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25 mLof household bleach solution was mixed with 30 mL of 0.50 M KI and 10 mL of 4 Nacetic acid in the titration of the liberated iodine 48 mL of 0.25 N Na_(2)S_(2)O_(3) was used to reach the end point the molarity of the household bleach solution is

Answer» <html><body><p>0.48 M<br/>0.96 M<br/>0.24 M<br/>0.24 M </p>Solution :The chemical equation involved in this problem are: <br/> `<a href="https://interviewquestions.tuteehub.com/tag/ca-375" style="font-weight:bold;" target="_blank" title="Click to know more about CA">CA</a>(OCI)_(2)+<a href="https://interviewquestions.tuteehub.com/tag/4h-318945" style="font-weight:bold;" target="_blank" title="Click to know more about 4H">4H</a>^(+)+2e^(-)rarrCa^(2)+2H_(2)O+CI_(2)` <br/> `2KI+CI_(2)rarr2KCI+I_(2)` <br/> `2Na_(2)S_(2)O_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)+I_(2)rarrNa_(2)S_(4)O_(6)+2` Nal <br/> `Ca(OCI_(2))+4H^(+)+2e^(-)+2Na_(2)S_(2)O_(3)rarrCa^(2+)+2H_(2)O+2KCI+Na_(2)S_(4)O_(6)+2NaI` <br/> Thus the normaility of`Ca(OCI)_(2)` solution =0.48 N <br/> Now since <a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a> (i) is `2e^(-)` change <br/> `therefore`Eq wt of `Ca(OcI)_(2)="Mol wt"//2` <br/> `=0.48//2` <br/> thus the molarity of the household <a href="https://interviewquestions.tuteehub.com/tag/bleach-399456" style="font-weight:bold;" target="_blank" title="Click to know more about BLEACH">BLEACH</a> =0.24 M</body></html>


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