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1.0 g NaHSO_3 and Na_2SO_3 was dissolved in water to prepare a 200 mL solution. Two separate experiments were carried out. (a). 25 " mL of " sample was mixed with 25 " mL of " I_2 solution and excess of I_2 left after the reaction with NaHSO_3 and Na_2SO_3 was back titrated with 0.1002 N Na_2S_2O_3, 1.34 " mL of " which was required (25 " mL of " I_2 solution is equivalent to 24.20" mL of " Na_2S_2O_3 solution). (b). 50 " mL of " sample was oxidised to Na_2SO_4 by the action of H_2O_2,H_2SO_4 sormed (from NaHSO_3) was titrated with 22.3 " mL of " 0.1 N NaOH. Find percentage of NaHSO_3 and Na_2SO_3 in the original sample.

Answer» <html><body><p></p>Solution :1 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> mixture is present in 200 mL <br/> (a). <a href="https://interviewquestions.tuteehub.com/tag/25-296426" style="font-weight:bold;" target="_blank" title="Click to know more about 25">25</a> " mL of " mixture] <br/> `impliesmEq` of `NaHSO_3+mEq` of `Na_2SO_3=mEq` of ` I_2` <br/> Used <br/> `mEq` of ` I_2` used`=24.20-1.34xx(0.1002)/(25)mL` <br/> `=22.905` <br/> m" Eq of "`I_2` in 200 mL`=(200)/(25)xx22.905=18.32` <br/> (b). 50 mL is oxidised to `Na_2SO_4` <br/> `H_2SO_4` formed by `NaHSO_3-=22.3mL` of `0.1` N `NaOH` <br/> `mEq` of `NaHSO_3=22.3xx0.1=(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>.23)/(25)mL` <br/> `mEq` of ` NaHSO_2 i n 200mL=2.23xx(200)/(50)=8.92` <br/> `[{:("Ew of "NaHSO_(3)" in experiment"," "),((i)=(Mw)/(2)," "),(underset(x=4)underset(1+x-6=-1)(HSO_(3)^(ɵ)),tounderset(x=6)underset(x-8=-2)(SO_(4)^(2-)(x=2))):}]` <br/> `[{:("Ew of "NaHSO_(3) " in experiment"," "),((ii)=(Mw)/(1)," "),(HSO_(3)^(ɵ),toH^(o+)+SO_(3)^(2-)):}]` <br/> M" Eq of "`Na_(2)SO_(3)` in the original solution <br/> `=18.32-2xx8.92=0.48` <br/> Weight of `Na_2SO_3=8.92mEqxx(1)/(1000)xxEw of "NaHSO_(3)` <br/> `["Ew of "NaHSO_(3)=(Mw)/(1)=<a href="https://interviewquestions.tuteehub.com/tag/104-266430" style="font-weight:bold;" target="_blank" title="Click to know more about 104">104</a>]` <br/> `=(8.92)/(1000)xx104=0.92` <br/> `therefore%` of `NaHSO_(3)=(0.927)/(1)xx100=92.7%` <br/> [<a href="https://interviewquestions.tuteehub.com/tag/equivalent-446407" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENT">EQUIVALENT</a> weight of `Na_(2)SO_(3)=(Mw)/(2)=(126)/(2)=63]` <br/> Weight of `Na_(2)SO_(3)=0.48mEqxx(1)/(1000)xx63=0.3g` <br/> `%` of `Na_(2)SO_(3)=(0.3)/(1)xx100=3%` <br/> `NaHSO_(3)=92.7%""Na_(2)SO_(3)=3%`</body></html>


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