100 ml, 0.1 M K_(2)SO_(4) is mixed with 100 ml, 0.1M Al_(2)(SO_(4))_(3) solution. The resultant solution is

100 ml, 0.1 M K_(2)SO_(4) is mixed with 100 ml, 0.1M Al_(2)(SO_(4))_(3

1.	100 ml, 0.1 M K_(2)SO_(4) is mixed with 100 ml, 0.1M Al_(2)(SO_(4))_(3) solution. The resultant solution is
Answer» <html><body><p>`0.1NK^(+)` ions <br/>`0.4NK^(+)` ions <br/>`0.2MSO_(4)^(-2)` ions<br/>`0.1MAl^(+<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)` ions </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^(+)]=(100xx0.1xx2)/(200)=0.1M=0.1N` <br/> `[SO_(4)^(2-)]=(100xx0.1xx1+100xx0.1xx3)/(200)` <br/> `=0.2M=0.4N` <br/> `[<a href="https://interviewquestions.tuteehub.com/tag/al-370666" style="font-weight:bold;" target="_blank" title="Click to know more about AL">AL</a>^(+3)]=(100xx0.1xx2)/(200)=0.1M=0.3N`</body></html>