(sp) od MnS, ZnS and HgS are 10^(-15),10^(-25),10^(-20)and10^(-54) respectively, which one will precipitate first ?

A solution which is 10^(-3) M each in Mn^(2+),Fe^(2+),Zn^(2+)andHg^(2+

1.	A solution which is 10^(-3) M each in Mn^(2+),Fe^(2+),Zn^(2+)andHg^(2+) is treated with 10^(-16)M sulphide ion. If K_(sp) od MnS, ZnS and HgS are 10^(-15),10^(-25),10^(-20)and10^(-54) respectively, which one will precipitate first ?
Answer» <html><body><p>FeS <br/>MnS<br/>HgS <br/>ZnS</p>Solution :Ionic product in the solution `=<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(-<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)xx10^(16)=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>)` <br/> The metal sulphide having the lowest solubility will precipitate <a href="https://interviewquestions.tuteehub.com/tag/first-461760" style="font-weight:bold;" target="_blank" title="Click to know more about FIRST">FIRST</a> provided the ionic product is higher than the `K_(<a href="https://interviewquestions.tuteehub.com/tag/sp-1219706" style="font-weight:bold;" target="_blank" title="Click to know more about SP">SP</a>)`. Here, all salts are of the same valence-type. So the sulphide having the lowest `K_(sp)` value will precipitate first provided `K_(sp)lt10^(-19)`, HgS has lowest `K_(sp)` value `(10^(-54))`, so will precipitate first.</body></html>